The lengths of lumber a machine cuts are normally distributed with a mean of 88 inches and a standard deviation of 0.6 inch. (a) What is the probability that a randomly selected board cut by the machine has a length greater than 88.28 inches?  (b) A sample of 35 boards is randomly selected. What is the probability that their mean length is greater than 88.28 inches?

a) The probability that a randomly selected board cut by the machine has a length greater than 88.28 inches is 0.3192.

b) The probability that the mean length of a sample of 35 randomly selected boards cut by the machine is greater than 88.28 inches is 0.0029.

Expert Answers

An illustration of the letter 'A' in a speech bubbles

From the given information, we know that the lengths of lumber (denoted by the random variable X) are normally distributed with a mean of 88 inches and a standard deviation of 0.6 inches. Thus, X ~ N(`\mu` = 88, `\sigma` = 0.6).

a) We want P( X > 88.28).

To...

View
This Answer Now

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

Start your 48-Hour Free Trial

From the given information, we know that the lengths of lumber (denoted by the random variable X) are normally distributed with a mean of 88 inches and a standard deviation of 0.6 inches. Thus, X ~ N(`\mu` = 88, `\sigma` = 0.6).

a) We want P( X > 88.28).

To find this probability using the standard normal distribution table, we have to find the Z score of X using the formula `Z = \frac{X - \mu}{\sigma}` .

Therefore, Z = `\frac{88.28 - 88}{0.6}` = `\frac{0.28}{0.6}` = 0.47 (to 2 decimal places).

Next, we find P(Z > 0.47) from the standard normal distribution table

P(Z > 0.47) = 1 - P(Z < 0.47) = 1 - 0.68082 = 0.3192 (to 4 decimal places).

Alternatively, we can use programs such as the R statistical program available here to calculate this probability. To achieve this in R, use the following code (the first line gives the code to be run, and the second line gives the solution received):

> pnorm(88.28, mean=88, sd=0.6, lower.tail = FALSE)
[1] 0.3203692

b) If X has a distribution with mean `\mu` and standard deviation `\sigma` , and it is approximately normally distributed, then `\bar{X}` is approximately normally distributed with mean `\mu` and standard error `\frac{\sigma}{\sqrt{n}}` .

Thus, to find P(`\bar{X}` > 88.28), where the size of the sample (n) is 35, we first find the Z score of `\bar{X}` using the formula `Z = \frac{\bar{X} - \mu}{\frac{\sigma}{\sqrt{n}}}` .

Thus, Z = `\frac{88.28 - 88}{\frac{0.6}{\sqrt{35}}}` = 2.76 (to 2 decimal places).

P(Z > 2.76) = 1 - p(Z < 2.76) = 1 - 0.99711 = 0.00289 (from tables).

Therefore, P(`\bar{X}` > 88.28) = 0.0029 (to 4 decimal places).

Alternatively, we can use the following R code to calculate this probability (the first line gives the code to be executed, and the second line gives the solution received):

> pnorm(88.28, mean=88, sd=(0.6/(sqrt(35))), lower.tail = FALSE)
[1] 0.00288267

Approved by eNotes Editorial Team