Question

Not sure how solve questions b, c, d, e ??

Answer 1

Given: Normal distribution

Values in thousands

Mean = 68

Standard deviation= 14

Sample size = 9

We will go ahead to calculate the z-scores for the sample value (69.4 and 72.3)

`\begin{gathered} z=\text{ }\frac{x\text{ - }\eta}{\delta} \\ z=\text{ }\frac{69.4\text{ - }68}{14} \\ z=\text{ }0.1 \\ \text{From z-score tables} \\ Pr( \end{gathered}`

Then for 72.3

`\begin{gathered} z=\text{ }\frac{x\text{ - }\eta}{\delta} \\ z=\text{ }\frac{72.3\text{ - }68}{14} \\ z=\text{ }0.30714 \end{gathered}`

Final answers:

The probability of between 69.6 and 72.3 is

0.080804 (Answer)