 # 1.2: Describing Distributions

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1.2: Describing Distributions
with Numbers Measures of Location (Center) Mean Median B. Measures of Spread (Variability) Quartiles (Quantiles) Variance and Standard deviation

Measures of Location 1. Mean (Average) How to find the mean (average):
1) Add the values together 2) Divide the total by the number of observations Example: Test Scores : 56, 65, 54, 55, 57, 54, 61, 62, 60, 55, 57, 56, 57, 61, 62, 60, 49, 66, 59, 80 Step 1 : …… = 1186 Step 2 : 1186 / 20 = 59.3 Mean

Mean To find the mean x of a set of observations, add their values and divide by the number of observations. If the n observations are x , x , x , ….. , x , their mean is : 1 2 3 n x = 1 n 3 2 + ... Or, in more compact notation: x = i 1 n

2. Median How to find the median M :
1) Arrange the observations in order from smallest to largest. 2) If the number of observations is odd, then the median is located at the center of the list. So, if there are n observations, then the median is located in spot (n + 1) / 2 3) If the number of observations is even, then the median is the average of the two terms in the middle spots. These are located in spots (n / 2) and (n / 2) + 1

Median Example of finding a Median :
List 1 : 2, 4, 6, 3, 5, 2, 6, 8, 10, 11, 1 Step 1: Order the list : 1, 2, 2, 3, 4, 5, 6, 6, 8, 10, 11 Step 2 : Find the middle term2 : (n+1) / 2 = (11 + 1) / 2 = 6 1, 2, 2, 3, 4, 5, 6, 6, 8, 10, 11 Median

Median Example of finding a Median :
List : 2, 4, 6, 3, 5, 2, 6, 8, 10, 11, 1, 12 Step 1: Order the list : 1, 2, 2, 3, 4, 5, 6, 6, 8, 10, 11, 12 Step 2 : Find the two middle terms : (n / 2) + 1 = (12 / 2) + 1 = 7 n / 2 = 12 / 2 = 6 Step 3 : Average the sixth and seventh terms : 1, 2, 2, 3, 4, 5, 6, 6, 8, 10, 11, 12 Median = (5 + 6) /2 = 5.5

In The Presence Of Outliers
Q: Do outliers affect the Mean and Median? Consider the list on numbers from 1 through 9 : 1, 2, 3, 4, 5, 6, 7 ,8 ,9 The Mean is : 5 The Median is : 5 What if we put the number 100 at the end of the list : 1, 2, 3, 4, 5, 6, 7 ,8 ,9, 100 The Mean is : 14.5 The Median is : 5.5 A: Outliers affect the Mean much more than the Median !

Distributions The mean is the point at which a histogram balances. For
symmetric distributions the mean and median will be nearly the same. However, since the mean is influenced by outliers, for skewed distributions the mean will be pulled in the direction of the long tail while the median will be resistant to the outliers and remain in nearly the same place.

Skewed Right M X

Skewed Left X M

Describing Spread The Five Number Summary : 1) The Median
2) First Quartile : 25% of the observations lie below the First Quartile 3) Third Quartile : 75% of the observations lie below the third quartile 4) Lowest Individual Observation (Minimum) 5) Highest Individual Observation (Maximum)

Quartiles Calculating the Quartiles :
1) Arrange the observations in increasing order and locate the Median M in the ordered list o’ observations. 2) The First Quartile Q1 is the median of the observations whose position in the ordered list is to the left of the location of the overall median. 3) The Third Quartile Q3 is the median of the observations whose position in the ordered list is to the right of the location of the overall median.

