Statistics And Probability

Statistics And Probability

REVIEW LESSONS Mean Median Mode Range Standard Deviation Variance
Measures of Central Tendency (Ungrouped and Grouped Data) Mean Median Mode Measures of Variability (Ungrouped and Grouped data Range Standard Deviation Variance Coefficient of Variation

THE MEAN Mean(x) is also known as arithmetic average. It is the sum of the item values divided by the number of items. Mean of Grouped Data If the number of items is too big, it is best to compute for the measures of central tendency (Mean, Median and Mode) using a frequency distribution.

𝒇 – frequency of the class interval
To determine the mean of a grouped data, use the formula: x = Ʃ𝒇𝒙 𝒏 where: 𝒇 – frequency of the class interval x – midpoint of the class interval n – total number of items

Example 1. Calculate the arithmetic mean of the given distribution on final scores of 100 Grade 11 students in Trigonometry. n = Ʃfx = 7,835

Solution: x = Ʃ𝒇𝒙 𝒏 = = 78.35 STATISTICS AND PROBABILITY SAMSUDIN N. ABDULLAH, Ph.D.

Example 2. What is mean of the given distribution of scores of 75 students in Statistics.
n = Ʃfx = 4,404 STATISTICS AND PROBABILITY SAMSUDIN N. ABDULLAH, Ph.D.

Solution: x = Ʃ𝒇𝒙 𝒏 = = 58.72 STATISTICS AND PROBABILITY SAMSUDIN N. ABDULLAH, Ph.D.

Problem. Compute for the mean of the ages of GSAT teachers.
STATISTICS AND PROBABILITY SAMSUDIN N. ABDULLAH, Ph.D.

Problem. Compute for the mean of the ages of ENHS teachers
Problem. Compute for the mean of the ages of ENHS teachers. Use the idea of ungrouped and grouped data. Then compare the results. 60, 62, 54, 40, 33, 35, 22, 23, 55, 57, 25, 26, 34, 44, 41, 44, 44, 44, 44, 45, 59, 58, 52, 50, 36, 33, 34, 37, 39, 22, 29, 30, 31, 32, 33, 34, 40, 45, 56, 63, 45, 25, 27, 28, 39, 34, 45, 37, 61, 60, 33, 32, 31, 22, 23, 24, 25, 26, 27, 28, 30, 50, 48, 52, 55, 62, 60, 33, 34, 44, 44, 44, 45, 46, 42, 37, 39, 40, 42, 44, 23, 24, 61, 50, 53, 55, 56, 57, 58, 59, 33, 30, 29, 33, 50, 28, 27, 45, 45, 44, 44, 56, 56, 57, 40, 44, 45, 24, 25, 26, 30, 31, 27, 27, 30, 24, 25, 41, 43, 42, 50, 53, 55, 54 STATISTICS AND PROBABILITY SAMSUDIN N. ABDULLAH, Ph.D.

Solution Using Ungrouped Data
x = Ʃ𝒙 𝒏 = 𝟔𝟎 + 𝟔𝟐 + 𝟓𝟒 + 𝟒𝟎 𝟓𝟒 𝟏𝟐𝟒 = 𝟓𝟎𝟏𝟕 𝟏𝟐𝟒 =𝟒𝟎.𝟒𝟔 STATISTICS AND PROBABILITY SAMSUDIN N. ABDULLAH, Ph.D.

Class Interval f x fx 62 – 65 3 63.5 190.5 58 – 61 9 59.5 535.5 54 – 57 13 55.5 721.5 50 – 53 51.5 463.5 46 – 49 2 47.5 95 42 – 45 24 43.5 1,044 38 – 41 39.5 355.5 34 – 37 10 35.5 355 30 – 33 17 31.5 26 – 29 27.5 357.5 22 – 25 15 23.5 352.5 n = 124 ∑fx = 5,006 Solution: x = Ʃ𝒇𝒙 𝒏 = 𝟓,𝟎𝟎𝟔 𝟏𝟐𝟒 =𝟒𝟎.𝟑𝟕 STATISTICS AND PROBABILITY SAMSUDIN N. ABDULLAH, Ph.D.

Class Interval f x fx 61 – 63 5 62 310 58 – 60 7 59 413 55 – 57 11 56
616 52 – 54 6 53 318 49 – 51 50 250 46 – 48 2 47 94 43 – 45 21 44 924 40 – 42 9 41 369 37 – 39 38 228 34 – 36 35 245 31 – 33 12 32 384 28 – 30 10 29 290 25 – 27 13 26 338 22 – 24 23 230 n = 124 ∑fx = 5,009 Solution: x = Ʃ𝒇𝒙 𝒏 = 𝟓,𝟎𝟎𝟗 𝟏𝟐𝟒 =𝟒𝟎.𝟒𝟎 STATISTICS AND PROBABILITY SAMSUDIN N. ABDULLAH, Ph.D.

THE MEDIAN Median of Grouped Data
Median (Md) is the value of the middle term when data are arranged in either ascending or descending order. Median of Grouped Data For large quantities of data, the median is computed using a frequency distribution with a cumulative frequency column. STATISTICS AND PROBABILITY SAMSUDIN N. ABDULLAH, Ph.D.

To determine the median of a grouped data, use the formula:
Md = L + 𝒏 𝟐 −𝑭 𝒇 𝒊 where: L – the exact lower limit of the median class n – total number of items F – “less than” or “equal to” cumulative frequency preceding the class interval containing the median f – frequency of the median class i – size of the class interval STATISTICS AND PROBABILITY SAMSUDIN N. ABDULLAH, Ph.D.

Example 1. Find the median score of students of Mr
Example 1. Find the median score of students of Mr. Dela Cruz Math class. STATISTICS AND PROBABILITY SAMSUDIN N. ABDULLAH, Ph.D.

Solution: Md = L + 𝒏 𝟐 −𝑭 𝒇 𝒊 L = 79.5 n = 100 F = 42 f = 25

Example 2. The ages of 115 ENHS teachers are given below
Example 2. The ages of 115 ENHS teachers are given below. Find the median age. STATISTICS AND PROBABILITY SAMSUDIN N. ABDULLAH, Ph.D.

Problem. Compute for the median of the ages of ENHS teachers
Problem. Compute for the median of the ages of ENHS teachers. Use the idea of ungrouped and grouped data. Then compare the results. 60, 62, 54, 40, 33, 35, 22, 23, 55, 57, 25, 26, 34, 44, 41, 44, 44, 44, 44, 45, 59, 58, 52, 50, 36, 33, 34, 37, 39, 22, 29, 30, 31, 32, 33, 34, 40, 45, 56, 63, 45, 25, 27, 28, 39, 34, 45, 37, 61, 60, 33, 32, 31, 22, 23, 24, 25, 26, 27, 28, 30, 50, 48, 52, 55, 62, 60, 33, 34, 44, 44, 44, 45, 46, 42, 37, 39, 40, 42, 44, 23, 24, 61, 50, 53, 55, 56, 57, 58, 59, 33, 30, 29, 33, 50, 28, 27, 45, 45, 44, 44, 56, 56, 57, 40, 44, 45, 24, 25, 26, 30, 31, 27, 27, 30, 24, 25, 41, 43, 42, 50, 53, 55, 54 STATISTICS AND PROBABILITY SAMSUDIN N. ABDULLAH, Ph.D.

Class Interval f F 62 – 65 3 124 58 – 61 9 121 54 – 57 13 112 50 – 53 99 46 – 49 2 90 42 – 45 24 88 38 – 41 64 34 – 37 10 55 30 – 33 17 45 26 – 29 28 22 – 25 15 n = 124 STATISTICS AND PROBABILITY SAMSUDIN N. ABDULLAH, Ph.D.

Note: CF – Cumulative Frequency & CP – Cumulative Percent
Example 3. Complete the table and compute for the median score of the Grade 11 students who took the Precalculus subject. Note: CF – Cumulative Frequency & CP – Cumulative Percent

THE MODE Mode of Grouped Data
MODE (Mₒ) is referred to as the most frequently occurring value in a given set. Mode of Grouped Data In a grouped distribution, the class interval where the value with the highest frequency is the modal class.

To determine the mode of a grouped data, use the formula:
Mo = Lmo + ( 𝒅₁ 𝒅₁ + 𝒅₂ )i where: Lmo – the exact lower limit of the modal class 𝑑₁ – the difference between the frequency of the modal class and that of the frequency below the modal class 𝑑₂ – the difference between the frequency of above the modal class i – the size of the class interval

Solution: Lmo = 34.5 d1 = 16 – 12 = 4 d2 = 16 – 15 = 1
Example 1. Determine the modal class and the modal value for the frequency distribution of ages of teachers in Esperanza NHS. Solution: Lmo = 34.5 d1 = 16 – 12 = 4 d2 = 16 – 15 = 1 i = 39 – = 5 STATISTICS AND PROBABILITY SAMSUDIN N. ABDULLAH, Ph.D.

Mo = ( 𝟒 𝟒 + 𝟏 )(5) = 𝟐𝟎 𝟓 = = 38.5 STATISTICS AND PROBABILITY SAMSUDIN N. ABDULLAH, Ph.D.

Example 2. Compute for the modal wage of the workers in a certain private school

Solution: Lmo = 1,319.5 d1 = 31 – 24 = 7 d2 = 31 – 12 = 19 i = 1,339 – 1, = 20 STATISTICS AND PROBABILITY SAMSUDIN N. ABDULLAH, Ph.D.

Mo = 1, ( 𝟕 𝟕 + 𝟏𝟗 )(20) = 1, 𝟏𝟒𝟎 𝟐𝟔 = 1, = 1, STATISTICS AND PROBABILITY SAMSUDIN N. ABDULLAH, Ph.D.

Problem. Find the modal age of ENHS teachers
Problem. Find the modal age of ENHS teachers. Use the idea of ungrouped and grouped data. Then compare the results. 60, 62, 54, 40, 33, 35, 22, 23, 55, 57, 25, 26, 34, 44, 41, 44, 44, 44, 44, 45, 59, 58, 52, 50, 36, 33, 34, 37, 39, 22, 29, 30, 31, 32, 33, 34, 40, 45, 56, 63, 45, 25, 27, 28, 39, 34, 45, 37, 61, 60, 33, 32, 31, 22, 23, 24, 25, 26, 27, 28, 30, 50, 48, 52, 55, 62, 60, 33, 34, 44, 44, 44, 45, 46, 42, 37, 39, 40, 42, 44, 23, 24, 61, 50, 53, 55, 56, 57, 58, 59, 33, 30, 29, 33, 50, 28, 27, 45, 45, 44, 44, 56, 56, 57, 40, 44, 45, 24, 25, 26, 30, 31, 27, 27, 30, 24, 25, 41, 43, 42, 50, 53, 55, 54 STATISTICS AND PROBABILITY SAMSUDIN N. ABDULLAH, Ph.D.

Class Interval f 61 – 63 5 58 – 60 7 55 – 57 11 52 – 54 6 49 – 51 46 – 48 2 43 – 45 21 40 – 42 9 37 – 39 34 – 36 31 – 33 12 28 – 30 10 25 – 27 13 22 – 24 n = 124 Solution: Mo = ( 𝟏𝟐 𝟏𝟐 + 𝟏𝟗 )(4) = 𝟒𝟖 𝟑𝟏 = = 44.05 STATISTICS AND PROBABILITY SAMSUDIN N. ABDULLAH, Ph.D.

