Question

A company produces steel rods. The lengths of the steel rods are normally distributed with a mean of 176.7-cm and a standard deviation of 1.6-cm. For shipment, 5 steel rods are bundled together.Find the probability that the average length of a randomly selected bundle of steel rods is greater than 178.8-cm.P(M > 178.8-cm) =

Answer 1

Let m be a random variable representing the lengths of the steel rods. Since the lengths are normally distributed, we would apply the formula for determining z score which is expressed as

z = (sample mean - population mean)/(population standard deviation/square root of number of samples)

From the information given,

population mean = 176.7

population standard deviation = 1.6

number of samples = 5

sample mean = 178.8

Thus, we have

`\begin{gathered} z\text{ = (178.8 - 176.7)/(1.6/}\sqrt[]{5})\text{ } \\ z\text{ = }2.9348 \end{gathered}`

We want to find P(M > 178.8-cm). This is same as 1 - P(M ≤ 178.8-cm). To find P(M ≤ 178.8-cm), we would find the probability value corresponding to a z score of 2.9348 from the normal distribution table. It is 0.9983

Thus,

P(M > 178.8-cm) = 1 - 0.9983 = 0.0017