Chapter 10.5, Problem 4BGP

To determine

To describe the center and spread the data using five-number summary.

Answer to Problem 4BGP

Median=10

Lower quartile=6

Upper quartile=16

Minimum=0

Maximum=21

Explanation of Solution

Given information: Given data is,

Junior season: 0, 2, 3, 3, 6, 6, 7, 8, 8, 10, 10, 12, 12, 13, 14, 15, 16, 17, 17, 21

Senior season: 3, 5, 6, 10, 12, 18, 20, 20, 21, 22, 22, 24, 24, 25, 26, 26, 28, 30, 32, 32

Formula used: Median is,

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

Lower quartile is,

  $Q1=(n+1)\frac{1}{4}$

Upper quartile is,

  $Q3=(n+1)\frac{3}{4}$

Calculation:

Re-arranging the data in an ascending order,

Junior Season(Points)

0,2,3,3,6,6,7,8,8,10,10,12,12,13,14,15,16,17,17,21

Senior Season(Points)

3,5,6,10,12,18,20,20,21,22,22,24,24,25,26,26,28,30,32,32

For Junior Season (Points)

For median substituting the values,

  $M=\frac{{(\frac{20}{2})}^{th}+{(\frac{20}{2}+1)}^{th}}{2}$

On solving,

  $M=\frac{{10}^{th}+{11}^{th}}{2}term$

Substituting the values,

  $M=\frac{10+10}{2}$

On solving,

  $M=10$

Hence, Median is 10

For lower quartile substituting the values,

  $Q1=(20+1)\frac{1}{4}$

On solving,

  $Q1=5.25$

On rounding off,

  $\approx 5$

Substituting the value,

  $Q1=6$

Hence, Lower quartile is 6

For higher quartile substituting the values,

  $Q3=(20+1)\frac{3}{4}$

On solving,

  $Q3=15.75$

Rounding off,

  $\approx 16$

Substituting the value,

  $Q3=15$

Hence, higher quartile is 15

Minimum is the smallest value of dataset.

Hence, minimum is 0.

Maximum is the largest value of the dataset.

Hence, maximum= 21.

The Junior Season (Points) five number summary is,

0, 6, 10, 16, 21

For Senior Season (Points)

For median substituting the values,

  $M=\frac{{(\frac{20}{2})}^{th}+{(\frac{20}{2}+1)}^{th}}{2}$

On solving,

  $M=\frac{{10}^{th}+{11}^{th}}{2}term$

Substituting the values,

  $M=\frac{22+24}{2}$

On solving,

  $M=23$

Hence, median is 23

For lower quartile substituting the values,

  $Q1=(20+1)\frac{1}{4}$

On solving,

  $Q1=5.25$

On rounding off,

  $\approx 5$

Substituting the value,

  $Q1=12$

Hence, Lower quartile is 12

For higher quartile substituting the values,

  $Q3=(20+1)\frac{3}{4}$

On solving,

  $Q3=15.75$

Rounding off,

  $\approx 16$

Substituting the value,

  $Q3=26$

Hence, upper quartile is 26

Minimum is the smallest value of dataset.

Hence, minimum= 3.

Maximum is the largest value of the dataset.

Hence, maximum= 32.

The Senior Season (Points) five number summary is,

3, 12, 23, 26, 32

Five number summary is less affected by the abnormalities. Hence, it is chosen.