Question

the lengths of lumber a machine cuts are normally distributed with a mean of 105 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 105.25 inches? (b) a sample of 39 boards is randomly selected. what is the probability that their mean length is greater than 105.25 inches? (a) the probability is. (round to four decimal places as needed.) (b) the probability is. (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 \(x = 105.25\), \(\mu=105\), and \(\sigma = 0.7\).

$$z=\frac{105.25 - 105}{0.7}=\frac{0.25}{0.7}\approx0.36$$

Step2: Find the probability for part (a)

We want \(P(X>105.25)\), which is equivalent to \(P(Z > 0.36)\). Using the property \(P(Z>z)=1 - P(Z\leq z)\). From the standard normal table, \(P(Z\leq0.36)=0.6406\). So \(P(Z > 0.36)=1 - 0.6406 = 0.3594\)

Step3: Calculate the z - score for part (b)

The formula for the z - score for a sample mean is \(z=\frac{\bar{x}-\mu}{\frac{\sigma}{\sqrt{n}}}\), where \(\bar{x}=105.25\), \(\mu = 105\), \(\sigma=0.7\), and \(n = 39\)

$$z=\frac{105.25-105}{\frac{0.7}{\sqrt{39}}}=\frac{0.25}{\frac{0.7}{6.245}}\approx\frac{0.25}{0.112}=2.23$$

Step4: Find the probability for part (b)

We want \(P(\bar{X}>105.25)\), which is equivalent to \(P(Z>2.23)\). Using the property \(P(Z > z)=1 - P(Z\leq z)\). From the standard normal table, \(P(Z\leq2.23)=0.9871\). So \(P(Z>2.23)=1 - 0.9871=0.0129\)

Answer:

(a) \(0.3594\)
(b) \(0.0129\)