Chapter 10.5, Problem 8PPS

To determine

To find mean median mode, range and standard deviation.

Answer to Problem 8PPS

Mean=81.7

Median=82.5

Mode=84

Range=13

Standard deviation=20.41

Explanation of Solution

Given information:

101, 99, 97,88,92,100,97,89,94,90

Constant= -13

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 88,86,84,75,79,87,84,76,81,77

Substituting the value,

Mean= $\frac{88+86+84+75+79+87+84+76+81+77}{10}=\frac{817}{10}=81.7$

Sort the number in ascending order,

75,76,77,79,81,84,84,86,87,88

Median= $\frac{{(\frac{N}{2})}^{th}ob+{(\frac{N}{2}+1)}^{th}ob}{2}=\frac{{(\frac{10}{2})}^{th}+{(\frac{10}{2}+1)}^{th}}{2}=\frac{5^{th}+6^{th}}{2}=\frac{81+84}{2}=82.5$

Mode=84(because it is most repetitive number)

Range= $\max -\min =88-75=13$

For standard deviation,

values	Difference= $x_{}^{}-\mu$	( $x_{}^{}-\mu {)}^2$
88	6.3	39.69
86	4.3	18.49
84	2.3	5.29
75	-6.7	44.89
79	-2.7	7.29
87	5.3	28.09
84	2.3	5.29
76	-5.7	32.49
81	-0.7	0.49
77	-4.7	22.09

  ${\displaystyle \sum _{}^{}{(x_{}^{}-\mu )}^2}$ =204.1

  ${\sigma }_{}=\sqrt{\frac{{\displaystyle \sum _{}^{}{(x_i^{}-\mu )}^2}}{N}}$ = ${\sigma }_{}=\sqrt{\frac{204.1}{10}=}20.41$