Chapter 10.5, Problem 19PPS

i.

To determine

To calculate mean, median, mode, range and standard deviation of Rhonda’s earnings

Answer to Problem 19PPS

Mean is 52.95.

Median is 53.

Mode is 53.

Standard deviation is 6.21

Range is 19

Explanation of Solution

Given information:

Rhonda’s Pay
45	55	53
47	53	54
44	56	59
63	47	53
60	57	62
44	50	45
60	53	49
62	47

Formula used: Mean formula is,

  $\overline{X}=\frac{{\displaystyle \sum fX}}{N}$

Where, f is frequency

N is total number of items

X is item.

Median formula is,

  where N cumulative frequency

Mode is the item with the maximum frequency.

Standard deviation is,

  $SD=\sqrt{\frac{{\displaystyle \sum }{(x-\overline{x})}^2}{n}}$

Where, SD= standard deviation,

x= value

The Range is the difference between the lowest and highest values.

Calculation:

For Mean,

Substituting the values in mean formula,

  $=\frac{45+55+53+47+53+54+44+56+59+63+47+53+60+57+62+44+50+45+60+53+49+62+47}{23}$

On solving,

  $=\frac{1218}{23}$

On solving,

  $=52.95$

Hence, Mean is 52.95

For median,

Sorting the frequency

44,44,45,45,47,47,47,49,50,53,53,53,53,54,55,56,57,59,60,60,62,62,63,

Substituting the values,

  $={\left(\frac{23+1}{2}\right)}^{th}term$

On solving,

  $={\left(12\right)}^{th}term$

Substituting the values,

  $median=53$

Median is 12^th term from the sorted frequency which is 53.

For mode,

Mode is the item with the maximum frequency. Which is 53

For standard deviation,

Substituting the values in standard deviation formula,

  $SD=\sqrt{\frac{\begin{array}{}$ {(45-52.95)}^2+{(55-52.95)}^2+{(53-52.95)}^2+{(47-52.95)}^2+{(53-52.95)}^2+{(54-52.95)}^2$$ +{(44-52.95)}^2+{(56-52.95)}^2+{(59-52.95)}^2+{(63-52.95)}^2+{(47-52.95)}^2+{(53-52.95)}^2+$$ {(60-52.95)}^2+{(57-52.95)}^2+{(62-52.95)}^2+{(44-52.95)}^2+{(50-52.95)}^2+{(45-52.95)}^2$$ +{(60-52.95)}^2+{(53-52.95)}^2+{(49-52.95)}^2+{(62-52.95)}^2+{(47-52.95)}^2$\end{array}}{23}}$

On solving,

  $SD=\sqrt{\frac{\begin{array}{}$ {(-7.95)}^2+{(2.05)}^2+{(0.05)}^2+{(-5.95)}^2+{(0.05)}^2+{(1.05)}^2$$ +{(-8.95)}^2+{(3.05)}^2+{(6.05)}^2+{(10.05)}^2+{(-5.95)}^2+{(0.05)}^2+$$ {(7.05)}^2+{(4.05)}^2+{(9.05)}^2+{(-8.95)}^2+{(-2.95)}^2+{(-7.95)}^2$$ +{(7.05)}^2+{(0.05)}^2+{(-3.95)}^2+{(9.05)}^2+{(-5.95)}^2$\end{array}}{23}}$

So,

  $SD=\sqrt{38.58}$

On solving,

  $SD=6.21$

Hence, Standard deviation is 6.21

For Range,

Range is the difference between the lowest and highest values.

Range $=\max -\min$

=63-44

=19

Hence, range is 19.

ii.

To determine

To calculate mean, median, mode, range and standard deviation of Rhonda’s earnings after deducting the $5 bonus.

Answer to Problem 19PPS

Mean is 47.91.

Median is 48.

Mode is 48.

