Chapter 12.2, Problem 4ACYP

To determine

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

Answer to Problem 4ACYP

  $\mu =30.6\text{ and }\sigma =2.74$ .

Explanation of Solution

Given:

The given data is:

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

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 total number of observations is 25. Then,

  $\begin{array}{l}\mu =\frac{\text{sum of total number of data}}{\text{number of data}}\\ \mu =\frac{\begin{array}{l}28+34+33+33+31+33+29+34+36\\ +31+30+29+32+28+36+29+33\\ +29+28+28+26+31+27+28+31\end{array}}{25}\\ \mu =\frac{765}{25}=30.6\end{array}$

Standard deviation:

To calculate the standard deviation of those numbers:

1. Work out the Mean (the simple average of the numbers)
2. Then for each number: subtract the Mean and square the result.
3. Then work out the mean of those squared differences.
4. Take the square root of that and we are done!

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:

  $\begin{array}{l}\sigma =\sqrt{\frac{{\left(36-30.6\right)}^2+{\left(34-30.6\right)}^2+\mathrm{.....}+{\left(27-30.6\right)}^2+{\left(266-30.6\right)}^2}{25}}\\ \sigma =\sqrt{\frac{188}{25}}\\ \sigma =\sqrt{7.52}=2.74\end{array}$

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

  $\mu =30.6\text{ and }\sigma =2.74$