Question

the lengths of lumber a machine cuts are normally distributed with a mean of 90 inches and a standard deviation of 0.4 inch. (a) what is the probability that a randomly selected board cut by the machine has a length greater than 90.13 inches? (b) a sample of 43 boards is randomly selected. what is the probability that their mean length is greater than 90.13 inches? (a) the probability is . (round to four decimal places as needed.)

Explanation:

Step1: Calculate z - score for part (a)

The z - score formula is $z=\frac{x-\mu}{\sigma}$, where $x = 90.13$, $\mu=90$, and $\sigma = 0.4$. So, $z=\frac{90.13 - 90}{0.4}=\frac{0.13}{0.4}=0.325$.

Step2: Find the probability for part (a)

We want $P(X>90.13)$, which is equivalent to $1 - P(X\leq90.13)$. Looking up the z - score of $0.325$ in the standard normal distribution table, $P(Z\leq0.325)\approx0.6277$. So, $P(X > 90.13)=1 - 0.6277 = 0.3723$.

Step3: Calculate z - score for part (b)

The standard deviation of the sample mean (also known as the standard error) is $\sigma_{\bar{x}}=\frac{\sigma}{\sqrt{n}}$, where $\sigma = 0.4$ and $n = 43$. So, $\sigma_{\bar{x}}=\frac{0.4}{\sqrt{43}}\approx\frac{0.4}{6.5574}\approx0.061$. The z - score for the sample mean is $z=\frac{\bar{x}-\mu}{\sigma_{\bar{x}}}=\frac{90.13 - 90}{0.061}=\frac{0.13}{0.061}\approx2.1311$.

Step4: Find the probability for part (b)

We want $P(\bar{X}>90.13)$, which is equivalent to $1 - P(\bar{X}\leq90.13)$. Looking up the z - score of $2.1311$ in the standard normal distribution table, $P(Z\leq2.1311)\approx0.9835$. So, $P(\bar{X}>90.13)=1 - 0.9835=0.0165$.

Answer:

(a) $0.3723$
(b) $0.0165$