Chapter 7.2, Problem 7.36E

(a)

To determine

To Explain: the individuals and the variables which are measured.

Answer to Problem 7.36E

Individuals: All the students in the school.

Variables: Eat breakfast on regular basis (Yes/No) and Sex (Male/Female)

Explanation of Solution

Given:

	Male	Female	Total
Eat breakfast regularly	190	110	300
Don't eat breakfast regularly	130	165	295
Total	320	275	595

On the basis of given data

Individuals are all the students in the school.

Variables are eat breakfast on regular basis (Yes/No) and Sex (Male/Female)

(b)

To determine

To Calculate: the probability if randomly selects a student from the school.

Answer to Problem 7.36E

  $\begin{array}{c}\text{P}\left(\text{female}\right)\text{=0}\text{.4622}\\ \text{P}\left(\text{regularly}\right)\text{=0}\text{.5042}\\ \text{P}\left(\text{female}\text{and}\text{regularly}\right)\text{=0}\text{.1849}\\ \text{P}\left(\text{female}\text{or}\text{regularly}\right)\text{=0}\text{.7815}\end{array}$

Explanation of Solution

Given:

	Male	Female	Total
Eat breakfast regularly	190	110	300
Don't eat breakfast regularly	130	165	295
Total	320	275	595

Formula used:

  $\text{Probability}\text{=}\frac{\text{Favourable Cases}}{\text{Total}\text{Cases}}$

Calculation:

	Male	Female	Total
Eat breakfast regularly	190	110	300
Don't eat breakfast regularly	130	165	295
Total	320	275	595

Required Probability using this table

  $\begin{array}{c}\text{P}\left(\text{female}\right)\text{=}\frac{\text{275}}{\text{595}}\\ \text{=0}\text{.4622}\\ \text{P}\left(\text{regularly}\right)\text{=}\frac{\text{300}}{\text{595}}\\ \text{=0}\text{.5042}\end{array}$

  $\begin{array}{c}\text{P}\left(\text{female}\text{and}\text{regularly}\right)\text{=}\frac{\text{110}}{\text{595}}\\ \text{=0}\text{.1849}\\ \text{P}\left(\text{female}\text{or}\text{regularly}\right)\text{=}\frac{\text{190+110+165}}{\text{595}}\\ \text{=0}\text{.7815}\end{array}$