Question

The circumferences of WNBA basketballs follow a normal distribution where mean = 29.0 ; and standard deviation = 0.1 inches.

What proportion of WNBA basketballs are SMALLER than 29.1 inches? Round to the nearest percent

Answer 1

`P(X<29.1)=P((X-\mu)/(\sigma)<(29.1-\mu)/(\sigma))=P(Z<(29.1-29)/(0.1))=P(z<1)`

And we can find this probability using the normal standard distribution, excel or a calculator and we got:

`P(z<1) =0.8413`

And if we convert this into percent we got 84.13% rounded to the nearest percent would be:

84%

Explanation:

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

`X \sim N(29,0.1)`

Where
`\mu=29` and
`\sigma=0.1`

We are interested on this probability

`P(X<29.1)`

And the best way to solve this problem is using the normal standard distribution and the z score given by:

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

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

`P(X<29.1)=P((X-\mu)/(\sigma)<(29.1-\mu)/(\sigma))=P(Z<(29.1-29)/(0.1))=P(z<1)`

And we can find this probability using the normal standard distribution, excel or a calculator and we got:

`P(z<1) =0.8413`

And if we convert this into percent we got 84.13% rounded to the nearest percent would be:

84%