Range = highest value – lowest value
MEASURES OF VARIABILITY describe the spread of the values about the mean. 1. Range 2. Standard Deviation 3. Variance THE RANGE The difference between the highest and the lowest values in a given set of data is the RANGE. Range = highest value – lowest value Example 1. Find the range for each set of data given below. a) 3, 8, 16, 12, 4, 5, 7, 15 b) 25, 32, 9 18, 12, 30, 28, 22 Answer: 13 Answer: 23 STATISTICS AND PROBABILITY SAMSUDIN N. ABDULLAH, Ph.D.

Example 2. Determine the range of data presented in a frequency distribution below.

Population STANDARD DEVIATION is the measure of the variation of a set of data in terms of the amounts by which the individual values differ from their mean. It is the most stable measure of spread. Population Standard Deviation of Ungrouped Data s = Ʃ𝒅² 𝒏 where: s – standard deviation d – deviation from the mean Ʃ𝒅² – sum of squared deviations n – number of items

Example 1. Calculate the standard deviation of the given scores in a quiz: 18, 20, 22, 15, 16, 12, 17, 21, 10, 19. s = Ʃ𝒅² 𝒏 = 𝟏𝟑𝟒 𝟏𝟎 = 𝟏𝟑.𝟒 s = 3.661 STATISTICS AND PROBABILITY SAMSUDIN N. ABDULLAH, Ph.D.

Standard Deviation of Grouped Data s = Ʃ𝒇𝒅² 𝒏
where: s – population standard deviation d – deviation from the mean Ʃ𝒇𝒅² – sum of product of frequency and squared deviations n – number of items STATISTICS AND PROBABILITY SAMSUDIN N. ABDULLAH, Ph.D.

Example 1. Calculate the standard deviation of the data presented below.

Class Intervals f x fx d d2 fd2
Solution: Class Intervals f x fx d d2 fd2 252 – 260 3 256 768 43 1849 5547 243 – 251 5 247 1235 34 1156 5780 234 – 242 9 238 2142 25 625 5625 225 – 233 12 229 2748 16 3072 216 – 224 220 1100 7 49 245 207 – 215 4 211 844 -2 198 – 206 2 202 404 -11 121 242 189 – 197 10 193 1930 -20 400 4000 180 – 188 8 184 1472 -29 841 6728 171 – 179 175 350 -38 1444 2888 162 – 170 166 830 -47 2209 11045 n = 65 Ʃfx = 13823 Ʃfd2 = 45188

Assignment. Find the mean, median, mode and standard deviation of the given data.
Class Intervals f 355 – 365 13 344 – 354 5 333 – 343 11 322 – 332 12 311 – 321 15 300 – 310 4 289 – 299 20 278 – 288 8 267 – 277 9 256 – 266 245 – 255 234 – 244 3 223 – 233 2 212 – 222 201 – 211

STATISTICS AND PROBABILITY SAMSUDIN N. ABDULLAH, Ph.D.

deviation from the mean
Class Interval Frequency (f) Class Mark (x) fx Cumulative Frequency (F) deviation from the mean (d) d2 fd2 99 – 105 3 102 306 68 23 529 1587 92 – 98 10 95 950 65 16 256 2560 85 – 91 88 1408 55 9 81 1296 78 – 84 8 648 39 2 4 32 71 – 77 11 74 814 31 -5 25 275 64 – 70 67 536 20 -12 144 1152 57 – 63 60 540 12 -19 361 3249 50 – 56 53 159 -26 676 2028 n = 68 Ʃfx = 5361 Ʃfd2 =12179 STATISTICS AND PROBABILITY SAMSUDIN N. ABDULLAH, Ph.D.

7 56 8 16 78 – 84 85 – 91 78.838 80.125 88.5 13.383 77.5 84.5 BETTER OR CLOSER 16.975% HOMOGENEOUS STATISTICS AND PROBABILITY SAMSUDIN N. ABDULLAH, Ph.D.

deviation from the mean
Class Interval Frequency (f) Class Mark (x) fx Cumulative Frequency (F) deviation from the mean (d) d2 fd2 60 – 64 3 62 186 92 – 98 10 85 – 91 16 78 – 84 8 71 – 77 11 64 – 70 57 – 63 9 25 – 29 27 81 n = Ʃfx = Ʃfd2 =

Course Outline in Grade 11 Statistics and Probability
CHAPTER I. Random Variables and Probability Distributions Random Variables Probability of an Event Probability Distribution Mean of a Discrete Probability Variance of a Discrete Probability STATISTICS AND PROBABILITY SAMSUDIN N. ABDULLAH, Ph.D.

CHAPTER II. Normal Distribution Normal Curve Distribution The z-scores
Regions of Areas Under the Normal Curve Determining Probabilities Percentiles Under Normal Curve Applying the Normal Curve Concepts in Problem Solving STATISTICS AND PROBABILITY SAMSUDIN N. ABDULLAH, Ph.D.

CHAPTER III. Sampling and Sampling Distribution
Sampling Techniques Commonly Used in Research Sampling Distribution of Sample Means Mean and Variance of the Sampling Distribution of Means Solving Problems Involving Sampling Distribution of the Sample Means

CHAPTER IV. Estimation of Parameters
Point Estimation of a Population Confidence Interval Estimates for the Population Mean Confidence Intervals for the Population Mean when σ is Unknown Point Estimate for the Population Proportion Interval Estimates of Population Proportions Interpreting Interval Estimates of Population Proportions Confidence Level and Sample Size

CHAPTER V. Conducting Hypothesis Testing Hypothesis Testing
Elements of Hypothesis Testing Hypothesis Testing Using the Traditional Method Small-Sample Tests About a Population Mean μ Significance Tests Using the Probability Value Approach Testing Hypothesis Involving Proportions STATISTICS AND PROBABILITY SAMSUDIN N. ABDULLAH, Ph.D.

One Way Analysis of Variance (ANOVA) Pearson r (Correlation Analysis)
CHAPTER VI. Commonly Utilized Inferential Statistical Tools (Application of Hypothesis Testing) z-test t-test One Way Analysis of Variance (ANOVA) Pearson r (Correlation Analysis) STATISTICS AND PROBABILITY SAMSUDIN N. ABDULLAH, Ph.D.

RANDOM VARIABLES AND PROBABILITY DISTRIBUTIONS
CHAPTER I RANDOM VARIABLES AND PROBABILITY DISTRIBUTIONS STATISTICS AND PROBABILITY SAMSUDIN N. ABDULLAH, Ph.D.

What is a random variable?
Random Variable is a function that associates a real number to each element in the sample space. It is a variable whose values are determined by chance. A random variable is discrete random variable if its set of possible outcomes is countable. Mostly, discrete random variables represent count data, such as the number of defective chairs produced in a factory. A random variable is a continuous random variable if it takes values on a continuous scale. Often, continuous random variables represent measured data, such as heights, weights, and temperatures.

A. Classify the following random variables as discrete or continuous.
The number of defective computers produced by a manufacturer The weight of newborns each year in a hospital The number of siblings in a family The amount of paint utilized in a building project The number of dropouts in a school The speed of a car The number of female athletes The time needed to finish the test The amount of sugar in a cup of coffee The number of people who are playing lotto each day 11. The number of accidents per year in an accident prone area

12. The amount of salt and ice to preserve ice cream
13. The number of all public school students in the world 14. The magnitude of several earthquakes 15. The number of private school teachers in the Philippines 16. The body temperature of a patient 17. The size of a Flat TV screen 18. The number of households in a subdivision 19. The heights of students 20. The vital statistics a female candidate 21. The number of used clothes for the refugees 22. The number of eggs in one tray STATISTICS AND PROBABILITY SAMSUDIN N. ABDULLAH, Ph.D.

23. The length of the top of a table
24. The amount of sugar needed to bake 25. The number of students in the TVL track 26. The width of a blackboard 27. The sticks of chalk in a box 28. The number of coins in my pocket 29. The number of Korean teachers here at ENHS 30. The kilogram of fruits in a table 31. The storm signals of typhoons 32. The distance between school and market 33. The angle of elevation 34. The height of flagpole 35. The thickness of a book STATISTICS AND PROBABILITY SAMSUDIN N. ABDULLAH, Ph.D.

A. Classify the following random variables as discrete or continuous.
The number of defective computers produced by a manufacturer The weight of newborns each year in a hospital The number of siblings in a family The amount of paint utilized in a building project The number of dropouts in a school The speed of a car The number of female athletes The time needed to finish the test The amount of sugar in a cup of coffee The number of people who are playing lotto each day 11. The number of accidents per year in an accident prone area Discrete Continuous Discrete Continuous Discrete Continuous Discrete Continuous Continuous Discrete Discrete

12. The amount of salt and ice to preserve ice cream
13. The number of all public school students in the world 14. The intensity of several earthquakes striking Mindanao 15. The number of private school teachers in the Philippines 16. The body temperature of a patient 17. The size of a Flat TV screen 18. The heights of students 19. The number of households in a subdivision 20. The vital statistics a female candidate 21. The number of used clothes for the refugees 22. The number of eggs in one tray Continuous Discrete Continuous Discrete Continuous Continuous Continuous Discrete Continuous Discrete Discrete STATISTICS AND PROBABILITY SAMSUDIN N. ABDULLAH, Ph.D.

23. The length of the top of a table
24. The amount of sugar needed to bake 25. The number of students in the TVL track 26. The width of a blackboard 27. The sticks of chalk in a box 28. The coins in my pocket 29. The Korean teachers here at ENHS 30. The kilogram of fruits in a table 31. The storm signals of typhoons 32. The distance between school and market 33. The angle of elevation 34. The height of flagpole 35. The thickness of a book Continuous Continuous Discrete Continuous Discrete Discrete Discrete Continuous Continuous Continuous Continuous Continuous Continuous STATISTICS AND PROBABILITY SAMSUDIN N. ABDULLAH, Ph.D.

Value of the Random Variable X (number of defective cell phones)
B.1. Suppose three cell phones are tested at random. Let D represent the defective cell phones and N represent the non-defective cell phones. Assume X be the random variable representing the number of defective cell phones. Complete the table below to show the values of the random variable. Possible Outcomes Value of the Random Variable X (number of defective cell phones) NNN NND 1 NDN DND 2 DDN DNN NDD DDD 3 The values of a random variable X are 0, 1, 2 and 3.

Value of the Random Variable Y
2. Suppose three coins are tossed. Let Y be the random variable representing the number of tails that occur. Find the values of the random variable Y. Complete the table below. Possible Outcomes Value of the Random Variable Y (number of tails) HHH THH 1 HTH HHT HTT 2 THT TTH TTT 3 The values of the random variable Y are 0, 1, 2 and 3.

Quiz # 2 Suppose four coins are tossed. Let X be the random variable representing the number of HEADS that occur. Find the values of the random variable X. Complete the table. STATISTICS AND PROBABILITY SAMSUDIN N. ABDULLAH, Ph.D.

