Question

The length of time students needed in order to complete a statistics test followed a distribution that was approximately normal. The mean was 74 minutes, the standar deviation was 8 minutes. What proportion of students took more than one hour to complete the test?

Answer 1

`P(X>60)=P((X-\mu)/(\sigma)>(60-\mu)/(\sigma))=P(Z>(60-74)/(8))=P(z>-1.75)`

And we can find this probability on this way:

`P(z>-1.75)=1-P(z<-1.75)`

And using the normal standard distribution table or excel and we got:

`P(z>-1.75)=1-P(z<-1.75)=1-0.04=0.960`

Explanation:

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

`X \sim N(74,8)`

Where
`\mu=74` and
`\sigma=8`

We are interested on this probability

`P(X>60)`

We can solve the problem using the z score formula given by:

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

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

`P(X>60)=P((X-\mu)/(\sigma)>(60-\mu)/(\sigma))=P(Z>(60-74)/(8))=P(z>-1.75)`

And we can find this probability on this way:

`P(z>-1.75)=1-P(z<-1.75)`

And using the normal standard distribution table or excel and we got:

`P(z>-1.75)=1-P(z<-1.75)=1-0.04=0.960`