Question

In a production process, the diameter measures of manufactured o-ring gaskets are known to be normally distributed with a mean diameter of 80 mm and a standard deviation of 3 mm. Any o-ring measuring 75 mm or less in diameter is defective and cannot be used. Using Excel, determine the percent or proportion of defective o-rings that will be produced.

Answer 1

DIRECT WAY EXCEL

"=NORM.DIST(75,80,3,TRUE)"

And we got:
`P(X\leq 75)= 0.04779`

OTHER WAY

`P(X\leq 75)=P((X-\mu)/(\sigma)\leq (75-\mu)/(\sigma))=P(Z\leq (75-80)/(3))=P(Z<-1.67)`

And we can find this probability using the normal standard table or excel:

`P(Z<-1.67)=0.04779`

Explanation:

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 diameter of a population, and for this case we know the distribution for X is given by:

`X \sim N(80,3)`

Where
`\mu=80` and
`\sigma=3`

And we know that if the diameter is 75 or less the ring would be considered defective , so then in order to find the proportion of defective we need to find the following probability:

`P(X\leq 75)`

One way to do this in excel is with the following formula:

"=NORM.DIST(75,80,3,TRUE)"

And we got:
`P(X\leq 75)= 0.04779`

And the other way is use the z score formula given by:

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

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

`P(X\leq 75)=P((X-\mu)/(\sigma)\leq (75-\mu)/(\sigma))=P(Z\leq (75-80)/(3))=P(Z<-1.67)`

And we can find this probability using the normal standard table or excel:

`P(Z<-1.67)=0.04779`