Question

When Colton commutes to work, the amount of time it takes him to arrive is normally distributed with a mean of 41 minutes and a standard deviation of 3 minutes. What percentage of his commutes will be between 33 and 35 minutes, to the nearest tenth?

Answer 1

`P(33<X<35)=P((33-\mu)/(\sigma)<(X-\mu)/(\sigma)<(35-\mu)/(\sigma))=P((33-41)/(3)<Z<(35-41)/(3))=P(-2.67<z<-2)`

And we can find the probability of interest with this difference

`P(-2.67<z<-2)=P(z<-2)-P(z<-2.67)`

And if we use the normal standard table or excel we got:

`P(-2.67<z<-2)=P(z<-2)-P(z<-2.67)=0.02275-0.00379=0.01896`

And if we convert the probability to a % we got 1.896% and rounded to the nearest tenth we got 1.9 %

Explanation:

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

`X \sim N(41,3)`

Where
`\mu=41` and
`\sigma=3`

We are interested on this probability

`P(33<X<35)`

And we can solve the problem using the z score formula given by:

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

And using this formula we got:

`P(33<X<35)=P((33-\mu)/(\sigma)<(X-\mu)/(\sigma)<(35-\mu)/(\sigma))=P((33-41)/(3)<Z<(35-41)/(3))=P(-2.67<z<-2)`

And we can find the probability of interest with this difference

`P(-2.67<z<-2)=P(z<-2)-P(z<-2.67)`

And if we use the normal standard table or excel we got:

`P(-2.67<z<-2)=P(z<-2)-P(z<-2.67)=0.02275-0.00379=0.01896`

And if we convert the probability to a % we got 1.896% and rounded to the nearest tenth we got 1.9 %