Question

The lengths of lumber a machine cuts are normally distributed with a mean of 93 inches and a standard deviation of 0.5 inch.(a) What is the probability that a randomly selected board cut by the machine has a length greater than 93.11 inches?(b) A sample of 42 boards is randomly selected. What is the probability that their mean length is greater than 93.11 inches?(a) The probability is a(Round to four decimal places as needed.)(b) The probability is(Round to four decimal places as needed.)Enter your answer in each of the

Answer 1

`\begin{gathered} a)0.4129 \\ b)\text{ 0.0764} \end{gathered}`

We proceed as follows;

The formula we have to use is the formula for z-scores as follows;

`\begin{gathered} z\text{ = }\frac{x-\mu}{\frac{\sigma}{\sqrt[]{n}}} \\ \\ \mu\text{ = mean = 93 inches} \\ \sigma\text{ = standard deviation = 0.5 in} \\ n\text{ = number of samples } \end{gathered}`

a) Probability that the board is greater in length than 93.11

Since there is no number of samples, then n will be 1

We have this as;

`z\text{ = }(93.11-93)/(0.5)\text{ = 0.22}`

We will have to use standard normal distribution table for P (z > 0.22) = 0.4129

b) A sample of 42 boards

We have it as;

`z\text{ = }\frac{93.11-93}{\frac{0.5}{\sqrt[]{42}}}\text{ = }(0.11)/(0.077)\text{ = 1.43}`

So, we use the standard normal distribution table to calculate the P (z > 1.43)

From the table, we have the answer as 0.0764