Question

The fracture strength of a certain type of manufactured glass is normally distributed with a mean of 509 MPa with a standard deviation of 17 MPa. (a) What is the probability that a randomly chosen sample of glass will break at less than 509 MPa

Answer 1

0.5 = 50% probability that a randomly chosen sample of glass will break at less than 509 MPa

Explanation:

Normal Probability Distribution

Problems of normal distributions can be solved using the z-score formula.

In a set with mean
`\mu` and standard deviation
`\sigma`, the z-score of a measure X is given by:

`Z = (X - \mu)/(\sigma)`

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the p-value, we get the probability that the value of the measure is greater than X.

Mean of 509 MPa with a standard deviation of 17 MPa.

This means that
`\mu = 509, \sigma = 17`

What is the probability that a randomly chosen sample of glass will break at less than 509 MPa?

This is the p-value of Z when X = 509. So

`Z = (X - \mu)/(\sigma)`

`Z = (509 - 509)/(17)`

`Z = 0`

`Z = 0` has a p-value of 0.5

0.5 = 50% probability that a randomly chosen sample of glass will break at less than 509 MPa