Question

The amount of time people spend exercising in a given week follows a normal distribution with a mean of 3.8 hours per week and a standard deviation of 0.8 hours per week.

i) Which of the following shows the shaded probability that a person picked at random exercises less than 2 hours per week?
ii) What is the probability that a person picked at random exercises less than 2 hours per week? (round to 4 decimal places)
iii) Which of the following shows the shaded probability that a person picked at random exercises between 2 and 4 hours per week?
iv) What is the probability that a person picked at random exercises between 2 and 4 hours per week? (round to 4 decimal places)

Answer 1

i and iii) In the figure attached part a we have the illustration for the area required for the probability of less than 2 hours and in b the illustration for the probability that X would be between 2 and 4

ii)
`P(X<2)=P((X-\mu)/(\sigma)<(2-\mu)/(\sigma))=P(Z<(2-3.8)/(0.8))=P(z<-2.25)`

And using the normal standard table or excel we got:

`P(z<-2.25)=0.0122`

iv)
`P(2<X<4)=P((2-\mu)/(\sigma)<(X-\mu)/(\sigma)<(4-\mu)/(\sigma))=P((2-3.8)/(0.8)<Z<(4-3.8)/(0.8))=P(-2.25<z<0.25)`

And we can find the probability with the following difference and usint the normal standard distirbution or excel and we got:

`P(-2.25<z<0.25)=P(z<0.25)-P(z<-2.25)= 0.5987-0.0122= 0.5865`

Explanation:

Let X the random variable that represent amount of time people spend exercising in a given week, and for this case we know the distribution for X is given by:

`X \sim N(3.8,0.8)`

Where
`\mu=3.8` and
`\sigma=0.8`

Part i and iii

In the figure attached part a we have the illustration for the area required for the probability of less than 2 hours and in b the illustration for the probability that X would be between 2 and 4

Part ii

We are interested on this probability:

`P(X<2)`

We can use the z score formula given by:

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

Using this formula we have:

`P(X<2)=P((X-\mu)/(\sigma)<(2-\mu)/(\sigma))=P(Z<(2-3.8)/(0.8))=P(z<-2.25)`

And using the normal standard table or excel we got:

`P(z<-2.25)=0.0122`

Part iv

We want this probability:

`P(2<X<4)`

Using the z score formula we got:

`P(2<X<4)=P((2-\mu)/(\sigma)<(X-\mu)/(\sigma)<(4-\mu)/(\sigma))=P((2-3.8)/(0.8)<Z<(4-3.8)/(0.8))=P(-2.25<z<0.25)`

And we can find the probability with the following difference and usint the normal standard distirbution or excel and we got:

`P(-2.25<z<0.25)=P(z<0.25)-P(z<-2.25)= 0.5987-0.0122= 0.5865`