Question

If a specimen is acceptable only if its hardness is between 63 and 75, what is the probability that a randomly chosen specimen has an acceptable hardness? (Round your answer to four decimal places.)

Answer 1

`P(63<X<75)=P((63-\mu)/(\sigma)<(X-\mu)/(\sigma)<(75-\mu)/(\sigma))=P((63-68)/(3)<Z<(75-68)/(3))=P(-1.667<z<2.333)`

And we can find this probability with this difference:

`P(-1.667<z<2.333)=P(z<2.333)-P(z<-1.667)`

And in order to find these probabilities we can using tables for the normal standard distribution, excel or a calculator.

`P(-1.667<z<2.333)=P(z<2.333)-P(z<-1.667)=0.9902-0.0478=0.9424`

Explanation:

Asuming this previous info: The Rockwell hardness of a metal is determined by impressing a hardened point into the surface of the metal and then measuring the depth of penetration of the point. Suppose the Rockwell hardness of a particular alloy is normally distributed with mean 68 and standard deviation 3.

Previous concepts

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".

Solution to the problem

Let X the random variable that represent the hardness of a population, and for this case we know the distribution for X is given by:

`X \sim N(68,3)`

Where
`\mu=68` and
`\sigma=3`

We are interested on this probability

`P(63<X<75)`

And the best way to solve this problem is using the normal standard distribution and the z score given by:

`z=(x-\mu)/(\sigma)`

If we apply this formula to our probability we got this:

`P(63<X<75)=P((63-\mu)/(\sigma)<(X-\mu)/(\sigma)<(75-\mu)/(\sigma))=P((63-68)/(3)<Z<(75-68)/(3))=P(-1.667<z<2.333)`

And we can find this probability with this difference:

`P(-1.667<z<2.333)=P(z<2.333)-P(z<-1.667)`

And in order to find these probabilities we can using tables for the normal standard distribution, excel or a calculator.

`P(-1.667<z<2.333)=P(z<2.333)-P(z<-1.667)=0.9902-0.0478=0.9424`