Answer:
Given Data:
The Population Mean $(\mu)= 76 $
The Population Standard Deviation $(\sigma) = 7.6 $
The Sample Size $(n)=32 $
Solution:
The probability that average final exam grade of this sample is between 75 and 80:
$P(75 < \bar{x} < 80) = P\left( \frac{75 – 76}{\frac{7.6}{\sqrt{32}}} < \frac{\bar{x} – \mu}{\frac{\sigma}{\sqrt{n}}} < \frac{80 – 76}{\frac{7.6}{\sqrt{32}}} \right)$
$P(75 < \bar{x} < 80) = P(-0.744 < z < 2.977)$
$P(75 < \bar{x} < 80) = P(z < 2.977) – P(z < -0.744)$
$P(75 < \bar{x} < 80) = 0.9985 – 0.2284$
$P(75 < \bar{x} < 80) = 0.7701$
Final Answer:
The probability that average final exam grade of this sample is between 75 and 80 $ = 0.7701 $
