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

Answer:

$P(X<29.1)=P(\frac{X-\mu}{\sigma}<\frac{29.1-\mu}{\sigma})=P(Z<\frac{29.1-29}{0.1})=P(z<1)$

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%

Step-by-step 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:

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

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

$P(X<29.1)=P(\frac{X-\mu}{\sigma}<\frac{29.1-\mu}{\sigma})=P(Z<\frac{29.1-29}{0.1})=P(z<1)$

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%