Chapter 13, Problem 18SGR

To determine

Calculate measures of variation and outlier for given earning by Claire.

$5, $7, $10, $6, $8

Answer to Problem 18SGR

Measures of variation for given earning by Claire are as follows,

Range = 5

Upper quartile=9

Lower quartile=5.5

Interquartile range=3.5

Explanation of Solution

Given:

Earning by Claire = $5, $7, $10, $6, $8

Calculations:

Here, we have to calculate measures of variation and outlier for given set of data.

Measures of variation include range, upper quartile, lower quartile and interquartile range of given data.

Range is the difference between higher and lower data value.

Here, range = 10-5=5

Median for given set of data is 7

Upper quartile is the median of upper half set of data.

Here, Upper quartile is, $\frac{8+10}{2}=9$

Lower quartile is the median of lower half set of data.

Here, lower quartile is, $\frac{5+6}{2}=5.5$

Interquartile range is the difference between upper quartile and lower quartile.

Here, interquartile range =9-5.5=3.5

Now, to find outlier, we need to multiply interquartile range by 1.5 and then subtracting it from lower quartile and adding it to upper quartile.

Thus, multiplying 1.5 with interquartile range, $3.5\times 1.5=5.25$

Then, subtracting it from lower quartile and adding it to upper quartile,

5.5-5.25=0.25 and 9+5.25=14.25

Therefore, limits of outlier are 0.25 to 14.25. But there is nothing beyond outlier limits. Hence there is no outlier for given earning by Claire.

Conclusion:

Therefore, we are able to calculate measures of variation and possible outlier for given set of data.