Value of the Random Variable Z
3. Two balls are drawn in succession without replacement from an urn containing 5 red balls and 6 blue balls. Let Z be the random variable representing the number of blue balls. Find the values of the random variables Z. Complete the table. Possible Outcomes Value of the Random Variable Z (number of blue balls) RR RB 1 BR BB 2 Note: Using the idea of a combination (₁₁C₂ = 55), there are 55 outcomes of the sample space. In that combinations, Blue doesn’t occur if you pick up all RED. Sometimes, BLUE occurs only once or twice. Thus, the values of the random variable Y are 0, 1 and 2.

n(S) = nCr = 𝑛! 𝑟! 𝑛−𝑟 ! = ₃C₂ = 3! 2! 3−2 ! = 3(2!) 2!1! = 3
4. A random experiment consists of selecting two balls in succession from an urn containing two black balls and one white ball. Specify the sample space for this experiment. Let K be the random variable that represents the number of black balls. What are the values of K? Solution: n(S) = nCr = 𝑛! 𝑟! 𝑛−𝑟 ! = ₃C₂ = 3! 2! 3−2 ! = 3(2!) 2!1! = 3 S = {(Black, Black), (Black, White), (White, Black)} No Black 0 1 Black 2 2 Black 1 The random variable K has values of 0, 1 and 2. STATISTICS AND PROBABILITY SAMSUDIN N. ABDULLAH, Ph.D.

n(S) = nCr = 𝑛! 𝑟! 𝑛−𝑟 ! = ₆C₂ = 6! 2!4! = 6(5)(4!) 2(1)(4!) = 15
5. A random experiment consists of selecting two balls in succession from an urn containing four black balls and two white balls. Specify the sample space for this experiment. Let M be the random variable that represents the number of black balls. What are the values of M? Solution: n(S) = nCr = 𝑛! 𝑟! 𝑛−𝑟 ! = ₆C₂ = 6! 2!4! = 6(5)(4!) 2(1)(4!) = 15 S = {W₁W₂, W₁B₁, W₁B₂, W₁B₃, W₁B₄, W₂B₁, W₂B₂, W₂B₃, W₂B₄, B₁B₂, B₁B₃, B₁B₄, B₂B₃, B₂B₄, B₃B₄} 0 Back 1 1 Black 8 2 Black 6 The random variable M has values of 0, 1 and 2. STATISTICS AND PROBABILITY SAMSUDIN N. ABDULLAH, Ph.D.

PROBABILITY OF AN EVENT

The values of a random variable X are 0, 1, 2, 3, 4 and 5.
6. Five coins are tossed. Let X be the random variable that represents the number of TAILS. Enumerate the outcomes of the sample space and determine the possible values of the random variable X. 0 TAIL HHHHH 1 TAIL THHHH HTHHH HHTHH HHHTH HHHHT 2 TAILS TTHHH THTHH THHTH THHHT HTTHH HHTTH HHHTT HTHTH HHTHT HTHHT 3 TAILS HHTTT HTHTT HTTHT HTTTH THHTT TTHHT TTTHH THTHT TTHTH THTTH 4 TAILS TTTTH THTTT TTHTT TTTHT TTTTH 5 TAILS TTTTT The values of a random variable X are 0, 1, 2, 3, 4 and 5. Note: There are 32 outcomes of the sample space since tossing five coins will give you an equation 2⁵ = 32. STATISTICS AND PROBABILITY SAMSUDIN N. ABDULLAH, Ph.D.

Exercise: A. List the outcomes of the sample space of the following experiments. Then find the cardinality of the sample space. Tossing three coins S = {TTT, TTH, THH, THT, HHT, HTH, HTT, HHH} n(S) = 8 2. Rolling a die and tossing a coin simultaneously. S = {1H, 2H, 3H, 4H, 5H, 6H, 1T, 2T, 3T, 4T, 5T, 6T} n(S) = 12 3. Tossing a coin and spinning the spinner with 8 numbers. S = {H1, H2, H3, H4, H5, H6, H7, H8, T1, T2, T3, T4, T5, T6, T7, T8} n(S) = 16 4. Getting a defective item when two items are randomly selected from a box of two defective and three non-defective items. S = {D₁D₂, D₁N₁, D₁N₂, D₁N₃, D₂N₁, D₂N₂, D₂N₃, N₂N₃, N₁N₂, N₁N₃} n(S) = 10 5. Drawing a spade from a standard deck of cards n(S) = 52 6. Drawing a card greater than 7 from a deck of cards 2³ = 8 6¹(2¹) = 6(2) = 12 2(8) = 16 ₅C₂ = 10

7. Rolling a Die and Tossing a Coin Simultaneously
B. Find the cardinality of the sample of each experiment. Tossing a Coin n(S) = 2¹ = 2 2. Tossing Two Coins n(S) = 2² = 4 3. Tossing Three Coins n(S) = 2³ = 8 4. Rolling a Die n(S) = 6¹ = 6 5. Rolling Two Dice n(S) = 6² = 36 6. Rolling Three Dice n(S) = 6³ = 216 7. Rolling a Die and Tossing a Coin Simultaneously n(S) = 6¹(2¹) = 6(2) = 12 STATISTICS AND PROBABILITY SAMSUDIN N. ABDULLAH, Ph.D.

8. Rolling Two Dice and Two Coins Simultaneously
9. Rolling a Die and Tossing Three Coins Simultaneously n(S) = 6¹(2³) = (6)(8) = 48 10. Rolling Three Dice and Tossing Three Coins n(S) = (6³)(2³) = (216)(8) = 1,728 11. Drawing a Standard Deck of Cards n(S) = 52 12. Drawing Three Balls from a Box Containing Ten Balls n(S) = ₁₀C₃ = 10! 3!7! = 10(9)(8)(7!) (3)(2)(1)(7!) = = 120 13. Drawing Four Marbles from an Urn Containing 15 Marbles n(S) = ₁₅C₄ = 15! 4!11! = 15(14)(13)(12)(11!) (4)(3)(2)(1)(11!) = 32, = 1,365 14. Drawing Two Apples from a Basket Containing 8 Apples n(S) = ₈C₂ = 𝟖! 𝟐!𝟔! = (𝟖)(𝟕)(𝟔!) (𝟐)(𝟏)(𝟔!) = 𝟓𝟔 𝟐 = 28 STATISTICS AND PROBABILITY SAMSUDIN N. ABDULLAH, Ph.D.

Review Problems on Probability of an Event

Find the probability of the following events.
Event (E) P(E) 1 Getting an even number in a single roll of a die 1 2 2 Getting a sum of 6 when two dice are rolled 5 36 3 Getting an ace when a card is drawn from a deck 1 13 4 The probability that all children are boys if a couple has three children 1 4 5 Getting an odd number and a tail when a die is rolled and a coin is tossed simultaneously 6 Getting a sum of 11 when two dice are rolled 1 18 7 Getting a black card and 10 when a card is drawn from a deck 1 26 8 Getting a red queen when a card is drawn from a deck 9 Getting doubles when two dice are rolled 1 6 10 Getting a red ball from a box containing 3 red and 6 black balls 1 3 STATISTICS AND PROBABILITY SAMSUDIN N. ABDULLAH, Ph.D.

Find the probability of the following events.
Event (E) P(E) 1 Getting an even number in a single roll of a die 1 2 2 Getting a sum of 6 when two dice are rolled 5 36 3 Getting an ace when a card is drawn from a deck 1 13 4 The probability that all children are boys if a couple has three children 1 4 5 Getting an odd number and a tail when a die is rolled and a coin is tossed simultaneously 6 Getting a sum of 11 when two dice are rolled 1 18 7 Getting a black card and 10 when a card is drawn from a deck 1 26 8 Getting a red queen when a card is drawn from a deck 9 Getting doubles when two dice are rolled 1 6 10 Getting a red ball from a box containing 3 red and 6 black balls 1 3 E = {2, 4, 6} E = {(1, 5), (2, 4), (5, 1), (4, 2), (3, 3)} E = {A of Spade, A of Club, A of Heart, A of Diamond}} S = {GGG, GBG, BBG, BBB} E = {1T, 3T, 5T} E = {(5, 6), (6, 5)} E = {10 of Spade, 10 of Club} E = {Q of Diamond, Q of Heart} E = {(1, 1), (2, 2), (3, 3), (4, 4), (5, 5), (6, 6)} STATISTICS AND PROBABILITY SAMSUDIN N. ABDULLAH, Ph.D.

Review Problems on Probability
From a standard deck of 52 cards, what is the probability of 1. picking a black card? 2. picking a face card? 3. not picking a face card? 4. picking a black and face card? 5. not picking a black and face card? 6. picking a red and nonface cards? 7. picking an ace card? 8. not picking an ace card? 1/2 3/13 10/13 3/26 23/26 5/13 1/13 12/13 STATISTICS AND PROBABILITY SAMSUDIN N. ABDULLAH, Ph.D.

Quiz (1/4 sheet of paper) A. Find the cardinality of each sample space. 1. Tossing six coins 2. Tossing a pair of coins and spinning a spinner with 10 numbers simultaneously 3. Rolling a pair of dice and drawing a card from standard deck simultaneously 4. Tossing three coins and rolling two dice simultaneously 5. Drawing five balls in a box containing 12 balls 64 40 1,872 288 792 STATISTICS AND PROBABILITY SAMSUDIN N. ABDULLAH, Ph.D.

B. On rolling a die, what is the probability of having 1. a 3?
2. an even number? 3. zero? 4. a number greater than 4? 5. a number lying between 0 and 7? 6. a number less than 4? 7. an odd number? 8. a prime number? 9. a composite number? 10. a multiple of 3? 1/6 1/2 1/3 1 1/2 1/2 1/2 1/3 1/3 STATISTICS AND PROBABILITY SAMSUDIN N. ABDULLAH, Ph.D.

C. From standard deck of cards, what is the probability of:
1. picking a red card? 2. picking a face card? 3. picking a nonface card? 4. picking a black and 9 card? 5. not picking a black and 9 card? 6. picking a club card? 7. not picking a club card? 8. picking a red face card? 9. not picking a red face card? 10. picking any card? 1/2 3/13 10/13 1/26 25/26 1/13 12/13 3/26 23/26 1 STATISTICS AND PROBABILITY SAMSUDIN N. ABDULLAH, Ph.D.

2. two are red and one is green? 3. exactly two are green?
A box contains 4 white balls, 3 red balls, and 3 green balls. If three balls are drawn at random, what is the probability that 1. they are all white? 2. two are red and one is green? 3. exactly two are green? 4. none is white? 5. they are of different colors? 6. none is red? 1/30 3/40 7/40 1/6 3/10 7/24 STATISTICS AND PROBABILITY SAMSUDIN N. ABDULLAH, Ph.D.

Solutions of C n(S) = ₁₀C₃ = 10! 3!7! = 10(9)(8)(7!) (3)(2)(1)(7!) = = 120 n(E) = ₄C₃ = 4! 3!1! = (4)(3!) 3!(1) = 4 P(E) = = 1 30 n(E) = (₃C₂)( ₃C₁) = ( 3! 2!1! )( 3! 1!2! ) = (3)(2!) 2!(1) • (3)(2!) 1!(2!) = 3(3) = 9 P(E) = = 3 40 3. n(E) = (₃C₂)( ₇C₁) = ( 3! 2!1! )( 7! 1!6! ) = (3)(2!) 2!(1) • (7)(6!) 1!(6!) = 3(7) = 21 P(E) = = 7 40 STATISTICS AND PROBABILITY SAMSUDIN N. ABDULLAH, Ph.D.

5. n(E) = (₄C₁)( ₃C₁)(₃C₁ ) = ( 4! 1!3! )( 3! 1!2! )( 3! 1!2! )
Solutions of C n(S) = ₁₀C₃ = 10! 3!7! = 10(9)(8)(7!) (3)(2)(1)(7!) = = 120 4. n(E) = ₆C₃ = 6! 3!3! = (6)(5)(4)(3!) (3!)(3)(2)(1) = = 20 P(E) = = 1 6 5. n(E) = (₄C₁)( ₃C₁)(₃C₁ ) = ( 4! 1!3! )( 3! 1!2! )( 3! 1!2! ) = (4)(3!) (1)(3!) • (3)(2!) 1!(2!) • (3)(2!) 1!(2!) = 4(3)(3) =36 P(E) = = 3 10 6. n(E) = ( ₇C₃) = 7! 3!4! = (7)(6)(5)(4!) 3(2)(1)(4!) = = 35 P(E) = = 7 24 STATISTICS AND PROBABILITY SAMSUDIN N. ABDULLAH, Ph.D.

