Chapter 13, Problem 16SGR

To determine

Calculate measures of variation and outlier for given number of minutes spent on reading.

31, 33, 32, 34, 35, 33

Answer to Problem 16SGR

Measures of variation for given number of minutes spent on reading are as follows,

Range = 4

Upper quartile=34

Lower quartile=32

Interquartile range=2

Explanation of Solution

Given:

Number of minutes spent on reading = 31, 33, 32, 34, 35, 33

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 = 35-31 = 4

Median for given set of data is,

  $\frac{33+33}{2}=33$

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

Here, Upper quartile = 34

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

Here, lower quartile = 32

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

Here, interquartile range =34-32 = 2

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, $2\times 1.5=3.0$

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

32-3=29 and 34+3=37

Therefore, limits of outlier are 29 to 37. But there is nothing beyond outlier limits. Hence ther is no outlier for given data of minutes spent on reading

Conclusion:

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