Question

the lengths of lumber a machine cuts are normally distributed with a mean of 97 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 97.12 inches? (b) a sample of 43 boards is randomly selected. what is the probability that their mean length is greater than 97.12 inches? (a) the probability is \\( \square \\). (round to four decimal places as needed.)

Explanation:

Step1: Calculate the z - score for part (a)

The formula for the z - score is \(z=\frac{x-\mu}{\sigma}\), where \(x = 97.12\), \(\mu=97\), and \(\sigma = 0.5\).

$$z=\frac{97.12 - 97}{0.5}=\frac{0.12}{0.5}=0.24$$

Step2: Find the probability for part (a)

We want \(P(X>97.12)\), which is equivalent to \(P(Z > 0.24)\).
Since \(P(Z>z)=1 - P(Z\leq z)\), and from the standard normal table \(P(Z\leq0.24)=0.5948\).

$$P(Z>0.24)=1 - 0.5948 = 0.4052$$

Step3: Calculate the standard error for part (b)

The formula for the standard error is \(\sigma_{\bar{x}}=\frac{\sigma}{\sqrt{n}}\), where \(\sigma = 0.5\) and \(n = 43\).

$$\sigma_{\bar{x}}=\frac{0.5}{\sqrt{43}}\approx\frac{0.5}{6.557}\approx0.0763$$

Step4: Calculate the z - score for part (b)

Using the formula \(z=\frac{\bar{x}-\mu}{\sigma_{\bar{x}}}\), with \(\bar{x}=97.12\), \(\mu = 97\), and \(\sigma_{\bar{x}}\approx0.0763\)

$$z=\frac{97.12-97}{0.0763}=\frac{0.12}{0.0763}\approx1.57$$

Step5: Find the probability for part (b)

We want \(P(\bar{X}>97.12)\), which is equivalent to \(P(Z>1.57)\)
Since \(P(Z > z)=1 - P(Z\leq z)\), and from the standard normal table \(P(Z\leq1.57)=0.9418\)

$$P(Z>1.57)=1 - 0.9418=0.0582$$

Answer:

(a) \(0.4052\)
(b) \(0.0582\)