It cannot It cannot It can Exercises:
A. Determine whether the given values can serve as the values of a probability distribution of the random variable X that can take on only the values 1, 2, 3, and 4. Explain your answer. P(1) = , P(2) = , P(3) = , P(4) = 5 19 = >1 2. P(1) = 0.25, P(2) = 0.75, P(3) = 0.25, P(4) = 0.25 = 1.5 3. P(1) = 0.15, P(2) = 0.27, P(3) = 0.29, P(4) = 0.29 = 1 It cannot It cannot It can STATISTICS AND PROBABILITY SAMSUDIN N. ABDULLAH, Ph.D.

Exercises: A. Determine whether the given values can serve as the values of a probability distribution of the random variable X that can take on only the values 1, 2, 3, and 4. Explain your answer. P(1) = , P(2) = , P(3) = , P(4) = 5 19 = >1 They cannot 2. P(1) = 0.25, P(2) = 0.75, P(3) = 0.25, P(4) = 0.25 = 1.5 They cannot STATISTICS AND PROBABILITY SAMSUDIN N. ABDULLAH, Ph.D.

3. P(1) = 0.15, P(2) = 0.27, P(3) = 0.29, P(4) = 0.29 = They can 4. P(1) = 1 5 , P(2) = 2 5 , P(3) = , P(4) = 1 5 = 1 They can 5. P(1) = 0.35, P(2) = 0.15, P(3) = 0.05, P(4) = 0.45 = 1 They can

6. P(1) = 0.25, P(2) = 0.21, P(3) = 0.19, P(4) = 0.18 = 0.83 7. P(1) = , P(2) = , P(3) = , P(4) = 1 8 = 1 8. P(1) = , P(2) = , P(3) = , P(4) = 1 17 = 38/34 They cannot They can They cannot

9. P(1) = 0.22, P(2) = 0.11, P(3) = 0.17, P(4) = 0.50 10. P(1) = 0.05, P(2) = 0.11, P(3) = 0.18, P(4) = 0.18

B. For each of the following, determine whether it can serve as the probability distribution of a random variable X. Explain your answer. P(X) = for x = 1, 2, 3, …, 8 It can = = 1 2. P(X) = for x = 1, 2, 3, …, 9 3. P(X) = 3+𝑥 3 −𝑥 for x = 1, 2, 3, 4 4. P(X) = 𝑥 for x = 1, 2, 3, 4 5. P(X) = 𝑥 −2 5 for x = 1, 2, 3, 4, 5

2. three are brown and one is black? 3. exactly two are orange?
Seatwork (1 whole) (Show your solution). A box contains 5 yellow ball, 4 brown balls, 4 orange balls and 3 black balls. If four balls are drawn at random, what is the probability that 1. they are all yellow? 2. three are brown and one is black? 3. exactly two are orange? 4. none is black? 5. they are of different colors? STATISTICS AND PROBABILITY SAMSUDIN N. ABDULLAH, Ph.D.

Decision-making is an important aspect in business, education, insurance, and other real-life situations. Many decisions are made by assigning probabilities to all possible outcomes pertaining to the situation and then evaluating the results. This situation requires the use of random variable and probability distribution. Discrete Probability Distribution or Probability Function consists of the values a random variable can assume and the corresponding probabilities of the values. STATISTICS AND PROBABILITY SAMSUDIN N. ABDULLAH, Ph.D.

Properties of a Probability Distribution
1. The probability of each value of the random variable must be between or equal to 0 and 1. In symbol, we write it as 0 ≤ P(E) ≤ 1. 2. The sum of the probabilities of all values of the random variables must be equal to 1. In symbol, we write it as Ʃ P(E) = 1. STATISTICS AND PROBABILITY SAMSUDIN N. ABDULLAH, Ph.D.

CONSTRUCTING PROBABILITY DISTRIBUTION and ITS CORRESPONDING HISTOGRAM
Example 1. Four coins are tossed. Let Z be the random variable representing the number of heads that occur. Construct probability distribution of Discrete Random Variable Z. STATISTICS AND PROBABILITY SAMSUDIN N. ABDULLAH, Ph.D.

Possible Outcomes of Each Event
Solution: n(S) = 2⁴ = 16 Random Variable Z Possible Outcomes of Each Event P(Z) 0 HEAD TTTT 1 16 1 HEAD HTTT THTT TTHT TTTH 1 4 2 HEADS HHTT HTHT TTHH THHT THTHT HTTH 3 8 3 HEADS HHHT HHTH HTHH THHH 4 HEADS HHH

Probability Distribution
2 4 Number of Heads P(Z) 1 3 1 16 1 4 3 8 1 4 1 16

Number of Tails Y 1 2 3 Probability P(Y) 𝟏 𝟖 𝟑 𝟖
Example 2. Three coins are tossed. Let Y be the random variable representing the number of tails that occur. Construct probability distribution of a discrete random variable. Number of Tails Y 1 2 3 Probability P(Y) 𝟏 𝟖 𝟑 𝟖 STATISTICS AND PROBABILITY SAMSUDIN N. ABDULLAH, Ph.D.

Number of Defective Computer (X)
Solution: n(S) = 5C3 = 10 S = {N1N2N3, D1N1N2, D1N1N3, D1N2N3, D2N1N2, D2N1N3, D2N2N3, D1D2N1, D1D2N2, D1D2N3} Number of Defective Computer (X) 1 2 Probability P(x) 𝟏 𝟏𝟎 𝟑 𝟓 𝟑 𝟏𝟎

Determine whether the table presents a probability distribution
Determine whether the table presents a probability distribution. Explain your answer. 1) 2) 3) 4) 5)

Solve the following problems.
The daily demand for copies of a movie magazine at a variety store has the probability distribution as follows. Number of Copies X Probability P(X) 0.06 1 0.14 2 0.16 3 4 0.12 5 0.10 6 0.08 7 0.07 8 9 0.04 10 0.03

Questions: 1. What is the probability that three or more copies will be demanded in a particular day? 0.64 2. What is the probability that the demand will be at least two but not more than six? 0.60 3. What is the probability that the demand is between four and eight? 0.25 4. What is the probability that the demand is less than nine? 0.93 5. What is the probability that the number of demand is even number? 0.45 6. What is the probability that the demand is more than five? 0.28

Mean of a Discrete Probability Distribution Preparatory Lessons:
A. Given the values of the variables x and y, evaluate the following summations: x₁ = 4, x₂ = 2, x₃ = 5, x₄ = 1 y₁ = 2, y₂ = 1, y₃ = 0, y₄ = 2 Ʃx = = 12 Ʃy = = 5 Ʃxy = 4(2) + 2(1) + 5(0) + 1(2) = 12 Ʃ(x + y) = (4 + 2) + (2 + 1) + (5 + 0) + (1 + 2) = 17 Ʃ4xy = 4(4)(2) + 4(2)(1) + 4(5)(0) + 4(1)(2) = 48 STATISTICS AND PROBABILITY SAMSUDIN N. ABDULLAH, Ph.D.

Score Number of Students
B. The following are the scores of 40 students in a test. Compute the mean score. Score Number of Students 42 8 50 12 53 9 38 7 46 4 Solution: x = 𝟒𝟐(𝟖) + 𝟓𝟎(𝟏𝟐) + 𝟓𝟑(𝟗) + 𝟑𝟖(𝟕) + 𝟒𝟔(𝟒) 40 = 𝟏𝟖𝟔𝟑 𝟒𝟎 = STATISTICS AND PROBABILITY SAMSUDIN N. ABDULLAH, Ph.D.

C. Consider rolling a die
C. Consider rolling a die. What is the average number of spots that would appear? Number of Spots X Probability P(x) x·P(X) 1 1 6 2 2 6 3 3 6 4 4 6 5 5 6 6 6 6 Mean = 𝟐𝟏 𝟔 = 3.5

Solution: x = 𝟏𝟎 + 𝟖 +𝟕 + 𝟏𝟓 + 𝟐𝟎 + 𝟖 + 𝟖 𝟕 = 𝟕𝟔 𝟕 𝑴 𝒅 = 8
I. Find the mean, median and mode of each set of data. Show your solution if any. Round off your answers in 4 decimal palaces. 1) 10, 8, 7, 15, 20, 8, 8 Solution: x = 𝟏𝟎 + 𝟖 +𝟕 + 𝟏𝟓 + 𝟐𝟎 + 𝟖 + 𝟖 𝟕 = 𝟕𝟔 𝟕 𝑴 𝒅 = 8 x = 𝑴 𝒐 = 8

I. Find the mean, median and mode of each set of data
I. Find the mean, median and mode of each set of data. Show your solution if any. Round off your answers in 4 decimal palaces. 2) 150, 80, 95, 115, 250, 300, 125, 130, 150, 150 Solution: x = 𝟏𝟓𝟎+𝟖𝟎+𝟗𝟓+𝟏𝟏𝟓+𝟐𝟓𝟎+𝟑𝟎𝟎+𝟏𝟐𝟓+𝟏𝟑𝟎+𝟏𝟓𝟎+𝟏𝟓𝟎 𝟏𝟎 = 𝟏𝟓𝟒𝟓 𝟏𝟎 𝑴 𝒅 = 𝟏𝟑𝟎+𝟏𝟓𝟎 𝟏𝟎 = 𝟐𝟖𝟎 𝟐 = 140 = 𝑴 𝒐 = 150

Solution: = 𝟔𝟏𝟎 𝟑𝟎 𝑴 𝒅 = 𝟐𝟑+𝟐𝟎 𝟐 = 𝟒𝟑 𝟐 = 21.5000 = 20.3333 𝑴 𝒐 = 15
II. Solve for x , 𝑴 𝒅 and 𝑴 𝒐 of the following set of scores. Solution: x = 𝟐𝟓(𝟏𝟎)+𝟐𝟑(𝟓)+𝟐𝟎(𝟒)+𝟏𝟓(𝟏𝟏) 𝟑𝟎 = 𝟔𝟏𝟎 𝟑𝟎 𝑴 𝒅 = 𝟐𝟑+𝟐𝟎 𝟐 = 𝟒𝟑 𝟐 = = 𝑴 𝒐 = 15

Formula for the Mean of the Probability Distribution
µ = Ʃx · P(x) Examples: 1. The probabilities that a customer will buy 1, 2, 3, 4, or 5 items in a grocery store are 𝟑 𝟏𝟎 , 𝟏 𝟏𝟎 , 𝟏 𝟏𝟎 , 𝟐 𝟏𝟎 , 𝟑 𝟏𝟎 . What is the average number of items that customer will buy? Solution: µ = 1( 𝟑 𝟏𝟎 ) + 2( 𝟏 𝟏𝟎 ) + 3( 𝟏 𝟏𝟎 ) + 4( 𝟐 𝟏𝟎 ) + 5( 𝟑 𝟏𝟎 ) = 𝟑 𝟏𝟎 + 𝟐 𝟏𝟎 + 𝟑 𝟏𝟎 + 𝟖 𝟏𝟎 + 𝟏𝟓 𝟏𝟎 µ = 3.1

2. The probabilities that a surgeon operates on 3, 4, 6, 7 or 8 patients in any day are 0.15, 0.10, 0.20, 0.25, and 0.30, respectively. Find the average number of patients that a surgeon operates on a day. 3. Suppose the casino realizes that it is losing money in the long term and decides to adjust the payout levels by subtracting $1.00 from each price. The new probability distribution for each outcome is provided by the following table.: Outcome -$2.00 -$1.00 $2.00 $3.00 Probability STATISTICS AND PROBABILITY SAMSUDIN N. ABDULLAH, Ph.D.

