Question

Calculate the sample standard deviation for the following data set. If necessary, round to one more decimal place than the largest number of decimal places given in the data.
Slugging Percentages for Leading Major League Baseball Players

0.612 0.612 0.523 0.523
0.606 0.606 0.631 0.631
0.584 0.584 0.592 0.592
0.644 0.644 0.597 0.597
0.639 0.639 0.607 0.607
0.564 0.564 0.673 0.673

Answer 1

Answer:

$s = 0.0394$

Step-by-step explanation:

Given:

0.612 0.523  0.606 0.631  0.584  0.592  0.644  0.597  0.639  0.607  0.564  0.673

Required

Calculate the sample standard deviation

First, calculate the mean

$\bar x = \frac{\sum x}{n}$

$\bar x = \frac{0.612 +0.523 +0.606 +0.631+ 0.584 + 0.592+ 0.644 +0.597 +0.639 +0.607 +0.564+ 0.673}{12}$

$\bar x = \frac{7.272}{12}$

$\bar x = 0.606$

The sample standard deviation is then calculated using:

$s = \sqrt{\frac{\sum(x - \bar x)^2}{n-1}}$

$\sum (x - \bar x)^2 = (0.612 -0.606)^2+(0.523 -0.606)^2+(0.606 -0.606)^2+(0.631 -0.606)^2+(0.584 -0.606)^2+(0.592 -0.606)^2+(0.644 -0.606)^2+(0.597 -0.606)^2+(0.639 -0.606)^2+(0.607 -0.606)^2+(0.564 -0.606)^2+(0.673 -0.606)^2$

$\sum (x - \bar x)^2 = 0.017098$

So, we have:

$s = \sqrt{\frac{\sum(x - \bar x)^2}{n-1}}$

$s = \sqrt{\frac{0.017098}{12-1}}$

$s = \sqrt{\frac{0.017098}{11}}$

$s = \sqrt{0.00155436363}$

$s = 0.0394$ -- approximated