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. 


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.