Question

for a normal population with a nean of u =80 and a standard devuatiin of 10, what is the probabilitu of obtaining a sample mean greater than M= 75 for a sample of n= 25 scores?

Answer 1

Let us consider the random variable $X\sim N(\mu=80,\sigma=10)$ , then its sample mean $\bar{X}\sim N(\mu=80,\sigma/\sqrt{n}=10/\sqrt{25}=2)$ , where n=25

We know that the standard normal variate Z~N(0,1)

$P(\bar{X}>75)=P(\frac{\bar{X}-\mu }{\sigma /\sqrt{n}}>\frac{75-80}{2})=P(Z>-2.5)=1-P(Z\leq -2.5)$

$1-P(Z\leq -2.5)=1-(1-P(Z<2.5))=P(Z<2.5)=0.9938$.

Therefore, the probability of obtaining a sample mean greater than M=75 is 0.9938.