The scores of 12-grade students on the National Assessment of Education Progress year 2000 mathematics test have a distribution that is approximately normal with mean 300 and standard deviation of 35.
a) Choose one twelfth-grader at random. What is the probability that his or her score is higher than 300? Higher than 335?
b) Now choose an SRS of four twelfth-graders. What is the probability that his or her mean score is higher than 300? Higher than 335?
My work:
a) P(X>300) = 0.5 and P(X>335) = 0.1587
b) For this step, I think since we are using an SRS with 4 students, we have to divide the standard deviation by the square root of 4: 35/sqt4 = 35/2 = 17.5 Than we use the normalcdf for 300 and 335. For above 300 = 1-0.5 = 0.5 For above 335 = 1-0.97725 = 0.02275
I just learned this. Am I correct?
Yes, you are correct. For a sample of size $n$ drawn from a normal distribution with mean $\mu$ and standard deviation $\sigma$, the sampling distribution of the sample mean is also normal with mean $\mu_{\bar x} = \mu$ and standard deviation $\sigma_{\bar x} = \sigma/\sqrt{n}$.