Chapter 12.2, Problem 4BCYP

To determine

To find: the mean and standard deviation of the population of the given data.

Answer to Problem 4BCYP

  $\mu =32.2\text{ and }\sigma =8.18$ .

Explanation of Solution

Given:

The given data from previous question:

28	34	33	33	31
33	29	34	36	31
30	29	32	28	36
29	33	29	28	28
26	31	28	27	29

Change the one number 30 to 70 from the data.

Calculation:

Mean: The mean (average) of a data set is found by adding all numbers in the data set and then dividing by the number of values in the set. 

According to the question, the new data table is:

28	34	33	33	31
33	29	34	36	31
70	29	32	28	36
29	33	29	28	28
26	31	28	27	29

the total number of observations is 25. Then,

  $\begin{array}{l}\mu =\frac{\text{sum of total number of data}}{\text{number of data}}\\ \mu =\frac{\begin{array}{l}28+34+33+33+31+33+29+34+36\\ +31+70+29+32+28+36+29+33\\ +29+28+28+26+31+27+28+31\end{array}}{25}\\ \mu =\frac{805}{25}=32.2\end{array}$

Standard deviation:

The formula for standard deviation is $\sigma =\sqrt{\frac{n{\displaystyle \sum _{k=1}^n{\left(x_k-{\mu }_k\right)}^2}}{n}}$ .

Thus, the standard deviation for the given data is:

  $\begin{array}{l}\sigma =\sqrt{\frac{{\left(76-30.6\right)}^2+{\left(36-30.6\right)}^2+\mathrm{.....}+{\left(27-30.6\right)}^2+{\left(266-30.6\right)}^2}{25}}\\ \sigma =\sqrt{\frac{1676}{25}}\\ \sigma =\sqrt{67.04}=8.18\end{array}$

Thus, the mean and standard deviation of the population of the given data is

  $\mu =32.2\text{ and }\sigma =8.18$