Quartiles Example of calculating First Quartile :
List of quiz scores: 10, 8, 9, 4, 6, 6, 8, 9, 2, 7 1) Order the list: 2, 4, 6, 6, 7, 8, 8, 9, 9, 10 Find the median: (7 + 8) / 2 = 7.5 2) Find all the observations whose position in the list is to the left of the median : 2, 4, 6, 6, 7 , 8, 8, 9, 9, 10 Find the median of these values : 6

Quartiles Example of calculating Third Quartile :
List of quiz scores: 10, 8, 9, 4, 6, 6, 8, 9, 2, 7, 11 1) Order the list: 2, 4, 6, 6, 7, 8, 8, 9, 9, 10, 11 Find the median: 8 2) Find all the observations whose position in the list is to the right of the median : 2, 4, 6, 6, 7, 8, 8, 9, 9, 10, 11 8, 9, 9, 10, 11 Find the median of these values : 9

Interquartile Range Determining Outliers
The interquartile range , IQR, is the distance between the first quartile and the third quartile. Determining Outliers Call an observation a suspected outlier if it falls more than 1.5 * IQR above the third quartile or below the first quartile. Example : Imagine we have a bunch of test scores with Q1 = 50 and Q3 = 80. The IQR = = 30 So, 1.5 * IQR = 1.5 * 30 = 45 This means that if there are any scores above Q = 125 or any scores Q = 5, then these scores are suspected outliers.

Boxplot Example: Low = 47, High = 98, Median = 77, Q1 = 65, Q3 = 85
A Boxplot is a graph of the five number summary. A central box spans the quartiles, with a line marking the median. Whiskers extend out from the box to the extremes. Highest Observation (98) 30 10 50 70 90 Q3 (85) Median (77) Q1 (65) Lowest Observation (47)

Describing Spread 2. The Standard Deviation
Variance: The variance of a set of observations is an “average” of the deviations of the observations from the mean. Note: You divide by (n - 1) instead of n. Standard Deviation: The SD is the square root of the variance.

Example : Test Scores : 65, 77, 83, 80, 95 1) Find the average : 80 2) Find the deviations from the mean, and their squares Obs Deviation from Mean Deviations Squared 65 -15 225 77 -3 9 83 3 9 80 95 15 225

3) Determine the mean of the squares: ( ) (5 - 1) = 117 Variance 4) Determine the Standard Deviation: = 10.8

More Fancy Notation The variance of a set of observations is the average of the squares of the deviations of the observations from their mean. In symbols, the variance on n observations , , is : s 2 x n 1 ... s 2 = (x - x ) n 1 + ... n - 1 or, in more compact notation : s 2 = (x - x ) i 1 n-1 The standard deviation s is the square root of the variance : s 2 (x - x ) i 2 1 s = n-1

Another Example of Standard Deviation
Consider the following years in our past : 1792, 1666, 1362, 1614, 1460, 1867, 1439 Find the standard deviation of these years. The Mean = 1600 s 2 = (x - x ) i 1 n-1 x i - ( ) 2 1792 1666 1362 1614 1460 1867 1439 192 66 -238 14 -140 267 -161 36864 4356 56644 196 19600 71298 25921 = 1 6 ( ) = s =

Why Do We Square The Deviations ?
1) The sum of the squared deviations of any set of observations from their mean is the smallest that the sum of squared deviations from any number can possibly be. Why use the Standard Deviation and not the Variance ? 1) The standard deviation is the measure of spread for an important class of symmetric unimodal distributions called the normal distribution. 2) The standard deviation is used by the normal distribution. 3) The variance uses squared deviations, which gives a different unit from the original data. Why use n - 1 ? 1) The sum of the deviations is *always* zero. So, if we know n-1 of the deviations, then the last deviation can be calculated. So, only n-1 of the deviations can vary freely. These are called degrees of freedom.

Properties of Standard Deviations
1) The standard deviation measures spread about the mean and should be used only when the mean is chosen as the measure of center. 2) s = 0 only when there is no spread. This happens only when all observations have the same value. Otherwise, s > 0. As the observations get more spread out from the mean, then s gets larger. 3) s, like the mean, is not resistant. A few outliers can make s very large.

Which Measure To Use ? Rules Of Thumb
Q: When is the mean better than median? When is the five number summary better than the standard deviation? Rules Of Thumb A1: If outliers appear, or if your distribution is skewed, then the mean could be affected, so use the median and the five number summary. A2: If the distribution is reasonably symmetric and is free of outliers, then the mean and standard deviation should be used.

