Chapter 10.3, Problem 27PFA

a.

To determine

The calculate the range and interquartile range

Answer to Problem 27PFA

Range is 433 and interquartile range is 215

Explanation of Solution

Given information:Weekly pay of 10 employees,

$54, $278, $70, $159, $482, $49, $205, $70, $386, $63

Formula Used:Range is,

  $Range=MaxValue-MinValue$

Lower quartile is

  $Q1=(n+1)\frac{1}{4}$

Upper quartile is,

  $Q3=(10+1)\frac{3}{4}$

Interquartile range is,

  $IQR=Q3-Q1$

Calculation:

For range substituting values in the range formula,

  $Range=482-49$

On solving,

  $Range=433$

Rearranging the data in ascending order,

49, 54, 63, 70, 70, 159, 205, 278, 386, 482.

For lower quartile substituting the value,

  $Q1=(10+1)\frac{1}{4}$

On solving,

  $Q1=2.75$

On rounding off,

  $Q1\approx 3$

Substituting the value of 3^rd observation,

  $Q1=63$

Hence, lower quartile is 63.

For upper quartile substituting values,

  $Q3=(10+1)\frac{3}{4}$

So,

  $Q3=\frac{33}{4}$

On solving,

  $Q3=8.25$

On rounding off,

  $Q3\approx 8$

Substituting the value of 8^th term,

  $Q3=278$

Hence, upper quartile is 278.

For interquartile range substituting the values,

  $IQR=278-63$

On solving,

  $IQR=215$

Hence, interquartile range is 215

Conclusion:Range is 433 and interquartile range is 215

b.

To determine

To calculate mean, median and mode.

Answer to Problem 27PFA

Mean is 181.6, median is 114.5 and mode is 482.

Explanation of Solution

Given information: Weekly pay of 10 employees,

49, 54, 63, 70, 70, 159, 205, 278, 386, 482.

Formula Used:

Mean is,

  $\begin{array}{l}\overline{x}=\frac{{\displaystyle \sum x}}{n}\end{array}$

Median is,

  $M=\frac{(\frac{n}{2})+(\frac{n}{2}+1)}{2}term$

Mode is the maximum value of a dataset

Calculation:

For mean, substituting the values,

  $\overline{x}=\frac{49+54+63+70+70+159+205+278+386+482}{10}$

So,

  $\overline{x}=\frac{1816}{10}$

On solving,

  $\overline{x}=181.6$

Hence, Mean is 181.6

For median, substituting the values,

  $M=\frac{(\frac{10}{2})+(\frac{10}{2}+1)}{2}term$

On solving,

  $M=\frac{5^{th}+6^{th}}{2}term$

Substituting the value,

  $M=\frac{70+159}{2}$

On solving,

  $M=114.5$

Hence, Median is 114.5

For mode,

482 is the mode, since it’s the maximum value.

Conclusion:Mean is 181.6, median is 114.5 and mode is 482.

c.

To determine

To identify any outliers, if present.

Answer to Problem 27PFA

There are no outliers present in the given data.

Explanation of Solution

Given information:

Weekly pay of 10 employees,

49, 54, 63, 70, 70, 159, 205, 278, 386, 482.

Concept Used:Outliers definition.

Calculation:

Evaluating the given data, no observation lies at an abnormal distance.

Hence, there are no outliers in the given set of data.

Conclusion:There are no outliers present in the given data.

d.

To determine

To calculate standard deviation.

Answer to Problem 27PFA

Standard deviation is

Explanation of Solution

Given information:Weekly pay of 10 employees,

49, 54, 63, 70, 70, 159, 205, 278, 386, 482.

Mean is 181.6

Formula Used:Standard deviation is,

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

Calculation:

Substituting the values, $sd=\sqrt{\frac{\begin{array}{}$ {(49-181)}^2+{(54-181)}^2+{(63-181)}^2+{(70-181)}^2+{(70-181)}^2+{(159-181)}^2+$$ {(205-181)}^2+{(278-181)}^2+{(386-181)}^2+{(482-181)}^2$\end{array}}{10}}$

On solving,

  $sd=\sqrt{\frac{{(-132)}^2+{(-127)}^2+{(-118)}^2+{(-111)}^2+{(-111)}^2+{(-22)}^2+{(24)}^2+{(97)}^2+{(205)}^2+{(301)}^2}{10}}$

So,

  $sd=\sqrt{\frac{215214}{10}}$

On solving,

  $sd=146.70$

Hence, Standard deviation is 146.70

Since, standard deviation is high it implies that the observations are scattered over a large range of values.

Conclusion:Standard deviation is 146.70