Question

A company produces steel rods. The lengths of the steel rods are normally distributed with a mean of 259.2-cm and a standard deviation of 2.1-cm. For shipment, 17 steel rods are bundled together. Find the probability that the average length of rods in a randomly selected bundle of steel rods is greater than 259-cm.

Answer 1

The probability that the average length of rods in a randomly selected bundle of steel rods is greater than 259 cm is 0.65173.

Explanation:

We are given that a company produces steel rods. The lengths of the steel rods are normally distributed with a mean of 259.2 cm and a standard deviation of 2.1 cm. For shipment, 17 steel rods are bundled together.

Let
`\bar X` = the average length of rods in a randomly selected bundle of steel rods

The z-score probability distribution for the sample mean is given by;

Z =
`(\bar X-\mu)/((\sigma)/(√(n) ) )` ~ N(0,1)

where,
`\mu` = population mean length of rods = 259.2 cm

`\sigma` = standard deviaton = 2.1 cm

n = sample of steel rods = 17

Now, the probability that the average length of rods in a randomly selected bundle of steel rods is greater than 259 cm is given by = P(
`\bar X` > 259 cm)

P(
`\bar X` > 259 cm) = P(
`(\bar X-\mu)/((\sigma)/(√(n) ) )` >
`(259-259.2)/((2.1)/(√(17) ) )` ) = P(Z > -0.39) = P(Z < 0.39)

= 0.65173

The above probability is calculated by looking at the value of x = 0.39 in the z table which has an area of 0.65173.