Chapter 3, Problem 1E

The content of liquid detergent bottles is being analyzed. Twelve bottles, randomly selected from the process, are measured, and the results are as follows (in fluid ounces): 16.05, 16.03, 16.02, 16.04, 16.05, 16.01, 16.02, 16.02, 16.03, 16.01, 16.00, 16.07

1. (a) Calculate the sample average.
2. (b) Calculate the sample standard deviation.

To determine

Calculate the sample average.

Answer to Problem 1E

The sample average is 16.029.

Explanation of Solution

Calculation:

Sample average:

Let $x_1,x_2,\mathrm{...},x_n$ are the observations in the sample, then the sample average is,

$\overline{x}=\frac{{\displaystyle \sum _{i=1}^n{x}_i}}{n}$

Consider the sample average for the given observations,

$\begin{array}{c}\overline{x}=\frac{{\displaystyle \sum _{i=1}^n{x}_i}}{n}\\ =\frac{\left(\begin{array}{l}16.05+16.03+16.02+16.04+16.05+16.01+\\ 16.02+16.02+16.03+16.01+16.00+16.07\end{array}\right)}{12}\\ =\frac{192.35}{12}\\ =16.029\end{array}$

To determine

Calculate the sample standard deviation.

Answer to Problem 1E

The sample standard deviation is 0.0202.

Explanation of Solution

Calculation:

Sample standard deviation:

Let $x_1,x_2,\mathrm{...},x_n$ are the observations in the sample, then the sample standard deviation is,

$s=\sqrt{\frac{{\displaystyle \sum _{i=1}^n{\left(x_i-\overline{x}\right)}^2}}{n-1}}$

Consider,

xi	$x_i-\overline{x}=x_i-16.029$	${\left(x_i-16.029\right)}^2$
16.05	0.021	0.0004
16.03	0.001	0.0000
16.02	̶ 0.009	0.0001
16.04	0.011	0.0001
16.05	0.021	0.0004
16.01	̶ 0.019	0.0004
16.02	̶ 0.009	0.0001
16.02	̶ 0.009	0.0001
16.03	0.001	0.0000
16.01	̶ 0.019	0.0004
16	̶ 0.029	0.0008
16.07	0.041	0.0017
		${\displaystyle \sum _{i=1}^n{\left(x_i-16.029\right)}^2}=0.0045$

Therefore,

$\begin{array}{c}s=\sqrt{\frac{{\displaystyle \sum _{i=1}^n{\left(x_i-16.029\right)}^2}}{n-1}}\\ =\sqrt{\frac{0.0045}{12-1}}\\ =\sqrt{\frac{0.0045}{11}}\\ =0.0202\end{array}$