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...

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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