Question

A company produces steel rods. The lengths of the steel rods are normally distributed with a mean of 208.6-cm and a standard deviation of 0.6-cm. For shipment, 12 steel rods are bundled together. Find the probability that the mean length of a randomly selected bundle of steel rods is greater than 208.2-cm.

Answer 1

Answer: 0.9896

Explanation:

Given : A company produces steel rods. The lengths of the steel rods are normally distributed with

Mean :
`\mu=208.6\text{ -cm}`

Standard deviation :
`\sigma=0.6\text{ -cm}`

Sample size : = 12

Let x be the random variable that represents the lengths of the steel rods .

To find probability , first we find z-score

Z-score :
`z=(x-\mu)/((\sigma)/(√(n)))`

For x= 208.2-cm.

`z=(208.2-208.6)/((0.6)/(√(12)))\approx-2.31`

By using standard normal distribution table , the probability that the mean length of a randomly selected bundle of steel rods is greater than 208.2-cm :-

`P(X>208.2)=P(z>-2.31)=1-P(z<2.01)\\\\=1-0.0104441=0.9895559\appprox0.9896`

Hence, the required probability is 0.9896.