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

Answer:

$P(33<X<35)=P(\frac{33-\mu}{\sigma}<\frac{X-\mu}{\sigma}<\frac{35-\mu}{\sigma})=P(\frac{33-41}{3}<Z<\frac{35-41}{3})=P(-2.67<z<-2)$

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 %

Step-by-step 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:

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

And using this formula we got:

$P(33<X<35)=P(\frac{33-\mu}{\sigma}<\frac{X-\mu}{\sigma}<\frac{35-\mu}{\sigma})=P(\frac{33-41}{3}<Z<\frac{35-41}{3})=P(-2.67<z<-2)$

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 %