## Related questions with answers

A normal population has a mean of 75 and a standard deviation of 5. You select a sample of 40. Compute the probability the sample mean is: a. Less than 74. b. Between 74 and 76. c. Between 76 and 77. d. Greater than 77

Solution

VerifiedMean of the normal distribution is $\mu=75$ and the standard deviation is $\sigma=5$. Sample size is $n=40$.

Let $\bar{X}$ be the sample mean.

$\textbf{a. $\bar{X}$ is less than 74 }$

$\begin{align*} P(\bar{X}<74)&=P(\bar{X}-\mu<74-\mu)\\ &=P \left(\frac{\bar{X}-\mu}{\sigma/\sqrt{n}}<\frac{74-\mu}{\sigma/\sqrt{n}}\right)\\ &=P \left(z<\frac{74-75}{5/\sqrt{40}}\right)\\ &=P \left(z<\frac{-1}{0.79}\right)\\ &=P(z<-1.26)\\ &=P(z>1.26)\\ &=0.5-P(0<z<1.26)\\ &=0.5-0.3962 \text{ \ (from Appendix B1.) \ }\\ &=0.1038\\ \end{align*}$

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