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

Answer:

$P(X>1600)=P(\frac{X-\mu}{\sigma}>\frac{1600-\mu}{\sigma})=P(Z>\frac{1600-1518}{325})=P(z>0.252)$

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

$P(z>0.252)=1-P(z$

$z =\frac{1600-1518}{\frac{325}{\sqrt{81}}}= 2.27$

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

$P(z>2.27)=1-P(z$

Step-by-step 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:

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

Using this formula we got:

$P(X>1600)=P(\frac{X-\mu}{\sigma}>\frac{1600-\mu}{\sigma})=P(Z>\frac{1600-1518}{325})=P(z>0.252)$

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

$P(z>0.252)=1-P(z$

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:

$\bar X \sim N(\mu, \frac{\sigma}{\sqrt{n}})$

And we can use the z score formula given by:

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

And replacing we got:

$z =\frac{1600-1518}{\frac{325}{\sqrt{81}}}= 2.27$

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

$P(z>2.27)=1-P(z$