Answer to Question #135607 in Statistics and Probability for Nekee Michel

Question

The length of timber cuts are normally distributed with a mean of 95inches. In a random sample of 45 boards, what is the probability that the mean of the sample will be between 94.8 inches and 95.8 inches?

Answer 1

The following information about the mean and standard deviation has been provided:

μ=95, σ=0.52, n=45

We need to compute $\Pr(94.8 \le \bar X \le 95.8)$The corresponding z-values needed to be computed are:

$Z_1= \frac{X_1-\mu}{\sigma/\sqrt{n}} = \frac{ 94.8-95}{ 0.52/\sqrt{ 45}}= -2.5801\\Z_2 = \frac{X_2-\mu}{\sigma/\sqrt{n}} = \frac{ 95.8-95}{ 0.52/\sqrt{ 45}}= 10.3203$

Therefore, the following is obtained:

$Pr(94.8≤ X ˉ ≤95.8)$ $=Pr(−2.5801≤Z≤10.3203)$

$=Pr(Z≤10.3203)−Pr(Z≤−2.5801)=1−0.0049=0.9951$