Question

a company produces steel rods. the lengths of the steel rods are normally distributed with a mean of 261.4-cm and a standard deviation of 0.7-cm. for shipment, 18 steel rods are bundled together.
find the probability that the average length of a randomly selected bundle of steel rods is less than 261.7-cm.
p(m < 261.7-cm) =
enter your answer as a number accurate to 4 decimal places. answers obtained using exact z-scores or z-scores rounded to 3 decimal places are accepted.

Explanation:

Step1: Identify the distribution of the sample mean

The population is normally distributed with mean $\mu = 261.4$ cm and standard deviation $\sigma = 0.7$ cm. The sample size $n = 18$. The sampling distribution of the sample mean $\bar{X}$ (here denoted as $M$) is also normal with mean $\mu_{\bar{X}}=\mu = 261.4$ and standard deviation $\sigma_{\bar{X}}=\frac{\sigma}{\sqrt{n}}=\frac{0.7}{\sqrt{18}}$.

Step2: Calculate the z - score

The z - score is calculated using the formula $z=\frac{\bar{x}-\mu_{\bar{X}}}{\sigma_{\bar{X}}}$. We want to find $P(M < 261.7)$, so $\bar{x}=261.7$.

First, calculate $\sigma_{\bar{X}}=\frac{0.7}{\sqrt{18}}\approx\frac{0.7}{4.2426}\approx0.165$.

Then, calculate the z - score: $z=\frac{261.7 - 261.4}{0.165}=\frac{0.3}{0.165}\approx1.818$ (rounded to 3 decimal places).

Step3: Find the probability using the z - table

We want to find $P(Z < 1.818)$, where $Z$ is a standard normal random variable. Looking up the z - score of $1.818$ in the standard normal table (or using a calculator with normal distribution functions), we know that the cumulative probability for $z = 1.818$ is approximately $0.9656$.

Answer:

$0.9656$