Chapter 10.5, Problem 10PPS

To determine

To find mean median mode, range and standard deviation.

Answer to Problem 10PPS

Mean=84.2

Median=85.5

Mode=no mode

Range=21

Standard deviation=6.77

Explanation of Solution

Given information:

72,56,71,63,68,59,77,74,76,66

Constant=0.5

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 adding constant,

New data is 88,72,87,79,84,75,93,90,92,82

Substituting the value,

Mean= $\frac{88+72+87+79+84+75+93+90+92+82}{10}=\frac{842}{10}=84.2$

Sort the number in ascending order,

72,75,79,82,84,87,88,90,92,93

Median= $\frac{{(\frac{N}{2})}^{th}ob+{(\frac{N}{2}+1)}^{th}ob}{2}=\frac{{(\frac{10}{2})}^{th}+{(\frac{10}{2}+1)}^{th}}{2}=\frac{5^{th}+6^{th}}{2}=\frac{84+87}{2}=85.5$

Mode= no mode.

Range= $\max -\min =93-72=21$

For standard deviation,

values	Difference= $x_{}^{}-\mu$	( $x_{}^{}-\mu {)}^2$
72	3.8	14.44
75	-12.2	148.84
79	2.8	7.84
82	-5.2	27.04
84	0.2	0.04
87	-9.2	84.64
88	8.8	77.44
90	5.8	33.64
92	7.8	60.84
93	-2.2	4.84

  ${\displaystyle \sum _{}^{}{(x_{}^{}-\mu )}^2}$ =459.6

  ${\sigma }_{}=\sqrt{\frac{{\displaystyle \sum _{}^{}{(x_i^{}-\mu )}^2}}{N}}$ = ${\sigma }_{}=\sqrt{\frac{459.6}{10}=}6.77$