Changing Units Consider the following values : 30, 40, 50, 60, 70
The mean is 50 and the standard deviation is 15.8 What happens to these if we take every score, multiply it by 2 and add 10 We get these values : 70, 90, 110, 130, 150 The mean is 110 and the standard deviation is 31.6

Changing Units Old values : 30, 40, 50, 60, 70 mean = 50 and s = 15.8
What happens to these if we take every score, multiply it by 2 and add 10 New values : 70, 90, 110, 130, 150 mean = 110 and s = 31.6 30 50 90 70 110 130 150 30 50 90 70 110 130 150 30 50 90 70 110 130 150

Linear Transformations
A linear transformation changes the original variable x into the new variable given an equation of the form : x new = bx + a Note: The constant a shifts all values of x either up or down by the value a. The constant b changes the size of the unit of the distribution. Effects of Linear Transformations 1) To get the new spread, multiply the old spread by |b|. 2) To get the new mean, multiply the old mean by b and add the constant a.

1.3: The Normal Distributions
Density Curves A density curve is a curve that : 1) is always on or above the vertical axis, and 2) has area exactly 1 underneath it. A density curve describes the overall pattern of a distribution. The area under the curve and above any range of values is the relative frequency of all observations that fall in that range.

Density Curves

Normal and Skewed Curves
Mean Median

Why are Normal Distributions important
in stats? Normal distributions are good descriptions for some distributions of real data. 2) Normal distributions are good to the results of many kinds of chance outcomes. 3) Many statistical inference procedures based on normal distributions work well for other roughly symmetric distributions.

The Rule In the normal distribution with mean  and standard deviation  : 68 % of the observations fall within  of the mean  95 % of the observations fall within 2 of the mean  99.7 % of the observations fall within 3 of the mean 

Normal Curve Example John collected data on the heights of women ages 18 to 24. He found that the distribution was roughly normal, with a mean of 64.5 inches and a standard deviation of 2.5 inches.

Standardizing Observations
If x is an observation from a roughly symmetric distribution that has mean  and standard deviation , then the standard value of x is : z = x -  Note: A standardized score is often called a z-score. Example : Women’s IQ’s have a symmetric distribution with a mean of 97 and a standard deviation of 6. What is the standard score for a woman with an IQ of 106 ? 6 = 9 6 = z = 1.5

Standardizing Observations
If x is an observation from a roughly symmetric distribution that has mean  and standard deviation , then the standard value of x is : z = x -  Note: A standardized score is often called a z-score. Example : Men’s IQ’s have a roughly symmetric distribution with a mean of 72 and a standard deviation of 8. What is the standard score for a man with an IQ of 66 ? 8 = -6 8 = z = - .75

The Standard Normal Distribution
If x is an observation from a roughly symmetric distribution that has mean  and standard deviation , then the standard value of x is : z = x -  Note: A standardized score is often called a z-score. Example : Men’s IQ’s have a roughly symmetric distribution with a mean of 72 and a standard deviation of 8. What is the standard score for a man with an IQ of 66 ? 8 = -6 8 = z = - .75 Q: What percentage of people have a score below 66 ?

The Standard Normal Table
Table A is a table of areas under the standard normal curve. The table entry for each z value is the area under the curve to the left of z

The Standard Normal Table
Example : Imagine we have done an experiment, and we want to find what percentage of people fell under a score, namely x. We then proceed to find that the z-score for the value x is .1357

The Standard Normal Table
Example : The Graduate Record Examinations (GRE) are widely used to help predict the performance of applicants to graduate schools. The range of possible sores on a GRE is 200 to 900. The psychology department at a university finds the scores of its applicants on the quantitative GRE are approximately normal with mean  = 544 and standard deviation  = 103. Answer the following : 1) Find the percentage of people who scored 700 or higher on the test. 2) Find the percentage of people who scored below 500 on the test. 3) Find the percentage of people who scored between 500 and 800 on the test.

1) Find the percentage of people who scored 700 or higher on the test.
Find the percentage to the right of the 700 marker.

