## Mean and Variance | Introduction to Statistics and Data Analysis | Probability & Statistics for Engineers & Scientists | Walpole | Problem 1.10

### Solution:

For the company A:

We know, based on our answer in Exercise 1.4, that the sample mean for samples in Company A is $\displaystyle 7.950$.

To compute for the sample variance, we shall use the formula

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

The formula states that we need to get the sum of $\displaystyle \left(x_i-\overline{x}\right)^2$, so we can use a table to solve $\displaystyle \left(x_i-\overline{x}\right)^2$ for every sample.

Note: You can refer to the solution of Exercise 1.7 on how to use a table to solve for the variance.

The variance is

$\displaystyle \left(s^2\right)_A=\sum _{i=1}^{10}\:\frac{\left(x_i-7.950\right)^2}{10-1}$

$\displaystyle =1.2078$

The standard deviation is just the square root of the variance. That is

$\displaystyle s_A=\sqrt{1.2078}=1.099$

For Company B:

Using the same method employed for Company A, we can show that the variance and standard deviation for the samples in Company B are

$\displaystyle \left(s^2\right)_B=0.3249$ and $\displaystyle s_B=0.570$.

## Variance & Standard Deviation | Introduction to Statistics and Data Analysis | Probability & Statistics for Engineers & Scientists | Walpole | Problem 1.9

### Solution:

For the samples with NO AGING:

We know, based on our answer in Exercise 1.3, that the sample mean for samples with no aging is $\displaystyle \overline{x}_{no\:aging}=222.10$.

To compute for the sample variance, we shall use the formula

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

The formula states that we need to get the sum of $\displaystyle \left(x_i-\overline{x}\right)^2$, so we can use a table to solve $\displaystyle \left(x_i-\overline{x}\right)^2$ for every sample.

Note: You can refer to the solution of Exercise 1.7 on how to use a table to solve for the variance.

The variance is

$\displaystyle \left(s^2\right)_{no\:aging}=\sum _{i=1}^{10}\:\frac{\left(x_i-\overline{x}\right)^2}{n-1}$

$\displaystyle =\frac{1}{10-1}\left[\left(227-222.10\right)^2+\left(222-222.10\right)^2+...+\left(221-222.10\right)^2\right]$

$\displaystyle =42.12$

The standard deviation is just the square root of the variance. That is

$\displaystyle s_{no\:aging}=\sqrt{s^2}=\sqrt{42.12}=6.49$

For the samples with AGING:

We know, based on our answer in Exercise 1.3, that the sample mean for samples with aging is $\displaystyle \overline{x}_{no\:aging}=209.90$.

To compute for the sample variance, we shall use the formula

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

The formula states that we need to get the sum of $\displaystyle \left(x_i-\overline{x}\right)^2$, so we can use a table to solve $\displaystyle \left(x_i-\overline{x}\right)^2$ for every sample.

Note: You can refer to the solution of Exercise 1.7 on how to use a table to solve for the variance.

The variance is

$\displaystyle \left(s^2\right)_{aging}=\sum _{i=1}^{10}\:\frac{\left(x_i-\overline{x}\right)^2}{n-1}$

$\displaystyle =\frac{1}{10-1}\left[\left(219-209.90\right)^2+\left(214-209.90\right)^2+...+\left(205-209.90\right)^2\right]$

$\displaystyle =23.62$

The standard deviation is just the square root of the variance. That is

$\displaystyle s_{aging}=\sqrt{s^2}=\sqrt{23.62}=4.86$

## Sample Variance & Standard Deviation | Introduction to Statistics and Data Analysis | Probability & Statistics for Engineers & Scientists | Walpole | Problem 1.8

### Solution:

We know, based on our answer in Exercise 1.2, that the sample mean is $\displaystyle \overline{x}=20.768$.

To compute for the sample variance, we shall use the formula

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

The formula states that we need to get the sum of $\displaystyle \left(x_i-\overline{x}\right)^2$, so we can use a table to solve $\displaystyle \left(x_i-\overline{x}\right)^2$ for every sample.

Note: You can refer to the solution of Exercise 1.7 on how to use a table to solve for the variance.

The variance is

$s^2=\frac{1}{20-1}[\left(18.71-20.768\right)^2+\left(21.41-20.768\right)^2$

$\displaystyle +...+\left(21.12-20.768\right)^2]$

$\displaystyle s^2=2.5345$

The standard deviation is just the square root of the variance. That is

$\displaystyle s=\sqrt{s^2}=\sqrt{2.5345}=1.592$

## Solution:

We know, based on our answer in Exercise 1.1, that the sample mean is $\displaystyle \overline{x}=3.787$.

To compute for the sample variance, we shall use the formula

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

The formula states that we need to get the sum of $\displaystyle \left(x_i-\overline{x}\right)^2$, so we can use a table to solve $\displaystyle \left(x_i-\overline{x}\right)^2$ for every sample.

The table above shows that

$\displaystyle \sum _{i=1}^{15}=13.197$

Therefore, the variance is

$\displaystyle s^2=\sum _{i=1}^n\frac{\left(x_i-\overline{x}\right)^2}{n-1}=\frac{13.197}{15-1}=0.9426\:$

The standard deviation is just the square root of the variance. That is

$\displaystyle s=\sqrt{s^2}=\sqrt{0.9426}=0.971$