σ² = Ʃx² · P(x) - µ² Variance of a Discrete Probability Distribution
σ² = Ʃ(x - µ)² · P(x) or σ² = Ʃx² · P(x) - µ² Standard Deviation of a Discrete Probability Distribution σ = Ʃ(x − µ)² · P(x) or σ = Ʃx² · P(x) − µ²

ơ = 8.8896 = 2.9815 (Standard Deviation)
Example: Find the variance and standard deviation of a given Discrete Probability Distribution below. x P(x) x.P(x) x - µ (x - µ)² (x - µ)².P(x) 1 0.20 -4.48 3 0.15 0.45 -2.48 6.1504 5 0.13 0.65 -0.48 0.2304 7 0.25 1.75 1.52 2.3104 9 0.27 2.43 3.52 Ʃx.P(x) = 5.48 Ʃ(x - µ)².P(x) = ơ²= (Variance) ơ = = (Standard Deviation) STATISTICS AND PROBABILITY SAMSUDIN N. ABDULLAH, Ph.D.

Assignment (1 /2 CW) Complete the table. Then, find the mean, variance and standard deviation of a given Discrete Probability Distribution below. x P(x) x.P(x) x - µ (x - µ)² (x - µ)².P(x) 1 0.10 2 0.18 5 0.22 6 0.19 7 0.15 11 0.16 Ʃx.P(x) = Ʃ(x - µ)².P(x) = STATISTICS AND PROBABILITY SAMSUDIN N. ABDULLAH, Ph.D.

ơ = 9.5099 = 3.0838 (Standard Deviation)
Assignment (1 /2 CW) Complete the table. Then, find the variance and standard deviation of a given Discrete Probability Distribution below. x P(x) x.P(x) x - µ (x - µ)² (x - µ)².P(x) 1 0.10 -4.51 2 0.18 0.36 -3.51 5 0.22 1.10 -0.51 0.2601 6 0.19 1.14 0.49 0.2401 7 0.15 1.05 1.49 2.2201 11 0.16 1.76 5.49 Ʃx.P(x) = 5.51 Ʃ(x - µ)².P(x) = 9.5099 ơ²= (Variance) ơ = = (Standard Deviation) STATISTICS AND PROBABILITY SAMSUDIN N. ABDULLAH, Ph.D.

CHAPTER II NORMAL DISTRIBUTION

NORMAL CURVE is a bell-shaped curve which shows the probability distribution of a continuous random variable. It represents a normal distribution. It has a mean µ = 0 and standard deviation ơ = 1. Its skewness is 0 and its kurtosis is 3. = 1 STATISTICS AND PROBABILITY SAMSUDIN N. ABDULLAH, Ph.D.

Properties of the Normal Probability Distribution
1. The distribution curve is bell-shaped. 2. The curve is symmetrical about its center. 3. The mean, the median, and the mode coincide at the center. 4. The width of the curve is determined by the standard deviation of the distribution. 5. The tails of the curve flatten out indefinitely along the horizontal axis, always approaching the axis but never touching it. That is, the curve is asymptotic to the base line. 6. The area under the curve is 1. Thus, it represents the probability or proportion or the percentage associated with specific sets of measurement values. STATISTICS AND PROBABILITY SAMSUDIN N. ABDULLAH, Ph.D.

Skewness talks about the degree of symmetry of a curve
Skewness talks about the degree of symmetry of a curve. It is asymmetry in a statistical distribution, in which the curve appears distorted or skewed either to the left or to the right. It can be quantified to define the extent to which a distribution differs from a normal distribution. Kurtosis, on the other hand, talks about the degree of peakedness of a curve. It refers to the pointedness or flatness of a peak in the distribution curve. STATISTICS AND PROBABILITY SAMSUDIN N. ABDULLAH, Ph.D.

Skewness is less than zero (negative).
Skewed to the Left Skewed to the Right Skewness is less than zero (negative). Skewness is greater than zero (positive). STATISTICS AND PROBABILITY SAMSUDIN N. ABDULLAH, Ph.D.

Types of Kurtosis STATISTICS AND PROBABILITY SAMSUDIN N. ABDULLAH, Ph.D.

A normal distribution (normal curve) is said to be Mesokurtic.
If the kurtosis of a curve is greater than zero (positive), the distribution is said to be Leptokurtic. This means that the distribution is taller and thinner than the normal curve. If the kurtosis of a curve is less than zero (negative), the distribution is said to be Platykurtic. This indicates that the distribution is flatter and wider than the normal curve. A normal distribution (normal curve) is said to be Mesokurtic. STATISTICS AND PROBABILITY SAMSUDIN N. ABDULLAH, Ph.D.

The skewness of a normal curve is 0 and its kurtosis is 3.

Determine the area BELOW the following. 1. z = 2 2. z = 2.9

B. Determine the area ABOVE the following. 1. z = 2.5 2. z = -2.5

C. Determine the area of the region indicated by the following
C. Determine the area of the region indicated by the following. Draw a normal curve for each. < z < 1 < z < 2 < z < 2.5 < z < 3.2 < z < 1.65 < z < 1.47 < z < 1.64 < z < 3 < z < 1.96 < z < -0.01 STATISTICS AND PROBABILITY SAMSUDIN N. ABDULLAH, Ph.D.

Determine the area of the region indicated by the following.
< z < 1 < z < 2 < z < 2.5 < z < 3 < z < 1.65 B. Determine the area of the region indicated by the following. 1. Below z = -2.76 2. Above z = -1.27 3. Below z = 1.09 4. Above z = 1.55 5. Below z = 2.13 STATISTICS AND PROBABILITY SAMSUDIN N. ABDULLAH, Ph.D.

Find the area of the shaded region of the normal curve.
1. A = or 34.13% STATISTICS AND PROBABILITY SAMSUDIN N. ABDULLAH, Ph.D.

2. A = 2(0.4938) = or 98.76% STATISTICS AND PROBABILITY SAMSUDIN N. ABDULLAH, Ph.D.

3. 2. -1.25 A = 0.5 – = or 10.56% STATISTICS AND PROBABILITY SAMSUDIN N. ABDULLAH, Ph.D.

4. A = = or 76.72% STATISTICS AND PROBABILITY SAMSUDIN N. ABDULLAH, Ph.D.

5. A = (0.50 – ) + ( – ) = = or 39.13% STATISTICS AND PROBABILITY SAMSUDIN N. ABDULLAH, Ph.D.

A = 0.5 – = or 10.56% STATISTICS AND PROBABILITY SAMSUDIN N. ABDULLAH, Ph.D.

A = 0.5 – – = = or 26.69% STATISTICS AND PROBABILITY SAMSUDIN N. ABDULLAH, Ph.D.

A = 0.5 – – = = or 63.10% STATISTICS AND PROBABILITY SAMSUDIN N. ABDULLAH, Ph.D.

-2.75 A = 0.5 – = or 0.30% STATISTICS AND PROBABILITY SAMSUDIN N. ABDULLAH, Ph.D.

-2.75 A = 0.5 – = = or 39.74% STATISTICS AND PROBABILITY SAMSUDIN N. ABDULLAH, Ph.D.

-2.75 A = 0.5 – – = = or 45.80% STATISTICS AND PROBABILITY SAMSUDIN N. ABDULLAH, Ph.D.

A = 1 – 2(0.4750) = 1 – 0.95 = 0.05 or 5% STATISTICS AND PROBABILITY SAMSUDIN N. ABDULLAH, Ph.D.

Applications of Normal Curve

The following formula is used when sample size is not given:

The scores of students in the first quarter examination for Mathematics has a mean (µ) 32 and standard deviation (σ) of 5. Find the z-scores corresponding to each of the following. 1. 37 2. 22 3. 33 4. 28 5. 40 6. 27 7. 34 8. 30 9. 32

Solutions: 1. z = 𝒙 − 𝝁 σ = 𝟑𝟕 −𝟑𝟐 𝟓 = 𝟓 𝟓 = 1

6. z = 𝒙 − 𝝁 σ = 𝟐𝟕 −𝟑𝟐 𝟓 = −𝟓 𝟓 = -1 7. z = 𝒙 − 𝝁 σ = 𝟑𝟒 −𝟑𝟐 𝟓 = 𝟐 𝟓 = 0.4 8. z = 𝒙 − 𝝁 σ = 𝟑𝟎 −𝟑𝟐 𝟓 = −𝟐 𝟓 = -0.4 9. z = 𝒙 − 𝝁 σ = 𝟑𝟐 −𝟑𝟐 𝟓 = 𝟎 𝟓 = 0 10. z = 𝒙 − 𝝁 σ = 𝟐𝟓 −𝟑𝟐 𝟓 = −𝟕 𝟓 = -1.4 STATISTICS AND PROBABILITY SAMSUDIN N. ABDULLAH, Ph.D.

B. The scores of a group of students in a standardized test are normally distributed with a mean of 60 and standard deviation of 8. Answer the following. How many percent of the students got below 72? What part of the group scored between 58 and 76? If there were 250 students who took the test, about how many students scored higher than 64? How many percent of the students got above 65?

Solution: z = 𝒙 − 𝝁 σ = 𝟕𝟐 −𝟔𝟎 𝟖 = 𝟏𝟐 𝟖 = 1.5 Referring to the z-table, the area below z = 1.5 is Therefore, about 93.32% of the group got below 72. STATISTICS AND PROBABILITY SAMSUDIN N. ABDULLAH, Ph.D.

2. z = 𝒙 − 𝝁 σ = 𝟓𝟖 −𝟔𝟎 𝟖 = −𝟐 𝟖 = -0.25 z = 𝒙 − 𝝁 σ = 𝟕𝟔 −𝟔𝟎 𝟖 = 𝟏𝟔 𝟖 = 2 A = = or 57.59% Thus, there were 57.59% of the students who scored between 58 and 76. STATISTICS AND PROBABILITY SAMSUDIN N. ABDULLAH, Ph.D.

3. z = 𝒙 − 𝝁 σ = z = 𝟔𝟒 −𝟔𝟎 𝟖 = 𝟒 𝟖 = 0.5 A = 0.5 – = 250(0.3085) = or 77 Thus, there were 77 students who got higher than 64. STATISTICS AND PROBABILITY SAMSUDIN N. ABDULLAH, Ph.D.

4. z = 𝒙 − 𝝁 σ = z = 𝟔𝟓 −𝟔𝟎 𝟖 = 𝟓 𝟖 = 0.63 A = 0.5 – = or 26.43% Thus, there were 26.43% of the students who got above 65. STATISTICS AND PROBABILITY SAMSUDIN N. ABDULLAH, Ph.D.

Solution: A = (1 – 0.05 ) – 0.5 = 0.95 – 0.5 = 0.45 1.65 = 𝒙 −𝟐𝟖𝟓 𝟏𝟐
C. A highly selective university only admits the top 5% of the total examinees in their entrance examination. The results of this year’s entrance examination follow a normal distribution with a mean of 285 and a standard deviation of 12. What is the least score of an examinee who can be admitted to the university? Solution: z = 𝒙 −𝟐𝟖𝟓 𝟏𝟐 A = (1 – 0.05 ) – 0.5 = 0.95 – 0.5 = 0.45 1.65 = 𝒙 −𝟐𝟖𝟓 𝟏𝟐 x – 285 = 1.65(12) = X = or 305

Learning the Probability Notations Under the Normal Curve
P(a < z < b) denotes the probability that the z-score is between a and b. P(z >a) denotes the probability that the z-score is greater than a. P(z < a) denotes the probability that the z-score is less than a. P(a ≤ z ≤ b) = P(a < a < b) STATISTICS AND PROBABILITY SAMSUDIN N. ABDULLAH, Ph.D.

