# Chapter 6 Normal Probability Distributions

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Chapter 6 Normal Probability Distributions
Overview The Standard Normal Distribution Normal Distributions: Finding Probabilities Normal Distributions: Finding Values The Central Limit Theorem

Overview y = Continuous random variable Normal distribution ( ) 
Curve is bell shaped and symmetric Score Even though the equation appears daunting, with the knowledge of and , it is the same as any other equation relating x and y. Remind student that e and  are constant values. The result is bell shape curve. With a graphing calculator, the instructor can show the graph of the equation with  = 0 and  = 1. ( ) x - µ 2 1 2 y = e  2 p

Definitions Density Curve (or probability density   function) the graph of a continuous probability   distribution 1. The total area under the curve must equal 1. 2. Every point on the curve must have a vertical height that is 0 or greater. Remind students that the area of a probability density curve must equal 1 as the total of all probabilities of a probability distribution equals 1.

Because the total area under the density curve is equal to 1, there is a correspondence between area and probability. page 228 of text This is an important fact for students to understand.

Heights of Adult Men and Women
µ = 63.6  = 2.5 Men: µ = 69.0  = 2.8 Normal distributions that have different means will be positioned differently on the number line. Normal distributions that have different standard deviations will be ‘spread’ differently and thus will have different heights for the bell curve. 63.6 69.0 Height (inches)

Definition Standard Normal Deviation
a normal probability distribution that has a mean of 0 and a standard deviation of 1 Again showing this graph with a graphing calculator will be very a very effective illustration for students. Changing one or both of the parameters,  or , while leaving the standard normal curve in place, makes a good demonstration.

The Empirical Rule Standard Normal Distribution: µ = 0 and  = 1
99.7% of data are within 3 standard deviations of the mean 95% within 2 standard deviations 68% within 1 standard deviation These percentages will be verified by the concepts learned in Chapter 5. Emphasize the Empirical Rule is appropriate for data that is in a BELL-SHAPED distribution. 34% 34% 2.4% 2.4% 0.1% 0.1% 13.5% 13.5% x - 3s x - 2s x - s x x + s x + 2s x + 3s

Probability of Half of a Distribution
0.5 Students will need this explanation. It will not be immediately intuitive for some. Because the normal curve is ‘symmetric’ and the total probability (area) is equal to 1, from the mean to one side (or the other) of the curve will have a probability (area) equal to 0.5. This concept is necessary to complete the other types of probability situations that can occur.

denotes the probability that the z score is greater than a
Notation 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 page 235 of text A review of inequality signs that their meanings will most likely be necessary.

Finding the Area to the Right of z = 1.27
This area is 0.1020 Example on page 233 of text z = 1.27

To find: z Score the distance along horizontal scale of the standard normal distribution; refer to the leftmost column and top row of theTable Area the region under the curve; refer to the values in the body of the Table page 231 of text It will be important to distinguish between finding a probability (area) given a particular z score from that of finding a z score given a particular area. Being able to recognize the correct method for each of these processes will be critical.

Table E Standard Normal Distribution
µ =  = 1 Emphasize that Table A-2 gives the area (probability) from the mean (=0) to the z-score only. Students will need to be reminded of this often in the initial stages of this topic. x z

Definition Standard Normal Deviation
a normal probability distribution that has a mean of 0 and a standard deviation of 1 Area found in Table A2 Area = 1 2 3 -1 -2 -3 page 229 and 231 of text The areas have been calculated by mathematicians using the calculus of the area between the curve and the x-axis. Fortunately, for the students in this course, the calculus knowledge is not necessary with the inclusion of Table A-2 (z-distribution) in the book. 0.4429 z = 1.58 Score (z )

Example: If thermometers have an average (mean) reading of 0 degrees and a standard deviation of 1 degree for freezing water and if one thermometer is randomly selected, find the probability that it reads freezing water between 0 degrees and 1.58 degrees. page 231 of text

Example: If thermometers have an average (mean) reading of 0 degrees and a standard deviation of 1 degree for freezing water and if one thermometer is randomly selected, find the probability that it reads freezing water between 0 degrees and 1.58 degrees. Area = P ( 0 < x < 1.58 ) = Interpretation of numerical answer. The probability that the chosen thermometer will measure freezing water between 0 and 1.58 degrees is

