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mputing formula for s2 on page 44 is equivalent to the one used to define s2 on page 37

mputing formula for s2 on page 44 is equivalent to the one used to define s2 on page 37

A First Course in Probability (10th Edition)
A First Course in Probability (10th Edition)
10th Edition
ISBN:9780134753119
Author:Sheldon Ross
Publisher:Sheldon Ross
Chapter1: Combinatorial Analysis
Section: Chapter Questions
Problem 1.1P: a. How many different 7-place license plates are possible if the first 2 places are for letters and...
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show that the computing formula for s2 on page 44 is equivalent to the one used to define s2 on page 37

time plot made it clear that the trend was the key feature, not the average, which we a poor summary. The testing machine required more work. 44 Chapter 2 Organization and Description of Data was 2.7 The Calculation of x and s Here, we discuss methods for calculating i and s from data that are already grouped into intervals. These calculations are, in turn, based on the formulas for the mean and standard deviation for data consisting of all of the individual observations. In this latter case, we obtain ï by summing all of the observations and dividing by the sample size n. An alternative formula for s2 forms the basis of the grouped data formula for variance. It was originally introduced to simplify hand calculations. Variance (handheld calculator formula) E xi /n i=1 i=1 52 n - 1 (In Exercise 2.51 you will be asked to show that this formula is, in fact, equivalen to the one on page 37.) This expression for variance is without , which reduce roundoff error when using a handheld calculator. Calculating variance using the handheld calculator formula Find the mean and the standard deviation of the following miles per gallon (mpg obtained in 20 test runs performed on urban roads with an intermediate-size car: EXAMPLE 18 algaib 19.7 21.5 22.5 22.2 22.6 21.9 20.5 19.3 19.9 21.7
Transcribed Image Text:time plot made it clear that the trend was the key feature, not the average, which we a poor summary. The testing machine required more work. 44 Chapter 2 Organization and Description of Data was 2.7 The Calculation of x and s Here, we discuss methods for calculating i and s from data that are already grouped into intervals. These calculations are, in turn, based on the formulas for the mean and standard deviation for data consisting of all of the individual observations. In this latter case, we obtain ï by summing all of the observations and dividing by the sample size n. An alternative formula for s2 forms the basis of the grouped data formula for variance. It was originally introduced to simplify hand calculations. Variance (handheld calculator formula) E xi /n i=1 i=1 52 n - 1 (In Exercise 2.51 you will be asked to show that this formula is, in fact, equivalen to the one on page 37.) This expression for variance is without , which reduce roundoff error when using a handheld calculator. Calculating variance using the handheld calculator formula Find the mean and the standard deviation of the following miles per gallon (mpg obtained in 20 test runs performed on urban roads with an intermediate-size car: EXAMPLE 18 algaib 19.7 21.5 22.5 22.2 22.6 21.9 20.5 19.3 19.9 21.7
lmple size six days are 37 Sec 2.5 Descriptive Measures important aspect of a set of data-their "middle" or their "average"-but they tell us nothing about the extent of variation. We observe that the dispersion of a set of data is small if the values are closely bunched about their mean, and that it is large if the values are scattered widely about their mean. It would seem reasonable, therefore, to measure the variation of a set of data in terms of the amounts by which the values deviate from their mean. If a set of numbers x1, x2, ..., Xn has mean x, the differences 4 15 12.5 requests X1 - X, X2 - X,... , Xn – X are called the deviations from the mean. We might use the average of the deviations as a measure of variation in the data set. Unfortunately, this will not do. For instance, refer to the observations 11, 9, 17, 19, 4, 15, displayed above in Figure 2.12, where X = 12.5 is the balance point. The six deviations are -1.5, -3.5, 4.5, 6.5, -8.5, and 2.5. The sum of positive deviations st 7 19 est values, is 13 requests. 4.5 + 6.5 + 2.5 = 13.5 on as the balance point, c ram for the data on the mu the dot diagram, each ober istance along the horizon. weights and the horizontl r of inertia or balance pir balance point of the obsan exactly cancels the sum of the negative deviations -1.5 – 3.5 – 8.5 = -13.5 so the sum of all the deviations is 0. As you will be asked to show in Exercise 2.50, the sum of the deviations is always zero. That is, > (x; -x) = 0 | i=1 so the mean of the deviations is always zero. Because the deviations sum to zero, we need to remove their signs. Absolute value and square are two natural choices. If we take their absolute value, so each negative deviation is treated as positive, we would obtain a measure of variation. However, to obtain the most common measure of vari- ation, we square each deviation. The sample variance, s-, is essentially the average of the squared deviations from the mean, x, and is defined by the following formula. 12.5 15 de a single number to e reason for preferring Sample Variance n - 1 7 actually gives a mor contained in the obser in of estima
Transcribed Image Text:lmple size six days are 37 Sec 2.5 Descriptive Measures important aspect of a set of data-their "middle" or their "average"-but they tell us nothing about the extent of variation. We observe that the dispersion of a set of data is small if the values are closely bunched about their mean, and that it is large if the values are scattered widely about their mean. It would seem reasonable, therefore, to measure the variation of a set of data in terms of the amounts by which the values deviate from their mean. If a set of numbers x1, x2, ..., Xn has mean x, the differences 4 15 12.5 requests X1 - X, X2 - X,... , Xn – X are called the deviations from the mean. We might use the average of the deviations as a measure of variation in the data set. Unfortunately, this will not do. For instance, refer to the observations 11, 9, 17, 19, 4, 15, displayed above in Figure 2.12, where X = 12.5 is the balance point. The six deviations are -1.5, -3.5, 4.5, 6.5, -8.5, and 2.5. The sum of positive deviations st 7 19 est values, is 13 requests. 4.5 + 6.5 + 2.5 = 13.5 on as the balance point, c ram for the data on the mu the dot diagram, each ober istance along the horizon. weights and the horizontl r of inertia or balance pir balance point of the obsan exactly cancels the sum of the negative deviations -1.5 – 3.5 – 8.5 = -13.5 so the sum of all the deviations is 0. As you will be asked to show in Exercise 2.50, the sum of the deviations is always zero. That is, > (x; -x) = 0 | i=1 so the mean of the deviations is always zero. Because the deviations sum to zero, we need to remove their signs. Absolute value and square are two natural choices. If we take their absolute value, so each negative deviation is treated as positive, we would obtain a measure of variation. However, to obtain the most common measure of vari- ation, we square each deviation. The sample variance, s-, is essentially the average of the squared deviations from the mean, x, and is defined by the following formula. 12.5 15 de a single number to e reason for preferring Sample Variance n - 1 7 actually gives a mor contained in the obser in of estima
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