Chapter 10, Problem 8MCQ

To determine

To calculate the range, median, lower quartile and upper quartile.

Answer to Problem 8MCQ

Range is 19, upper quartile is 77, lower quartile is 70, and median is 75.5

Explanation of Solution

Given information:

The data set is,

77, 75, 72, 70, 79, 77, 70, 76

Formula used:

Range is,

  $range=\max -\min$

Median is,

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

Lower quartile is,

  $Q1=\frac{1}{4}{(n+1)}^{th}$

Upper quartile,

  $Q3=\frac{3}{4}{(n+1)}^{th}$

Calculation:

Re-arranging the data in an ascending order,

70, 70, 72, 75, 76, 77, 77, 79

The maximum is 79

The minimum is 70

For range, substituting the value,

  $range=79-70$

On solving,

  $range=9$

Therefore, range is 9

For lower quartile, substituting the value,

  $Q1=\frac{1}{4}{(8+1)}^{th}$

On solving,

  $Q1={2.25}^{nd}term$

On rounding off,

  $\approx 2$

Substituting the value,

  $Q1=70$

Therefore, lower quartile is 70.

For upper quartile, substituting the value,

  $Q3=\frac{3}{4}{(8+1)}^{th}$

On solving,

  $Q3={6.75}^{th}term$

On rounding off,

  $\approx 7$

Substituting the value,

  $Q3=77$

Therefore, upper quartile is 77.

For median, substituting the value,

  $M=\frac{{(\frac{8}{2})}^{th}+{(\frac{8}{2}+1)}^{th}}{2}$

On solving,

  $M=\frac{4^{th}+5^{th}}{2}$

Substituting the values,

  $M=\frac{75+76}{2}$

On solving,

  $M=75.5$

Hence, median is 75.5