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


For the data of Exercise 1.4 on page 13, compute both the mean and variance in “flexibility” for both company A and company B. 


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.


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


Exercise 1.3 on page 13 showed samples of tensile strength data, one for specimens that were exposed to an aging process and one in which there was no aging of the specimens. Calculate the sample variance as well as standard deviation in tensile strength for both samples. 


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


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


Compute the sample variance and standard deviation for the water absorbency data of Exercise 1.2 on page 13.


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


Sample Variance & Standard Deviation | Introduction to Statistics and Data Analysis | Probability & Statistics for Engineers & Scientists | Walpole | Problem 1.7


Consider the drying time data for Exercise 1.1 on page 13. Compute the sample variance and sample standard deviation.


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.

\displaystyle x\displaystyle x_i-\overline{x}\displaystyle \left(x_i-\overline{x}\right)^2
\displaystyle 3.4\displaystyle 3.4-3.787=-0.387\displaystyle \left(-0.387\right)^2=0.150
\displaystyle 2.5\displaystyle 2.5-3.787=-1.287\displaystyle \left(-1.287\right)^2=1.656
\displaystyle 4.8\displaystyle 4.8-3.787=1.013\displaystyle \left(1.013\right)^2=1.026
\displaystyle 2.9\displaystyle 2.9-3.787=-0.887\displaystyle \left(-0.887\right)^2=0.787
\displaystyle 3.6\displaystyle 3.6-3.787=-0.187\displaystyle \left(-0.187\right)^2=0.035
\displaystyle 2.8\displaystyle 2.8-3.787=-0.987\displaystyle \left(-0.987\right)^2=0.974
\displaystyle 3.3\displaystyle 3.3-3.787=-0.487\displaystyle \left(-0.487\right)^2=0.237
\displaystyle 5.6\displaystyle 5.6-3.787=1.813\displaystyle \left(1.813\right)^2=3.287
\displaystyle 3.7\displaystyle 3.7-3.787=-0.087\displaystyle \left(-0.087\right)^2=0.008
\displaystyle 2.8\displaystyle 2.8-3.787=-0.987\displaystyle \left(-0.987\right)^2=0.974
\displaystyle 4.4\displaystyle 4.4-3.787=0.613\displaystyle \left(0.613\right)^2=0.376
\displaystyle 4.0\displaystyle 4.0-3.787=0.213\displaystyle \left(0.213\right)^2=0.045
\displaystyle 5.2\displaystyle 5.2-3.787=1.413\displaystyle \left(1.413\right)^2=1.997
\displaystyle 3.0\displaystyle 3-3.787=-0.787\displaystyle \left(-0.787\right)^2=0.619
\displaystyle 4.8\displaystyle 4.8-3.787=1.013\displaystyle \left(1.013\right)^2=1.026
\displaystyle SUM \displaystyle 13.197

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