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

Answer:

$P(X>60)=P(\frac{X-\mu}{\sigma}>\frac{60-\mu}{\sigma})=P(Z>\frac{60-74}{8})=P(z>-1.75)$

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:

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

Step-by-step 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:

$z=\frac{x-\mu}{\sigma}$

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

$P(X>60)=P(\frac{X-\mu}{\sigma}>\frac{60-\mu}{\sigma})=P(Z>\frac{60-74}{8})=P(z>-1.75)$

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:

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