SOLUTION:
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)
[tex]\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}[/tex]
Then for 72.3
[tex]\begin{gathered} z=\text{ }\frac{x\text{ - }\eta}{\delta} \\ z=\text{ }\frac{72.3\text{ - }68}{14} \\ z=\text{ }0.30714 \end{gathered}[/tex]
Final answers:
The probability of between 69.6 and 72.3 is
0.080804 (Answer)