Standard deviation is 6.21

Range is 19

Explanation of Solution

Given information:

Rhonda’s pay after deducting $5 bonus

Rhonda’s Pay
40	50	47
42	48	49
39	51	54
58	42	48
55	52	57
39	45	40
55	48	44
57	42

Formula used: Mean formula is,

  $\overline{X}=\frac{{\displaystyle \sum fX}}{N}$

Where, f is frequency

N is total number of items

X is item.

Standard deviation is,

  $SD=\sqrt{\frac{{\displaystyle \sum }{(x-\overline{x})}^2}{n}}$

Where, SD= standard deviation,

x= value

The Range is the difference between the lowest and highest values.

Calculation:

For Mean,

Substituting the values in mean formula,

  $=\frac{\text{40+50+47+42+48+49+39+51+54+58+42+48+55+52+57+39+45+40+55+48+44+57+42}}{23}$

On solving,

  $=47.91$

Hence, Mean is 47.91

For median,

Sorting the frequency

39,39,40,40,42,42,42,44,45,48,48,48,48,49,50,51,52,54,55,55,57,57,58

Substituting the values,

  $={\left(\frac{23+1}{2}\right)}^{th}term$

On solving,

  $={\left(12\right)}^{th}term$

Substituting the value,

  $median=48$

For mode,

Mode is the item with the maximum frequency which is 48.

For standard deviation,

Substituting the values in standard deviation formula,

  $SD=\sqrt{\frac{\begin{array}{l}{\text{(40-47}}^{\text{.91)}}\text{+(50-47}{\text{.91)}}^2\text{+(47-47}{\text{.91)}}^2\text{+(42-47}{\text{.91)}}^2\text{+(48-47}{\text{.91)}}^2\text{+(49-47}{\text{.91)}}^2\\ \text{+(39-49}{\text{.91)}}^2\text{+(51-47}{\text{.91)}}^2\text{+(54-47}{\text{.91)}}^2\text{+(58-47}{\text{.91)}}^2\text{+(42-47}{\text{.91)}}^2\text{+(48-47}{\text{.91)}}^2\\ \text{+(55-47}{\text{.91)}}^2\text{+(52-47}{\text{.91)}}^2\text{+(57-47}{\text{.91)}}^2\text{+(39-47}{\text{.91)}}^2\text{+(45-47}{\text{.91)}}^2\text{+(40-47}{\text{.91)}}^2\\ \text{+(55-47}{\text{.91)}}^2\text{+(48-47}{\text{.91)}}^2\text{+(44-47}{\text{.91)}}^2\text{+(57-47}{\text{.91)}}^2\text{+(42-47}{\text{.91)}}^2\end{array}}{23}}$

On solving,

  $SD=\sqrt{\frac{\begin{array}{l}{\text{(-7}}^{\text{.91)}}\text{+(2}{\text{.9)}}^2\text{+(-0}{\text{.91)}}^2\text{+(-5}{\text{.91)}}^2\text{+(0}{\text{.09)}}^2\text{+(1}{\text{.09)}}^2\\ \text{+(-8}{\text{.91)}}^2\text{+(3}{\text{.09)}}^2\text{+(6}{\text{.09)}}^2\text{+(10}{\text{.09)}}^2\text{+(-5}{\text{.91)}}^2\text{+(0}{\text{.09)}}^2\\ \text{+(7}{\text{.9)}}^2\text{+(4}{\text{.09)}}^2\text{+(9}{\text{.09)}}^2\text{+(-8}{\text{.91)}}^2\text{+(-2}{\text{.91)}}^2\text{+(-7}{\text{.91)}}^2\\ \text{+(7}{\text{.09)}}^2\text{+(0}{\text{.09)}}^2\text{+(-3}{\text{.91)}}^2\text{+(9}{\text{.09)}}^2\text{+(-5}{\text{.91)}}^2\end{array}}{23}}$

So,

  $SD=\sqrt{38.62}$

On solving,

  $SD=6.21$

Hence, Standard deviation is 6.21

For Range,

Range is the difference between the lowest and highest values.

=58-39

=19

Hence, range is 19