Exercise 22 Page 900

We are given a table showing a family's daily water usage for $2$ weeks. We want to find the standard deviation of the given data set. The standard deviation is the average amount by which each individual value deviates or differs from the mean. In the formula, $x_1,\dots ,x_n$ are the values of the set of data, $\mu$ is the mean, and $n$ is the number of values. Recall that the mean is the sum of the values divided by the number of values. Let's start by calculating their sum. There are $14$ values in our set, so we have to divide the sum by $14.$ Let's use this value and apply the standard deviation formula to each value in the set.

$x_n$	$x_n-\mu$	$(x_n-\mu {)}^2$
$92.3$	$92.3-89.34=2.96$	$2.96^2\approx 8.76$
$81.3$	$81.3-89.34=\text{-}8.04$	$(\text{-}8.04{)}^2\approx 64.64$
$85.3$	$85.3-89.34=\text{-}4.04$	$(\text{-}4.04{)}^2\approx 16.32$
$81.6$	$81.6-89.34=\text{-}7.74$	$(\text{-}7.74{)}^2\approx 59.9$
$89.7$	$89.7-89.34=0.36$	$0.36^2\approx 0.13$
$76.9$	$76.9-89.34=\text{-}12.44$	$(\text{-}12.44{)}^2\approx 154.75$
$101.2$	$101.2-89.34=11.86$	$11.86^2\approx 140.66$
$94.0$	$94.0-89.34=4.66$	$4.66^2\approx 21.71$
$80.3$	$80.3-89.34=\text{-}9.04$	$(\text{-}9.04{)}^2\approx 81.72$
$89.6$	$89.6-89.34=0.26$	$0.26^2\approx 0.07$
$91.4$	$91.4-89.34=2.06$	$2.06^2\approx 4.24$
$96.3$	$96.3-89.34=6.96$	$6.96^2\approx 48.44$
$88.8$	$88.8-89.34=\text{-}0.54$	$(\text{-}0.54{)}^2\approx 0.29$
$102.1$	$102.1-89.34=12.76$	$12.76^2\approx 162.81$
Sum of Values		$\approx 764.43$

Finally, we need to divide by $14$ and then calculate the square root. Therefore, we have found that the standard deviation of the family's daily water usage is approximately $7.39$ gallons.