P(X>700)=P(Z>1.51)=1-P(Z<1.51)=1 - .9345 = .0655
1) Find the percentage of people who scored 700 or higher on the test. 103 = 156 103 = Find the z-score : z = 1.51 .9345 .0655 P(X>700)=P(Z>1.51)=1-P(Z<1.51)= = .0655

Find the percentage to the left of 500
2) Find the percentage of people who scored below 500 on the test. Find the percentage to the left of 500

2) Find the percentage of people who scored below 500 on the test.
103 = - 44 103 = Find the z-score : z = - 0.43 0.3336 Answer :

3) Find the percentage of people who scored between 500 and 800
on the test. Find the percentage between 500 and 800

3) Find the percentage of people who scored between 500 and 800
on the test. Find the first z-score : z = 103 = - 44 - 0.43 Find the second z-score : z = 103 = 256 2.49 0.9936 Area = 0.3336 = 0.66

The Soup Nazi charges, on the average, \$4.50
Example : The Soup Nazi charges, on the average, \$4.50 for a cup of soup, and if you’re lucky, some bread, with a standard deviation of \$0.45. 4.50 What is the probability that our check will be more than \$5.00 ?

What is the probability that our check will be more than \$5.00 ?
4.50 0.1335 0.8665 5.00 What is the probability that our check will be more than \$5.00 ? P (X > 5 ) =P(Z >1.11)=0.1335 13.35 % Z = 0.45 = 1.11

“Backward” Normal Calculations
We could find the observed value (x) of a given proportion in N( , ) by unstandardizing the z-score. State the problem Draw a picture Use the normal table to find the proportion closest to the one you need Read off the z-value Unstandardize x= + z

Example Find the value of z such that the probability of being less than z is 0.10. 1. z: P(Z < z) = .10 s = 1 .10

Example Closest is .1003 So z = -1.28
Find the value of z such that the probability of being less than z is .10. 1. z: P(Z < z) = .10 2. 3. In the body of the normal table, find the closest value to Once found, determine the z value. s = 1 .10 Closest is .1003 So z = -1.28 P(Z < -1.28) = .1003

Example Find the value of z such that the probability of being greater than z is .33. 1. z: P(Z > z) = .33 2. z: P(Z < z) = = .67 .67 s = 1 .33 ???

Example So z = .44 P(Z > .44) = .33
Find the value of z such that the probability of being greater than z is .33. 1. z: P(Z > z) = .33 2. z: P(Z < z) = = .67 s = 1 .33 .67 3. In the body of the normal table, find the closest value to Once found, determine the z value. I found .6700 So z = .44 P(Z > .44) = .33

Example X = time Americans stir sugar into their iced tea
X ~ N(12.3, 3.1) seconds Find the percent of Americans who spend between 20 to 22 seconds in stirring sugar into their iced tea? i.e. P(20 < X < 22)

Example X = time Americans stir sugar into their iced tea
X ~ N(12.3, 3.1) Find P(20 < X < 22) = P( < Z < ) = P(2.48 < Z < 3.13) = P(Z < 3.13) - P(Z < 2.48) = = .0057

Example X = time Americans stir sugar into their iced tea
X ~ N(12.3, 3.1) (2) About 18.4% of Americans spend more than how many seconds stirring sugar into their iced tea? i.e. Find the value of X such that the probability of being greater than this value is .184. (1) z: P(Z > z) = .184 (2) z: P(Z < z) = = .816 (3) From the normal table, z = 0.90 (4) So x = +z = (3.1) = = 15.09 The person would have to stir seconds.

Example X = IQ scores X ~ N(112, 9)
Find the IQ score that replaces you in the top 2% of all scores. 1. z: P(Z > z) = .02 2. z: P(Z < z) = = .98 3. From the normal table, z = 2.05 x = +z = (9) =

Exercise The distribution of SAT Math scores is approximately normally distributed with mean 500 and standard deviation 100. 1. In what range do the middle 95% of all SAT Math scores lie? 2. What proportion of SAT Math scores are between 450 and 650? 3. If high school students having SAT Math scores in the top 10% of all scores are eligible for a certain scholarship, what is the lowest score a person eligible for the scholarship can have?