The Central Limit Theorem is of fundamental importance in Statistics because it justifies the use of normal curve methods for a wide range of problems. This theorem applies automatically to sampling from infinite population. The following formula is used when sample is given. z = 𝒙 − 𝝁 𝓸 𝒏 where: 𝑥 = sample mean μ = population mean σ = population standard deviation n = sample size STATISTICS AND PROBABILITY SAMSUDIN N. ABDULLAH, Ph.D.

σ =population standard deviation
The following formula is used when sample is not given given. z = 𝒙 − 𝝁 σ where: 𝑥 = sample mean μ = population mean σ =population standard deviation STATISTICS AND PROBABILITY SAMSUDIN N. ABDULLAH, Ph.D.

Quiz (1/2 CW) Find the following: Draw a normal curve for each problem
P(z < -2.52) = P(z > 2.17) = P(1.23 < z < 2.21) = P(-0.23 < z < -1.41) = P(-2.03 < z < 1.08) = STATISTICS AND PROBABILITY SAMSUDIN N. ABDULLAH, Ph.D.

Problems 1. The average time it takes a group of college students to complete a certain examination is 46.2 minutes. The standard deviation is 8 minutes. Assume that the variable is normally distributed. a. What is the probability that a randomly selected college student will complete the examination in less than 43 minutes? b. If 50 randomly selected college students take the examination, what is the probability that the mean time it takes the group to complete the test will be more than 43 minutes? STATISTICS AND PROBABILITY SAMSUDIN N. ABDULLAH, Ph.D.

Given: x = 43 minutes μ = 46.2 minutes σ = 8 Solution: P(x < 43) = ? z = 𝒙 − 𝝁 σ = 43 − = −3.2 8 = -0.40 P(x < 43) = P(z < -0.40) = – = or 34.46% Thus, the probability that a randomly selected college student will complete the test in less than 43 minutes is 34.46%. STATISTICS AND PROBABILITY SAMSUDIN N. ABDULLAH, Ph.D.

b) Given: x = 43 minutes μ = 46.2 minutes σ = 8 n = 50 Solution: P(x > 43) = ? z = 𝒙 − 𝝁 𝓸 𝒏 = 43 − 46 𝟖 𝟓𝟎 = −3.2 𝟖 𝟕.𝟎𝟕 = −𝟑.𝟐 𝟏.𝟏𝟑 = -2.83 P(x > 43) = P(z > ) = = or 99.77% Thus, the probability that 50 randomly selected college students will complete the test in more than 43 minutes is 99.77%. STATISTICS AND PROBABILITY SAMSUDIN N. ABDULLAH, Ph.D.

2. The entrance examination scores of incoming freshmen in a state college are normally distributed with a mean of 78 and a standard deviation of 10. What is the probability that a randomly selected student has a score a. below 78? b. below 76? c. between 75 to 80? d. above 95? e. What is the probability that the 45 randomly selected freshmen can have a mean of greater than 76? STATISTICS AND PROBABILITY SAMSUDIN N. ABDULLAH, Ph.D.

Given: x = 78 μ = 78 σ = 10 Solution: P(x < 78) = ? z = 𝒙 − 𝝁 σ = 78 − = 0 10 = 0 P(x < 78) = P(z < 0) = 0.50 or 50% Thus, the probability of a randomly selected student to have a score of less than 78 is 50%. STATISTICS AND PROBABILITY SAMSUDIN N. ABDULLAH, Ph.D.

b) Given: x = 76 μ = 78 σ = 10 Solution: P(x < 76) = ? z = 𝒙 − 𝝁 σ = 76 − = −2 10 = -0.2 P(x < 78) = P(z < -0.2) = or 7.93% Thus, the probability of a randomly selected student to have a score less than 76 is 7.93%. STATISTICS AND PROBABILITY SAMSUDIN N. ABDULLAH, Ph.D.

c) Given: x₁ = 75 x₂ = 80 μ = 78 σ = 10 Solution:
P(75 < x < 80) = ? z = 𝒙 − 𝝁 σ = 75 − = −3 10 = -0.3 z = 80 − = 2 10 = 0.2 P(75 < x < 80) = = or 19.72% Thus, the probability of a randomly selected student to have a score between 75 and 80 is %. STATISTICS AND PROBABILITY SAMSUDIN N. ABDULLAH, Ph.D.

d) Given: x = 95 μ = 78 σ = 10 Solution: P(x > 95) = ? z = 𝒙 − 𝝁 σ
= 95 − = = 1.7 P(x > 95) = P(x > 1.7) = – = or 4.46% Thus, the probability of a randomly selected student to have a score above %. STATISTICS AND PROBABILITY SAMSUDIN N. ABDULLAH, Ph.D.

e) Given: x = 76 μ = 78 σ = 10 n = 45 Solution: P(x > 95) = ?
z = 𝒙 − 𝝁 ℴ 𝒏 = 76 − 78 𝟏𝟎 𝟒𝟓 = −2 𝟏𝟎 𝟔.𝟕𝟏 = −𝟐 𝟏.𝟒𝟗 = -1.34 P(x > 76) = P(x > 1.34) = = or 90.99% Thus, the probability that the 45 randomly selected freshmen can have a mean of greater than 76 is 90.99%. STATISTICS AND PROBABILITY SAMSUDIN N. ABDULLAH, Ph.D.

b. What scores comprise the middle 95% of all scores?
3. Suppose from the 1,000 incoming freshmen who took the entrance examination, it was found out that their mean score was 80 and the standard deviation was 12. a. How many students passed the test if the passing score is set at 75? b. What scores comprise the middle 95% of all scores? c. What scores comprise above 95% of all scores? d. What scores comprise below 89% of all scores? STATISTICS AND PROBABILITY SAMSUDIN N. ABDULLAH, Ph.D.

Given: x = 75 μ = 80 ℴ = 12 Solution: P(x > 75) z = 𝒙 − 𝝁 σ = 𝟕𝟓 − 𝟖𝟎 𝟏𝟐 = −𝟓 𝟏𝟐 = -0.42 P(x > 75) = P(z > -0.42) = = (1000) = or 663 Thus, there were 663 freshmen who passed the entrance examination.. STATISTICS AND PROBABILITY SAMSUDIN N. ABDULLAH, Ph.D.

SAMPLING AND SAMPLING DISTRIBUTION
CHAPTER III SAMPLING AND SAMPLING DISTRIBUTION STATISTICS AND PROBABILITY SAMSUDIN N. ABDULLAH, Ph.D.

POPULATION SAMPLE Sampling is a process of getting the sample.

Statistic versus Parameter
Statistics – a branch of Mathematics. It is a subject offered in a school. Statistic – a datum in a collection of statistics. It is a characteristic of a sample. It is used to estimate the value of a population. The average grade of students would be an example of a statistic.

Statistic versus Parameter
Sample Statistic – any quantity computed from a sample taken from a population with the intention of using this quantity to estimate same but unknown quantities of the population. The examples would be sample mean and sample variance. Parameter – a useful component of statistical analysis. It refers to the characteristics that are used to define a given population. Statistic describes a sample while parameter describes a population. In other words, statistic is used to estimate a parameter.

Examples of a Parameter Population mean (µ)
Population standard deviation (σ) Population variance (σ²) Examples of a Statistic Population mean (µ) Population standard deviation (σ) Population variance (σ²)

Say something about the following figures.
Sample Mean 10 8 34 40 34 29 33 32 17 32.7 35 32.7 32 38 24 33 31 26 33 30 55 80 Figure 2 Figure 1

Descriptive Statistics of the two given sets of sample data

Random Sampling refers to the sampling technique in which each member of the population is given equal chance from a population is called sample and the process of taking samples is called sampling.

Since survey research has a larger scope of respondents, sampling technique is very necessary. For instance, the population of the research is 6,033 students, teachers, parents and school administrators. It doesn’t mean that all of these 6,033 target respondents will be given a survey questionnaire. Sampling technique should be done systematically so that expenses and time will be minimized but the generality and reliability of the information will be maintained. STATISTICS AND PROBABILITY SAMSUDIN N. ABDULLAH, Ph.D.

Probability Sampling Methods 1. Simple Random Sampling
- Fishbowl method - Lottery Method 2. Systematic Sampling 3. Stratified Sampling 4. Cluster Sampling 5. Multistage Sampling STATISTICS AND PROBABILITY SAMSUDIN N. ABDULLAH, Ph.D.

Simple Random Sampling (SRS) is a basic sampling technique where a researcher selects a group of a sample for study from a larger group (population). Each individual is chosen entirely by chance and each member of the population has an equal chance of being included in the sample. STATISTICS AND PROBABILITY SAMSUDIN N. ABDULLAH, Ph.D.

Systematic Sampling is a statistical method involving the selection of elements from an ordered sampling frame. The most common form of systematic sampling is an equiprobability method. In this approach, progression through the list is treated circularly, with a return to the top once the end of the list is passed.

Stratified Sampling is a method of sampling in which the researcher divides the population into separate groups, called strata. Then, a probability sampling is drawn from each group.

Cluster Sampling is a sampling technique used when mutually homogeneous yet internally heterogeneous groupings are evident in a statistical population. It is often used in marketing research. In this sampling technique, the total population is divided into groups called clusters a simple random sample of the group is selected.

Multistage Sampling is the taking of samples in stages using smaller and smaller sampling units at each stage. It can be a complex form of cluster sampling since it is a type of sampling which involves dividing the populations into groups. A combination of stratified, cluster and simple random sampling is used in multistage sampling technique. STATISTICS AND PROBABILITY SAMSUDIN N. ABDULLAH, Ph.D.

Non-probability Sampling Methods 1. Quota Sampling
2. Convenience Sampling 3. Purposive Sampling 4. Self-Selection Sampling 5. Snowball Sampling 6. Judgemental Sampling STATISTICS AND PROBABILITY SAMSUDIN N. ABDULLAH, Ph.D.

Problem: A researcher is conducting a study about the effect of student absenteeism on academic performance of students. The main respondents of the study are the students from all grade levels. The number of sub-population per grade level is as follows: Grade 7 – 1209 Grade 8 – 1083 Grade 9 – 985 Grade 10 – 889 Grade 11 – 1087 Grade 12 – 780 What appropriate sampling technique can be applied? How many samples do we have? How many samples from each grade level? STATISTICS AND PROBABILITY SAMSUDIN N. ABDULLAH, Ph.D.

Stratified Random Sampling using Slovin’s Equation
where: n = desired sample N = population e = margin of error = 5% = 0.05 STATISTICS AND PROBABILITY SAMSUDIN N. ABDULLAH, Ph.D.

Proportional Percentage: 𝟑𝟕𝟓 𝟔𝟎𝟎𝟑 = 0.0622
Solution: Grade 7 – 1209 Grade 8 – 1083 Grade 9 – 985 Grade 10 – 889 Grade 11 – 1087 Grade 12 – 780 6033 n = 𝑵 𝟏 + 𝑵 𝒆 𝟐 = 𝟔𝟎𝟑𝟑 𝟏 + 𝟔𝟎𝟑𝟑 (𝟎.𝟎𝟓) 𝟐 = 𝟔𝟎𝟑𝟑 𝟏 + 𝟔𝟎𝟑𝟑(𝟎.𝟎𝟎𝟐𝟓) = 𝟔𝟎𝟑𝟑 𝟏 + 𝟏𝟓.𝟎𝟖𝟐𝟓 = 𝟔𝟎𝟑𝟑 𝟏𝟔.𝟎𝟖𝟐𝟓 n = 375 Proportional Percentage: 𝟑𝟕𝟓 𝟔𝟎𝟎𝟑 = STATISTICS AND PROBABILITY SAMSUDIN N. ABDULLAH, Ph.D.

