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Statistics / Elementary Statistics: Picturing the World 6 / Chapter 4.1 / Problem 30E

Elementary Statistics: Picturing the World | 6th Edition | ISBN: 9780321911216 | Authors: Ron Larson; Betsy Farber

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Solved: Finding the Mean, Variance, and Standard Deviation

Chapter 4, Problem 30E

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QUESTION:

Finding the Mean, Variance, and Standard Deviation In Exercise, (a) find the mean, variance, and standard deviation of the probability distribution, and (b) interpret the results.

Baseball The number of games played in the World Series from 1903 to 2012

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QUESTION:

Finding the Mean, Variance, and Standard Deviation In Exercise, (a) find the mean, variance, and standard deviation of the probability distribution, and (b) interpret the results.

Baseball The number of games played in the World Series from 1903 to 2012

ANSWER:

Step 1 of 2

Given the number of games played in the World Series from 1903 to 2012.

Then the table is given below.

Games played	Probability
4	0.176
5	0.241
6	0.213
7	0.333
8	0.037

a). Here we have to find the mean, variance and standard deviation.

$\mu{}=\sum\limits_{\ }^{\ }xp(x)\$

$=4\times{}0.176+5\times{}0.241+6\times{}0.213+7\times{}0.333+8\times{}0.037$

$=$ 0.704+1.205+1.278+2.331+0.296

$\mu{}=$ 5.814

Therefore, the mean is 5.814.

Then the variance is

${\sigma{}}^{2}=\sum\limits_{\ }^{\ }p(x)(x-\mu{}{)}^{2}$

$=0.176(4-5.814{)}^{2}+0.241(5-5.814{)}^{2}+0.213(6-5.814{)}^{2}+0.333(7-5.81{4}^{\ }{)}^{2}+0.037(8-5.814{)}^{2}$

${\sigma{}}^{2}=$ 1.391

Then the standard deviation is the square root of the variance.

$\sigma{}=\sqrt{variance}$

$=\sqrt{1.391}$

Therefore, the standard deviation is 1.1794.

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