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)

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) [tex]P(X<2)=P(\frac{X-\mu}{\sigma}<\frac{2-\mu}{\sigma})=P(Z<\frac{2-3.8}{0.8})=P(z<-2.25)[/tex]

And using the normal standard table or excel we got:

[tex]P(z<-2.25)=0.0122[/tex]

iv) [tex]P(2<X<4)=P(\frac{2-\mu}{\sigma}<\frac{X-\mu}{\sigma}<\frac{4-\mu}{\sigma})=P(\frac{2-3.8}{0.8}<Z<\frac{4-3.8}{0.8})=P(-2.25<z<0.25)[/tex]

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

[tex]P(-2.25<z<0.25)=P(z<0.25)-P(z<-2.25)= 0.5987-0.0122= 0.5865[/tex]

Step-by-step 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:

[tex]X \sim N(3.8,0.8)[/tex]

Where [tex]\mu=3.8[/tex] and [tex]\sigma=0.8[/tex]

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:

[tex]P(X<2)[/tex]

We can use the z score formula given by:

[tex]z=\frac{x-\mu}{\sigma}[/tex]

Using this formula we have:

[tex]P(X<2)=P(\frac{X-\mu}{\sigma}<\frac{2-\mu}{\sigma})=P(Z<\frac{2-3.8}{0.8})=P(z<-2.25)[/tex]

And using the normal standard table or excel we got:

[tex]P(z<-2.25)=0.0122[/tex]

Part iv

We want this probability:

[tex]P(2<X<4)[/tex]

Using the z score formula we got:

[tex]P(2<X<4)=P(\frac{2-\mu}{\sigma}<\frac{X-\mu}{\sigma}<\frac{4-\mu}{\sigma})=P(\frac{2-3.8}{0.8}<Z<\frac{4-3.8}{0.8})=P(-2.25<z<0.25)[/tex]

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

[tex]P(-2.25<z<0.25)=P(z<0.25)-P(z<-2.25)= 0.5987-0.0122= 0.5865[/tex]