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?

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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...

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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

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