Chapter 10.3, Problem 6PPS

To determine

To find minimum, lower quartile, median, upper quartile and maximum.

Answer to Problem 6PPS

Minimum is 16, Lower quartile is 19, Median is 21, Upper quartile is 24, Maximum is 45

Explanation of Solution

Given information:

Data is 45,17,16,22,25,19,20,21,32,37,19,21,24,20,18,22,23,19

Formula used:

Lower quartile= $\frac{(N+1)}{4}$

Median = $\frac{(\frac{N^{th}}{2})ob+{(\frac{N}{2}+1)}^{th}ob}{2}$

Upper quartile= $\frac{3(N+1)}{4}$

Where N is number of tiles

Calculation:

Sorting the data

16,17,18,19,19,19,20,20,21,21,22,22,23,24,25,32,37,45

Total data points=18

Minimum= 16

Maximum=45

For median substituting values,

  $=\frac{(\frac{{18}^{th}}{2})ob+{(\frac{18}{2}+1)}^{th}ob}{2}$

On solving,

  $=\frac{9^{th}ob+{10}^{th}ob}{2}$

Substituting the value of 9^th and 10^th term,

  $=\frac{21+21}{2}$

On solving,

  $=21$

Hence, Median is 21

As the data is even, pair 9^th and 10^th observation. Lower quartile is middle value of lower half and upper quartile is middle value of upper half

For Lower quartile substituting the values,

  $=\frac{{(9+1)}^{th}}{2}ob$

On solving,

  $=5^{th}ob$

Substituting the value,

  $=19$

For Upper quartile substituting the values,

  $=\frac{3{(18+1)}^{th}}{4}ob$

So,

  $={14.25}^{th}ob$

On rounding the values,

  $\approx {14}^{th}ob$

Substituting the value,

  $=24$

Hence, Minimum is 16, Lower quartile is 19, Median is 21, Upper quartile is 24, Maximum is 45