Chapter 10.5, Problem 14PPS

To determine

To find mean median mode, range and standard deviation.

Answer to Problem 14PPS

Mean=19.5

Median=16.25

Mode=6.5

Range=32.5

Standard deviation=11.62

Explanation of Solution

Given information:

Constant=6.5

Data set: 6.5, 32.5, 26, 13, 6.5, 19.5, 39, 13, 32.5, 6.5

Formula used:

Mean= $\frac{{\displaystyle \sum _{i=1}^n{X}_i}}{n}$

Where, n=total number of terms,

  ${\displaystyle \sum _{i=1}^n{X}_i}$ = sum of all terms

Mode= most repeating numbers.

Median= $\frac{{(n+1)}^{th}}{2}observation$

Range= maximum- minimum

Standard deviation, ${\sigma }_{}=\sqrt{\frac{{\displaystyle \sum _{}^{}{(x_i^{}-\mu )}^2}}{N}}$

Where $x_i^{}$ = each value from data

  $\mu$ =mean of data

  $N$ = size of data

Calculation:

After multiplying with constant,

New data is 6.5, 32.5, 26, 13, 6.5, 19.5, 39, 13, 32.5, 6.5

Substituting the value,

Mean= $\frac{6.5+32.5+26+13+6.5+19.5+39+13+32.5+6.5}{10}=\frac{195}{10}=19.5$

Sort the number in ascending order,

6.5, 6.5, 6.5, 13, 13, 19.5, 26, 32.5, 32.5, 39

Median= $\frac{{(\frac{N}{2})}^{th}ob+{(\frac{N}{2}+1)}^{th}ob}{2}=\frac{{(\frac{10}{2})}^{th}+{(\frac{10}{2}+1)}^{th}}{2}=\frac{5^{th}+6^{th}}{2}=\frac{13+19.5}{2}=16.25$

Mode= 6.5

Range= $\max -\min =39-6.5=32.5$

For standard deviation,

values	Difference= $x_{}^{}-\mu$	( $x_{}^{}-\mu {)}^2$
6.5	-13	169
32.5	13	169
26	6.5	42.25
13	-6.5	42.25
6.5	-13	169
19.5	0	0
39	19.5	380.25
13	-6.5	42.25
32.5	13	169
6.5	-13	169

  ${\displaystyle \sum _{}^{}{(x_{}^{}-\mu )}^2}$ =1352

  ${\sigma }_{}=\sqrt{\frac{{\displaystyle \sum _{}^{}{(x_i^{}-\mu )}^2}}{N}}$ = ${\sigma }_{}=\sqrt{\frac{1352}{10}=}11.62$