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

Question #135607
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?
1
Expert's answer
2020-09-29T18:10:38-0400

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

μ=95, σ=0.52, n=45

We need to compute Pr(94.8Xˉ95.8)\Pr(94.8 \le \bar X \le 95.8)The corresponding z-values needed to be computed are:

Z1=X1μσ/n=94.8950.52/45=2.5801Z2=X2μσ/n=95.8950.52/45=10.3203Z_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(94.8Xˉ95.8)Pr(94.8≤ X ˉ ≤95.8) =Pr(2.5801Z10.3203)=Pr(−2.5801≤Z≤10.3203)

=Pr(Z10.3203)Pr(Z2.5801)=10.0049=0.9951=Pr(Z≤10.3203)−Pr(Z≤−2.5801)=1−0.0049=0.9951


Need a fast expert's response?

Submit order

and get a quick answer at the best price

for any assignment or question with DETAILED EXPLANATIONS!

Comments

No comments. Be the first!

Leave a comment