The weights of college football players are normally distributed with a mean of 200 pounds and a standard deviation of 50 pounds. If a college football player is randomly select…

Question

asked Nov 16, 2021 74.1k views

1 Answer

Veldmuis · Answer 1 · 2021-11-23T08:43:33+0000

Answer:

$P(170<X<220)=P((170-\mu)/(\sigma)<(X-\mu)/(\sigma)<(220-\mu)/(\sigma))=P((170-200)/(50)<Z<(220-200)/(50))=P(-0.6<z<0.4)$

And we can find this probability with this difference:

P(-0.6<z<0.4)=P(z<0.4)-P(z<-0.6)=0.655-0.274= 0.381

Explanation:

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

$X \sim N(200,50)$

Where
$\mu=200$ and
$\sigma=50$

We want to find the following probability:

P(170<X<220)

And we can use the z score formula given by:

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

And using this formula we got:

$P(170<X<220)=P((170-\mu)/(\sigma)<(X-\mu)/(\sigma)<(220-\mu)/(\sigma))=P((170-200)/(50)<Z<(220-200)/(50))=P(-0.6<z<0.4)$

And we can find this probability with this difference:

P(-0.6<z<0.4)=P(z<0.4)-P(z<-0.6)=0.655-0.274= 0.381