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 selected, find the probability that he weighs between 170 and 220 pounds.

Answer :

Answer:

[tex]P(170<X<220)=P(\frac{170-\mu}{\sigma}<\frac{X-\mu}{\sigma}<\frac{220-\mu}{\sigma})=P(\frac{170-200}{50}<Z<\frac{220-200}{50})=P(-0.6<z<0.4)[/tex]

And we can find this probability with this difference:

[tex]P(-0.6<z<0.4)=P(z<0.4)-P(z<-0.6)=0.655-0.274= 0.381 [/tex]

Step-by-step 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:

[tex]X \sim N(200,50)[/tex]  

Where [tex]\mu=200[/tex] and [tex]\sigma=50[/tex]

We want to find the following probability:

[tex]P(170<X<220)[/tex]

And we can use the z score formula given by:

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

And using this formula we got:

[tex]P(170<X<220)=P(\frac{170-\mu}{\sigma}<\frac{X-\mu}{\sigma}<\frac{220-\mu}{\sigma})=P(\frac{170-200}{50}<Z<\frac{220-200}{50})=P(-0.6<z<0.4)[/tex]

And we can find this probability with this difference:

[tex]P(-0.6<z<0.4)=P(z<0.4)-P(z<-0.6)=0.655-0.274= 0.381 [/tex]

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