Example: If thermometers have an average (mean) reading of 0 degrees and a standard deviation of 1 degree for freezing water, and if one thermometer is randomly selected, find the probability that it reads freezing water between degrees and 0 degrees. Area = P ( < x < 0 ) = Example on page 232 of text Again emphasize that the answer - a probability (and area) - will not be negative, even though the z score was negative. The probability that the chosen thermometer will measure freezing water between and 0 degrees is

Finding the Area Between z = 1.20 and z = 2.30
Area A is A Example on page 234 of text z = 1.20 z = 2.30

Other Normal Distributions
If   0 or   1 (or both), we will convert values to standard scores using Formula 5-2, then procedures for working with all normal distributions are the same as those for the standard normal distribution. The z score formula was first presented in Chapter 2, Section 2-6, Measures of Position, page 85. x - µ z =

Converting to Standard Normal Distribution
P x (a)

Converting to Standard Normal Distribution
x -  z = Tell students to round z score answers to two decimal places if using Table A-2 from the text. P P x z (a) (b)

Probability of Weight between 143 pounds and 201 pounds
z = = 2.00 x = 143 29 s =29 page 242 of text Weight z

Probability of Weight between 143 pounds and 201 pounds
OR % of women have weights between 143 lb and 201 lb. x = 143 s =29 Alternative way of interpreting answer. Weight z

Finding a probability when given a z-score using the TI83
Distributions 2:normalcdf (lower bound, upper bound, mean, s.d.)

Finding a z - score when given a probability Using the Table
1. Draw a bell-shaped curve, draw the centerline, and identify the region under the curve that corresponds to the given probability. If that region is not bounded by the centerline, work with a known region that is bounded by the centerline. 2. Using the probability representing the area bounded by the centerline, locate the closest probability in the body of Table E and identify the corresponding z score. 3. If the z score is positioned to the left of the centerline, make it a negative. page 236 of text Some students get confused with this process. It is actually the ‘reverse’ process of that discussed in the first part of the section. The most common mistake students make with this process is to not make z-scores to the left of the mean negative. Checking the final answer for reasonableness - reminding students that the horizontal line is a ‘number line’ - is important.

Finding z Scores when Given Probabilities
95% 5% 5% or 0.05 0.45 0.50 1.645 (z score will be positive) Finding the 95th Percentile

Finding z Scores when Given Probabilities
90% 10% Bottom 10% Student will need to be reminded that Table A-2 will not indicate that the correct z score for this problem will be negative. This is something that the student will have to remember to do when writing the correct answer to the problem. Emphasize the number line aspect again and that numbers below 0 will be negative. 0.10 0.40 -1.28 (z score will be negative) Finding the 10th Percentile

Finding z Scores when Given Probabilities Using the TI83
Distributions 3:Invnorm (%,mean,s.d)

Cautions to keep in mind
1. Don’t confuse z scores and areas.     Z scores are distances along the horizontal  scale, but areas are regions under the  normal curve. Table A-2 lists z scores in the left column and across the top row, but  areas are found in the body of the table. 2. Choose the correct (right/left) side of the graph. 3. A z score must be negative whenever it is      located to the left of the centerline of 0. page 249 of text

Finding z Scores when Given Probabilities
95% 5% 5% or 0.05 0.45 0.50 1.645 (z score will be positive) Finding the 95th Percentile

Finding z Scores when Given Probabilities
90% 10% Bottom 10% Student will need to be reminded that Table A-2 will not indicate that the correct z score for this problem will be negative. This is something that the student will have to remember to do when writing the correct answer to the problem. Emphasize the number line aspect again and that numbers below 0 will be negative. 0.10 0.40 -1.28 (z score will be negative) Finding the 10th Percentile

Procedure for Finding Values Using the Table and By Formula
1. Sketch a normal distribution curve, enter the given probability or percentage in the appropriate region of the graph, and identify the x value(s) being sought. Use Table E to find the z score corresponding to the region bounded by x and the centerline of Cautions:     Refer to the BODY of Table E to find the closest area, then identify     the corresponding z score.     Make the z score negative if it is located to the left of the centerline. 3. Using the Formula, enter the values for µ, , and the z score found in step 2, then solve for x. x = µ + (z • ) 4. Refer to the sketch of the curve to verify that the solution makes sense in the context of the graph and the context of the problem. page of text

