# The lengths of lumber a machine's cuts are normally distributed with a mean of 89 inches and a standard deviation of 0.7 inch. (a) What is the probability that a randomly selected board cut by the machine has a length greater than 89.24 inches? (b) A sample of 37 boards is randomly selected. What is the probability that their mean length is greater than 89.24 inches?

From the information given, we know that the lengths of lumber (denoted by the random variable X) are normally distributed with a mean (\mu ) of 89 inches and a standard deviation (\sigma ) of 0.7 inches. Thus, X ~ N(\mu = 89, \sigma = 0.7).

a) We want P(X...

Start your 48-hour free trial to unlock this answer and thousands more. Enjoy eNotes ad-free and cancel anytime.

From the information given, we know that the lengths of lumber (denoted by the random variable X) are normally distributed with a mean (\mu ) of 89 inches and a standard deviation (\sigma ) of 0.7 inches. Thus, X ~ N(\mu = 89, \sigma = 0.7).

a) We want P(X > 89.24).

To find this probability using the standard normal distribution tables available here, we must first find the Z score of X using the following formula Z = \frac{X - \mu}{\sigma} = \frac{89.24 - 89}{0.7} = 0.34 (to 2 decimal places).

Thus, we want P(Z > 0.34) = 1 - P(Z < 0.34) = 1 - 0.63307 = 0.36693 (from tables).

Therefore, P(X > 89.24) = 0.3669 (to 4 decimal places).

Alternatively, this probability can be calculated using various programs such as the R statistical program available here.

Using R, the first line of code shown below can be executed to give the solution in the second line:

> pnorm(89.24, mean=89, sd=0.7, lower.tail = FALSE)
[1] 0.365853

b) If X is normally distributed with mean \mu and standard deviation \sigma , the \bar{X} is approximately normally distributed with mean \mu and standard error \frac{\sigma}{\sqrt{n}} .

We want P(\bar{X} >89.24).

The Z score of the sample mean \bar{X} is \frac{\bar{X} - \mu}{\frac{\sigma}{\sqrt{n}}} = \frac{89.24 - 89}{\frac{0.7}{\sqrt{37}}} = 2.09 (to 2 decimal places).

Thus, P(Z > 2.09) = 1 - p(Z < 2.09) = 1 - 0.98169 = 0.01831 (from tables).

Therefore, P(\bar{X} > 89.24) = 0.0183 (to 4 decimal places).

Alternatively, this probability can be directly calculated without finding its Z score using the following R code (the first line gives the code to be executed, and the second line gives the solution received):

> pnorm(89.24, mean=89, sd=(0.7/(sqrt(37))), lower.tail = FALSE)
[1] 0.01851113

Approved by eNotes Editorial Team