Question

Calculate the sample variance and sample standard deviation for the following frequency distribution of heights in centimeters for a sample of 8-year-old boys. If necessary, round to one more decimal place than the largest number of decimal places given in the data.
Class Frequency120.6 - 123.6 17123.7 - 126.7 49126.8 - 129.8 29129.9 - 132.9 41133.0 - 136.0 35

Answer 1

Answer:

$\bar X = \frac{\sum_{i=1}^n X_i f_i}{n}= \frac{22026.1}{171}=128.81$

$s^2 = \frac{2839948.85- \frac{(22026.1)^2}{171}}{171-1}= 16.59$

$s= \sqrt{16.587}=4.07$

Step-by-step explanation:

For this case we can calculate the sample variance and deviation with the following table

Class               Midpoint (Xi)    fi        Xi*fi       Xi^2 *fi

120.6-123.6        122.1             17      2075.7   253443

123.7-126.7        125.2            49      6134.8    768077

126.8-129.8        128.3            29     3720.7    477365.8

129.9-132.9        131.4             41       5387.4   707904.4

133.0-136.0        134.5            35      4007.5   633158.8

___________________________________________

Total                                        171   22026.1    2839948.85

For this case we can calculate the mean or expected value with the following formula:

$\bar X = \frac{\sum_{i=1}^n X_i f_i}{n}= \frac{22026.1}{171}=128.81$

Now we can calculate the sample variance with the following formula:

$s^2 =\frac{\sum f_i X^2_i -[\frac{(\sum X_i f_i)}{n}]^2}{n-1}$

And if we replace we got:

$s^2 = \frac{2839948.85- \frac{(22026.1)^2}{171}}{171-1}= 16.59$

And the standard deviation would be the square root of the variance and we got:

$s= \sqrt{16.59}=4.07$