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 Sampling Distribution

For a sample of size \( n = 18 \) from a normal population, the sampling distribution of the sample mean \( M \) is normal with mean \( \mu_M=\mu = 261.4 \) and standard deviation \( \sigma_M=\frac{\sigma}{\sqrt{n}}=\frac{0.7}{\sqrt{18}} \).

Step2: Calculate Z - Score

The z - score for \( M = 261.7 \) is \( z=\frac{M-\mu_M}{\sigma_M}=\frac{261.7 - 261.4}{\frac{0.7}{\sqrt{18}}} \).
First, calculate the numerator: \( 261.7-261.4 = 0.3 \).
Then, calculate the denominator: \( \frac{0.7}{\sqrt{18}}\approx\frac{0.7}{4.2426}\approx0.165 \).
So, \( z=\frac{0.3}{0.165}\approx1.818 \).

Step3: Find Probability from Z - Table

We want to find \( P(M < 261.7)=P(Z < 1.818) \). Using a standard normal table or calculator, \( P(Z < 1.818)\approx0.9656 \).

Answer:

\( 0.9656 \)