Question

Assume that SAT scores are normally distributed with mean mu equals 1518 and standard deviation sigma equals 325. If 1 SAT score is randomly selected, find the probability that it is greater than 1600. If 81 SAT scores are randomly selected, find the probability that they have a mean greater than 1600.

Answer 1

`P(X>1600)=P((X-\mu)/(\sigma)>(1600-\mu)/(\sigma))=P(Z>(1600-1518)/(325))=P(z>0.252)`

And we can find this probability using the z score formula and the complement rule and we got:

`P(z>0.252)=1-P(z<0.252) =1-0.599= 0.401`

`z =(1600-1518)/((325)/(√(81)))= 2.27`

And we can find this probability using the z score formula and the complement rule and we got:

`P(z>2.27)=1-P(z<2.27) =1-0.988=0.012`

Explanation:

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

`X \sim N(1518,325)`

Where
`\mu=1518` and
`\sigma=325`

We want to find this probability:

`P(X>1600)`

And we can use the z score formula given by:

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

Using this formula we got:

`P(X>1600)=P((X-\mu)/(\sigma)>(1600-\mu)/(\sigma))=P(Z>(1600-1518)/(325))=P(z>0.252)`

And we can find this probability using the z score formula and the complement rule and we got:

`P(z>0.252)=1-P(z<0.252) =1-0.599= 0.401`

For the other part we need to take in count that the distribution for the sampel mean if the sample size is large (n>30) is given by:

`\bar X \sim N(\mu, (\sigma)/(√(n)))`

And we can use the z score formula given by:

`z=(x-\mu)/((sigma)/(√(n)))`

And replacing we got:

`z =(1600-1518)/((325)/(√(81)))= 2.27`

And we can find this probability using the z score formula and the complement rule and we got:

`P(z>2.27)=1-P(z<2.27) =1-0.988=0.012`