Finding P10 for Weights of Women
10% 90% 40% 50% Drawing for example on page 250 of text Weight x = ? 143

Finding P10 for Weights of Women
The weight of 106 lb (rounded) separates the lowest 10% from the highest 90%. 0.10 0.40 0.50 Interpretation should be emphasized. Weight x = 106 143 -1.28

Forgot to make z score negative???
UNREASONABLE ANSWER! 0.10 0.40 0.50 Weight x = 180 143 1.28

REMEMBER! Make the z score negative if the value is located to the left (below) the mean. Otherwise, the z score will be positive.

Definition Sampling Distribution of the mean
the probability distribution of sample means, with all samples      having the same sample size n. page 256 of text

Central Limit Theorem Given:
1. The random variable x has a distribution (which may or may not be normal) with mean µ and standard deviation . 2. Samples all of the same size n are randomly selected from the population of x values. page 257 of text

Central Limit Theorem Conclusions: n
1. The distribution of sample x will, as the sample size increases, approach a normal distribution. 2. The mean of the sample means will be the population mean µ. 3. The standard deviation of the sample means will approach  n

Practical Rules Commonly Used:
1. For samples of size n larger than 30, the distribution of the sample means can be approximated reasonably well by a normal distribution. The approximation gets better as the sample size n becomes larger. 2. If the original population is itself normally distributed, then the sample means will be normally distributed for any sample size n (not just the values of n larger than 30).

the standard deviation of sample mean
Notation the mean of the sample means the standard deviation of sample mean  (often called standard error of the mean) µx = µ x = n

Distribution of 200 digits from Social Security Numbers
(Last 4 digits from 50 students) 20 Frequency 10 An in class project will solidify this new concept. See margin of page 256. Students should note the the standard deviation (‘spread’) of this data would be fairly large. Distribution of 200 digits

SSN digits x 1 5 9 4 7 8 3 2 6 4.75 4.25 8.25 3.25 5.00 3.50 5.25 2 6 5 7 8 3 4 1 9 4.00 5.25 4.25 4.50 4.75 3.75 6.00 Results of experiment by author. Page 258 of text.

Distribution of 50 Sample Means for 50 Students
15 Frequency 10 5 Students should note that the standard deviation of the means would be a smaller number than that of the population. Also discuss position of the mean of the means versus the mean of the population and the shape of the distribution of means versus the distribution shape of the population.

As the sample size increases, the sampling distribution of sample means approaches a normal distribution. page 259 of text

Example: Given the population of women has normally distributed weights with a mean of 143 lb and a standard deviation of 29 lb, a.) if one woman is randomly selected, find the probability that her weight is greater than 150 lb. b.) if 36 different women are randomly selected, find the probability that their mean weight is greater than 150 lb. Example on page of text

Example: Given the population of women has normally distributed weights with a mean of 143 lb and a standard deviation of 29 lb, a.) if one woman is randomly selected, the probability that her weight is greater than 150 lb. is = Interpretation of numerical answer. 0.0948  = 143 150 = 29 0.24

Example: Given the population of women has normally distributed weights with a mean of 143 lb and a standard deviation of 29 lb, b.) if 36 different women are randomly selected, the probability that their mean weight is greater than 150 lb is z = = 1.45 29 36 = Interpretation of numerical answer. 0.4265 x = 143 150 x= 1.45

Example: Given the population of women has normally distributed weights with a mean of 143 lb and a standard deviation of 29 lb, a.) if one woman is randomly selected, find the probability    that her weight is greater than 150 lb. P(x > 150) = b.) if 36 different women are randomly selected, their mean    weight is greater than 150 lb. P(x > 150) = It is much easier for an individual to deviate from the mean than it is for a group of 36 to deviate from the mean. The last statement on this slide should be discussed in detail.

Finding a z - score for a group Using TI83
Distributions 2:normalcdf (lower bound, upper bound, mean, s.e) Replace s.d. with standard error