Grade 7 – 1209 x = 75 Grade 8 – 1083 x = 67 Grade 9 – x = 61 Grade 10 – x = 55 Grade 11 – 1087 x = 68 Grade 12 – x = 49 375 Then, apply the simple random sampling technique in choosing the individual respondent per group. STATISTICS AND PROBABILITY SAMSUDIN N. ABDULLAH, Ph.D.

Quiz (1 whole): A researcher is conducting a study about the full implementation of Senior High School (SHS) curriculum in Sultan Kudarat. The following are the sub-population of the study: Students – 3050 Teachers – Parents – Principals – Compute for the total number of sample as well as the sample per group. STATISTICS AND PROBABILITY SAMSUDIN N. ABDULLAH, Ph.D.

Proportional Percentage: 𝟑𝟔𝟒 𝟒𝟎𝟕𝟎 = 0.0894
Solution: Students – 3050 Teachers – 550 Parents – 320 Principals – 150 4070 n = 𝑵 𝟏 + 𝑵 𝒆 𝟐 = 𝟒𝟎𝟕𝟎 𝟏 + 𝟒𝟎𝟕𝟎 (𝟎.𝟎𝟓) 𝟐 = 𝟒𝟎𝟕𝟎 𝟏 + 𝟒𝟎𝟕𝟎(𝟎.𝟎𝟎𝟐𝟓) = 𝟒𝟎𝟕𝟎 𝟏 + 𝟏𝟎.𝟏𝟕𝟓 = 𝟒𝟎𝟕𝟎 𝟏𝟏.𝟏𝟕𝟓 n = 364 Proportional Percentage: 𝟑𝟔𝟒 𝟒𝟎𝟕𝟎 = STATISTICS AND PROBABILITY SAMSUDIN N. ABDULLAH, Ph.D.

Students – 3050 x = 273 Teachers – x = 49 Grade 9 – x = 29 Grade 10 – x = 13 364 Then, apply the simple random sampling technique in choosing the individual respondent per group. STATISTICS AND PROBABILITY SAMSUDIN N. ABDULLAH, Ph.D.

Population USM - Kabacan – 1580 MSU - Maguindanao – 1398 CCSPC – 1409
SKSU – 1216 STATISTICS AND PROBABILITY SAMSUDIN N. ABDULLAH, Ph.D.

Proportional Percentage: 𝟑𝟕𝟑 𝟓𝟔𝟎𝟑 = 0.0666
Solution: USM - Kabacan – 1580 MSU - Maguindanao – 1398 CCSPC – 1409 SKSU – 1216 5603 n = 𝑵 𝟏 + 𝑵 𝒆 𝟐 = 𝟓𝟔𝟎𝟑 𝟏 + 𝟓𝟔𝟎𝟑 (𝟎.𝟎𝟓) 𝟐 = 𝟓𝟔𝟎𝟑 𝟏 + 𝟓𝟔𝟎𝟑(𝟎.𝟎𝟎𝟐𝟓) = 𝟓𝟔𝟎𝟑 𝟏 + 𝟏𝟒.𝟎𝟎𝟕𝟓 = 𝟓𝟔𝟎𝟑 𝟏𝟓.𝟎𝟎𝟕𝟓 n = 373 Proportional Percentage: 𝟑𝟕𝟑 𝟓𝟔𝟎𝟑 = STATISTICS AND PROBABILITY SAMSUDIN N. ABDULLAH, Ph.D.

USM - Kabacan – 1580x0.0666 = 105 MSU - Maguindanao – 1398x0.0666 = 93
CCSPC – 1409x = 94 SKSU – 1216x = 81 373 Then, apply the simple random sampling technique in choosing the individual respondent per group. STATISTICS AND PROBABILITY SAMSUDIN N. ABDULLAH, Ph.D.

II. A researcher is conducting a study about the implementation of Solid Waste Management in the City Divisions of Region XII. The following are the sub-population of the study: General Santos City – 4050 Koronadal City – Cotabato City – Tacurong City – Kidapawan City – Compute for the total number of sample as well as the sample per group. STATISTICS AND PROBABILITY SAMSUDIN N. ABDULLAH, Ph.D.

b. What is the probability that a child has an IQ score above 120?
Assignment (1 whole) Direction: Use the idea of a Normal Curve and the Central Limit Theorem to solve the following problems. Illustrate the shaded region of a normal curve representing your answer. The IQ scores of children in a special education class are normally distributed with a mean of 95 and a standard deviation of 10. a. What is the probability that one of the children has an IQ score below 100? b. What is the probability that a child has an IQ score above 120? c. What are the chances that a child has an IQ score of 140? d. How many children have IQ scores above 100 if there are 30 of them in class? STATISTICS AND PROBABILITY SAMSUDIN N. ABDULLAH, Ph.D.

Select your answers from the following: Mean Median Mode Range
Standard Deviation Variance Coefficient of Variation Kurtosis Skewness Scatteredness Frequency Percentage 13. t-distribution curve Normal Curve Statistics Zero Bell-Shaped Research Statistics and Probability Simple Random Sampling STATISTICS AND PROBABILITY SAMSUDIN N. ABDULLAH, Ph.D.

Two Types of Statistics
1. Descriptive Statistics is concerned with the gathering, classification and presentation of data and the collection of summarizing values to describe group characteristics of data. The most commonly used summarizing values to describe group characteristics of data are percentage, measures of central tendency (mean, mode, median); measures of variability (range, standard deviation, variance, coefficient of variation); of skewness and kurtosis. Examples of descriptive statistics are the class average of examination, range of student scores, average salary, means of managerial satisfaction and average return of investment. STATISTICS AND PROBABILITY SAMSUDIN N. ABDULLAH, Ph.D.

2. Inferential Statistics pertains to the methods dealing with making inference, estimates or prediction about a large set of data using the information gathered. Commonly used inferential statistical tools or techniques are testing hypothesis using the z-test, t-test, analysis of variance (ANOVA), simple linear correlation (Pearson r), Spearman’s Rho, chi-square (x²) and regression. STATISTICS AND PROBABILITY SAMSUDIN N. ABDULLAH, Ph.D.

Two Forms of Hypothesis
Null Hypothesis (Ho) is the hypothesis to be tested and it represents what the investigation doubts to be true. Alternative Hypothesis (Ha) is the operational statement of the theory that the experimenter or researcher believes to be true and wishes to be true. STATISTICS AND PROBABILITY SAMSUDIN N. ABDULLAH, Ph.D.

Two Types of Hypothesis Testing
One-tailed (directional) test occurs when the researcher has the prior expectation about the sample value he expects to observe. 2. Two-tailed (non-directional) test occurs when the alternative hypothesis does not specify a directional difference for the parameter of interest. This test is applied when the researcher doesn’t have the prior expectation regarding the value he expects to see in the sample. STATISTICS AND PROBABILITY SAMSUDIN N. ABDULLAH, Ph.D.

Two Types of Hypothesis Testing
One-tailed (directional) test occurs when the researcher has the prior expectation about the sample value he expects to observe. STATISTICS AND PROBABILITY SAMSUDIN N. ABDULLAH, Ph.D.

2. Two-tailed (non-directional) test occurs when the alternative hypothesis does not specify a directional difference for the parameter of interest. This test is applied when the researcher doesn’t have the prior expectation regarding the value he expects to see in the sample. STATISTICS AND PROBABILITY SAMSUDIN N. ABDULLAH, Ph.D.

What is a Hypothesis? A hypothesis is basically a statement about the target population. This is formulated as a result of years of observation and researches. New researches may result from one’s desire to determine whether or not a researcher’s hypothesis is supported when a sample data are subjected to rigorous scientific statistical methods. A statistical hypothesis is an assertion or conjecture concerning one or more populations STATISTICS AND PROBABILITY SAMSUDIN N. ABDULLAH, Ph.D.

Steps in Hypothesis Testing
Step 1. Formulate the null and alternative hypotheses. Step 2. Set the level of significance (α). Step 3. Select the appropriate test statistic (statistical tool). Step 4. Establish the critical (rejection) region. Step 5. Compute the value of the test statistic from the sample data. Step 6. State your conclusion. STATISTICS AND PROBABILITY SAMSUDIN N. ABDULLAH, Ph.D.

Testing a Hypothesis About a Single Mean Using Large Samples (z-test)

Examples: 1. In a recent survey of nurses in Region XII, it was found out that the average monthly net income of nurses is ₱ 8, Suppose a researcher wants to test this figure by a random sample of 158 nurses in Region XII to determine whether the monthly net income has changed. Suppose further the average net monthly income of the 158 nurses is ₱ 9, and the population standard deviation was found to be ₱ 1, STATISTICS AND PROBABILITY SAMSUDIN N. ABDULLAH, Ph.D.

z-test (right-tailed) The z-critical value = 1.65 Computation:
Solution: Ho: x = ₱8,048.25 Ha: x > ₱8,048.25 II. α = 0.05 z-test (right-tailed) The z-critical value = 1.65 Computation: z = 𝑥 − 𝜇 ℴ 𝑛 = − = = z = 12.22 VI. Decision Making/Conclusion Since that z-computed value of is greater than the z-critical value of 1.65, we have to reject the null hypothesis. Thus, the current average salary of nurses in Region XII which is ₱9, is significantly higher than ₱8, STATISTICS AND PROBABILITY SAMSUDIN N. ABDULLAH, Ph.D.

2. The owner of a factory that sells a particular bottled fruit juice claims that the average capacity of their product is 250 mL. To test the claim, a consumer group gets a sample of 100 such bottles, calculates the capacity of each bottle, and then finds the mean capacity to be 248 mL. The standard deviation is 5 mL. Is the claim true at 1% significant level? STATISTICS AND PROBABILITY SAMSUDIN N. ABDULLAH, Ph.D.

The z-critical value = -2.33 Computation: z = 𝑥 − 𝜇 ℴ 𝑛
Solution: Ho: x = 250 mL Ha: x < 250 mL II. α = 0.01 z-test (left-tailed) The z-critical value = Computation: z = 𝑥 − 𝜇 ℴ 𝑛 = 248 − = − = −2 0.5 z = -4 VI. Decision Making/Conclusion Since that z-computed value of -4 is less than the z-critical value of -2.33, we have to reject the null hypothesis. Thus, the 248 mL is significantly lower than 250 mL. The claim is not true. STATISTICS AND PROBABILITY SAMSUDIN N. ABDULLAH, Ph.D.

3. A researcher claims that there is a significant difference on the Mathematics performance of male and female students. A population of male students in Grade 10 has a mean of and a standard deviation of To prove his claim, a sample of 81 female students in the same grade level is found to have a mean of Is the claim of a researcher true? Use the 5% level of significance. STATISTICS AND PROBABILITY SAMSUDIN N. ABDULLAH, Ph.D.

Solution: Ho: x = 38.25 Ha: x ≠ 38.25 II. α = 0.05 z-test (two-tailed)
The z-critical value = 1.96 Computation: z = 𝑥 − 𝜇 ℴ 𝑛 = − = − = − z = -1.24 VI. Decision Making/Conclusion Since that z-computed value of is greater than the z-critical value of -1.65, we have to accept the null hypothesis. The claim of a researcher is not true. Thus, there is no significant difference on the Mathematics performance of male and female students.

Confidence Coefficients of z-Distribution (z-test)

Direction: Fill-in the boxes with the correct answers regarding hypothesis testing. Second row serves as your example. STATISTICS AND PROBABILITY SAMSUDIN N. ABDULLAH, Ph.D.

