Chapter 1.2, Problem 2.1CYU

To determine

To Explain: the few lines on the shape, spread, centre and outliers using the provided back-to-back stem plot.

Explanation of Solution

Given:

  

Calculation:

For male, the provided data from the stem plot

4, 5, 5, 5, 6, 7, 7, 7, 7, 8, 10, 10, 10, 10, 11, 12, 14, 22, 35, 38

For female, the provided data from the stem plot

13, 13, 13, 15, 19, 22, 23, 23, 24, 26, 26, 30, 31, 34, 38, 49, 50, 50, 51, 57

Shape:

7 and 10 were the maximum times for male data (4). So, in this info, there are two modes. The distribution of the male data is, in other words, bimodal.

Just 13 have reached the maximum time for female data (3). So, one mode exists in this info. The distribution of female data is, in other words, unimodal.

Centre:

For male information

Mean = (4+5+...+38)/20 = 233/20 = 11.65

For female information

Mean = (13+13+..+51+57)/20 = 607/20 = 30.3

Spread:

For male information Range = 38-4 = 34

Standard deviation = 9.18

1^stQuartile, $Q_1$  = 6.5

2^nd Quartile, $Q_2$  = 9

3^rd Quartile, $Q_3$  = 11.5

Interquartile Range,

IQR = $Q_3-Q_1$

= 11.5-6.5

= 5

For female information

Range = 57-13 = 44

Standard deviation = 13.88

1^st Quartile, Q₁ = 20.5

2^nd Quartile, Q₂ = 26

3^rd Quartile, Q₃ = 43.5

Interquartile Range,

IQR = $Q_3-Q_1$

= 43.5-20.5

= 23

Outlier:

For male information

Outliers will be the values if they fall outside $\left(Q_1-1.5IQR,Q_3+1.5IQR\right)$ therefore, (4, 14).

So, the outliers are: 22, 35, 38

For female information Outliers will be the values if they fall outside $\left(Q_1-1.5IQR,Q_3+1.5IQR\right)$ therefore, (9, 55).

So, the outliers are: 57