Chapter 10.5, Problem 12PPS

To determine

To find mean median mode, range and standard deviation.

Answer to Problem 12PPS

Mean=10.84

Median=11

Mode=11.6

Range=5.4

Standard deviation=1.705

Explanation of Solution

Given information:

64, 42, 58, 40, 61, 67, 58, 52, 51, 49

Constant=0.2

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 12.8, 8.4, 11.6, 8, 12.2, 13.4, 11.6, 10.4, 10.2, 9.8

Substituting the value,

Mean= $\frac{12.8+8.4+11.6+8+12.2+13.4+11.6+10.4+10.2+9.8}{10}=\frac{108.4}{10}=10.84$

Sort the number in ascending order,

8, 8.4, 9.8, 10.2, 10.4, 11.6, 11.6, 12.2, 12.8, 13.4

Median= $\frac{{(\frac{N}{2})}^{th}ob+{(\frac{N}{2}+1)}^{th}ob}{2}=\frac{{(\frac{10}{2})}^{th}+{(\frac{10}{2}+1)}^{th}}{2}=\frac{5^{th}+6^{th}}{2}=\frac{10.4+11.6}{2}=11$

Mode= 11.6

Range= $\max -\min =13.4-8=5.4$

For standard deviation,

values	Difference= $x_{}^{}-\mu$	( $x_{}^{}-\mu {)}^2$
8	1.96	3.846
8.4	-2.44	5.95
9.8	0.76	0.577
10.2	-2.84	8.0656
10.4	1.36	1.849
11.6	2.56	6.55
11.6	0.76	0.577
12.2	-0.44	0.193
12.8	-0.64	0.4096
13.4	-1.04	1.081

  ${\displaystyle \sum _{}^{}{(x_{}^{}-\mu )}^2}$ =29.104

  ${\sigma }_{}=\sqrt{\frac{{\displaystyle \sum _{}^{}{(x_i^{}-\mu )}^2}}{N}}$ = ${\sigma }_{}=\sqrt{\frac{29.104}{10}=}1.705$