Direction: Fill-in the boxes with the correct answer regarding hypothesis testing. Second row serves as your example. STATISTICS AND PROBABILITY SAMSUDIN N. ABDULLAH, Ph.D.

Another Problem on Hypothesis Testing
A researcher wants to prove that the average monthly salary of the private school teachers is significantly different from the average monthly salary of the public school teachers. The average salary of the public school teacher is Pph24,500 and a population standard deviation of Php4, A sample of 150 private school teachers was considered and found to have an average monthly salary of Php15,000. Is the claim of a researcher true? Use hypothesis testing to justify your answer. STATISTICS AND PROBABILITY SAMSUDIN N. ABDULLAH, Ph.D.

Solution: Ho: x = 24,500.00 Ha: x < 24,500 II. α = 0.05 z-test (Left-tailed) The z-critical value = Computation: z = 𝑥 − 𝜇 ℴ 𝑛 = 15,000−24,500 4, = −9,500 4, = −9, z = VI. Decision Making/Conclusion Since that z-computed value of is less than the z-critical value of -1.65, we have to reject the null hypothesis. The claim of a researcher is true. Thus, the monthly average salary of private school teachers is significantly lower than the monthly salary of private school teachers. STATISTICS AND PROBABILITY SAMSUDIN N. ABDULLAH, Ph.D.

Solution: Ho: x = 594.41 Ha: x ≠ 594.41 II. α = 0.05
Given: µ = ℴ = 87.16 samples: 578, 605, 599, 790, 554, 615, 568, 498, 598, 625, 618, 608, 589, 580, 589 Question: Is the sample mean significantly different from the population mean? V. Computation: x = = = z = − = = = Solution: Ho: x = Ha: x ≠ II. α = 0.05 z-test (Two-tailed) The z-critical value = 1.96 STATISTICS AND PROBABILITY SAMSUDIN N. ABDULLAH, Ph.D.

IV. Since that the z-comp = 0. 2899 is less than z-critical = 1
IV. . Since that the z-comp = is less than z-critical = 1.96, we must reject the null hypothesis. Thus, the sample mean is not significantly different from the population mean. STATISTICS AND PROBABILITY SAMSUDIN N. ABDULLAH, Ph.D.

Is the claim of a teacher true using 1% level of significance?
2. A teacher claims that the learning performance of male and females students in Mathematics is comparable. In a recently concluded standardized test in Mathematics , male students were found to have a population mean of and a standard deviation of To prove his claim, a teacher randomly chose his samples of female students and their scores were as follows: 35, 35, 44, 49, 50, 53, 54, 45, 35, 38, 29, 30, 38, 40, 30, 35, 36, 28, 36, 30. Is the claim of a teacher true using 1% level of significance? STATISTICS AND PROBABILITY SAMSUDIN N. ABDULLAH, Ph.D.

Solution: Ho: x = 48.25 Ha: x ≠ 48.25 II. α = 0.01 z-test (Two-tailed) The z-critical value = -2.58 Computation: z = 𝑥 − 𝜇 ℴ 𝑛 = − = − = −9, z = VI. Decision Making/Conclusion Since that z-computed value of is less than the z-critical value of -2.58, we have to reject the null hypothesis. The claim of a researcher is not true. Thus, the learning performance of male students is significantly higher than female students in Mathematics. STATISTICS AND PROBABILITY SAMSUDIN N. ABDULLAH, Ph.D.

3. In a recently concluded English proficiency examination, a population of male students was found to have a mean of and a standard deviation of A sample of female students registered the following raw scores: 90, 75, 68, 80, 68, 70, 68, 68, 78, 85, 83, 65, 71, 82, 58, 68, 76, 80, 85, 78, 78, 80, 85. Using the 5% level of significance, are female students more proficient in English compared with male students? STATISTICS AND PROBABILITY SAMSUDIN N. ABDULLAH, Ph.D.

VI. Decision Making/Conclusion
Solution: Ho: x = 70.08 Ha: x ≠ 70.08 II. α = 0.05 z-test (Two-tailed) The z-critical value = 1.96 Computation: z = 𝑥 − 𝜇 ℴ 𝑛 = − = = z = VI. Decision Making/Conclusion Since that z-computed value of is greater than the z-critical value of 1.96, we have to reject the null hypothesis. Female students are more proficient in English compared with male students. STATISTICS AND PROBABILITY SAMSUDIN N. ABDULLAH, Ph.D.

THE DIFFERENCE BETWEEN THE z-Distribution CURVE (NORMAL CURVE) AND t-Distribution Curve
The confidence coefficients of the z-distribution are constant with the given confidence level regardless of the number of sample while the confidence coefficients of the t-distribution change depending upon to the degrees of freedom. STATISTICS AND PROBABILITY SAMSUDIN N. ABDULLAH, Ph.D.

Testing a Hypothesis About a Single Mean Using Small Samples (t-test)

1. A certain brand of laundry soap is advertised to have a net weight of 500 grams. If the net weights of a random sample of 10 boxes are 495, 503, 507, 498, 490, 505, 510, 502, 493, and 506 grams, can it be concluded that the average net weight of the boxes is less than the advertised amount? Use 3% level of significance. STATISTICS AND PROBABILITY SAMSUDIN N. ABDULLAH, Ph.D.

Solution: Ho: x = 500 grams Ha: x ≠ 500 grams II. α = 0.01
t-test (two-tailed) tcritical (df = 9) = 3.250 Computation: x = = x = 500.9 t = 𝑥 − 𝜇 𝑠 𝑛 = − = − t = − = z = VI. Decision Making/Conclusion Since that t-computed value of is less than the t-critical value of 3.250, we have to accept the null hypothesis. Thus, the net weights of a sample of 10 boxes of soap are statistically equal to the advertised brand of soap.

Testing a Hypothesis About Two Sample Means (t-test)
; Where: x₁ = first sample mean x₂ = second sample mean s₁ = standard deviation of a first sample s₂ = standard deviation of a second sample n₁ = number of the first sample n₂ = number of the second sample STATISTICS AND PROBABILITY SAMSUDIN N. ABDULLAH, Ph.D.

Problems: 1. The pre-test results of the two sections in Mathematics are as follows: Section A: 25, 20, 24, 25, 26, 28, 20, 18 Section B: 23, 21, 23, 26, 25, 27, 19, 17, 19 Using 5% level of significance, is there a significant difference in the pre-test scores of Section A and Section B? STATISTICS AND PROBABILITY SAMSUDIN N. ABDULLAH, Ph.D.

Hₒ: x₁ = x₂ Hₐ: x₁ ≠ x₂ α = 0.05 t-test (two-tailed)
df = – 2 = 15 tcritical = V. Computation: x₁ = = 23.25 x₂ = = 22.22 STATISTICS AND PROBABILITY SAMSUDIN N. ABDULLAH, Ph.D.

VI. Decision Since that the t-computed value = is less than the t-critical value = , we have to accept the null hypothesis. Therefore, there is no significant difference on the pre-test scores of Section A and Section B. STATISTICS AND PROBABILITY SAMSUDIN N. ABDULLAH, Ph.D.

THE NATURE OF STATISTICS
Statistics refers to the methods in collection, presentation, analysis and interpretation of data. Data Gathering or Collection may be done through interview, questionnaires, tests, observation, registration and experiments. Presentation of Data refers to the organization of data into tables, graphs, charts or paragraphs. Hence, presentation of data may be tabular, graphical or textual. Analysis of Data pertains to the process of extracting from the given data relevant and noteworthy information and this uses statistical tools or techniques. Interpretation of Data refers to the drawing of conclusions or inferences from the analyzed data. STATISTICS AND PROBABILITY SAMSUDIN N. ABDULLAH, Ph.D.

IDENTIFYING THE STATISTICAL TOOL APPLICABLE FOR THE GIVEN STATEMENT OF THE PROBLEM

1. SOP: What is the profile of STEM teachers in terms of teaching experience and educational attainment? 2. SOP: To what extent is the problem solving skills of grade 7 students? 3. SOP: Is there a significant gender difference on the performance of students in their Geometry subjects? 4. SOP: What is the impact of the reading interest on students’ literary comprehension? STATISTICS AND PROBABILITY SAMSUDIN N. ABDULLAH, Ph.D.

5. SOP: What is the effect of teachers’ educational qualifications on the learning performance of students in Mathematics? 6. SOP: Is there a significant difference in the learning performance of the students exposed in the three different methods of teaching: Traditional, Game-Based, and Activity-Oriented? 7. SOP: Is there a significant difference between the responses of the women and men in the legalization of the divorce in the Philippines? 8. SOP: Are the public school teachers more competent compared to the private school teachers? STATISTICS AND PROBABILITY SAMSUDIN N. ABDULLAH, Ph.D.

9. SOP: What is the profile of the NQuESH takers in terms of administrative experience and educational attainment? 10. SOP: What is the level of the reading comprehension of grade 7 students? 11. SOP: Is there a significant difference between the performance of the students in the two previous grading periods? 12. SOP: Is there a significant relationship between the reading interest and literary comprehension of the students? STATISTICS AND PROBABILITY SAMSUDIN N. ABDULLAH, Ph.D.

13. SOP: Is the learning performance of the students in Mathematics significantly influenced by the educational qualification of their teachers? 14. SOP: Is there a significant difference in the learning performance of the students exposed in the three different methods of teaching: Traditional, CAI, and PWA? 15. SOP: Is there a significant relationship between the responses of the women and men in the legalization of the divorce in the Philippines? 16. SOP: Are the public school teachers more satisfied with their jobs compared to the private school teachers? STATISTICS AND PROBABILITY SAMSUDIN N. ABDULLAH, Ph.D.

1. To what degree is the student absenteeism in the following causes:
1.1 Physical/school factors, 1.2 Health problems, 1.3 Personal attitudes, 1.4 Family-related issues, 1.5 Teacher-related reasons, 1.6 Subject-related matters, 1.7 Classroom atmosphere, 1.8 Peer relationship, 1.9 Financial constraints, and 1.10 Obsession in the computer or online games/social networking sites? 17.

2. What is the level of academic performance of low performing students in the following tool subjects: 2.1 Filipino, 2.2 English, 2.3 Mathematics, and 2.4 Science? 3. Is there a significant difference in the attitudes of students towards absenteeism when they are grouped according to: 3.1 Grade 7, 3.2 Grade 8, 3.3 Grade 9, and 3.4 Grade 10? 18. 19. STATISTICS AND PROBABILITY SAMSUDIN N. ABDULLAH, Ph.D.

4. Is the assessment of the respondents towards absenteeism significantly different according to the following types of respondents: 4.1 Low performing students, 4.2 Their respective parents or guardians, and 4.3 Their close friends? 5. Does the academic performance of struggling students in the tool subjects significantly differ from each other? 20. 21. STATISTICS AND PROBABILITY SAMSUDIN N. ABDULLAH, Ph.D.

6. Is there a significant difference in the attitudes of male and female students towards absenteeism? 7. Is there a significant relationship between the causes of absenteeism and academic performance of the struggling students? 22. 23. STATISTICS AND PROBABILITY SAMSUDIN N. ABDULLAH, Ph.D.

8. What intervention programs can be proposed to minimize, if not totally eradicate absenteeism among the low performing students? 24. STATISTICS AND PROBABILITY SAMSUDIN N. ABDULLAH, Ph.D.

9. Is there a significant difference between the academic performance of TVL and HUMSS students?
25. STATISTICS AND PROBABILITY SAMSUDIN N. ABDULLAH, Ph.D.

SAMSUDIN N. ABDULLAH, Ph.D.
Thank you so much From SAMSUDIN N. ABDULLAH, Ph.D. Master Teacher II

Statistics And Probability

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