Question

A final exam in Sociology has a mean of 72 and a standard deviation of 9.2. If 35 students are randomly selected, find the probability that that the mean of their test scores will be greater than 76. (Round to tenth of a percent)

Answer 1

Answer: 0.5%

Step-by-step explanation:

We assume that the test scores in final exam is normally distributed.

Given : Mean :$\mu= 72$

Standard deviation : $\sigma=9.2$

Sample size : n= 35

Let x be the random variable that represents the test scores  of the students.

The formula for the z-score :-

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

For x=76

$z=\dfrac{76-72}{\dfrac{9.2}{\sqrt{35}}}\approx2.57$

By using the standard normal distribution table ,

The p-value =$P(x>76)=P(z>2.57)=1-P(z\leq2.57)$

$=1-0.994915=0.005085=0.5085\%\approx0.5\%$

Hence, the probability that that the mean of their test scores will be greater than 76 =0.5%