Molly earned a score of 940 on a national achievement test. the mean test score was 850 with a standard deviation of 100. what proportion of students had a higher score than molly? (assume that test scores are normally distributed.)

Solution: We are given:

μ=850,σ=100

We have to find P (x>940)

We need to find the z score:

z = (x-μ) / σ

= (940-850) / 100

=0.9

Now we need to find P (z>0.9)

Using the standard normal table, we have:

P (z>0.9) = 0.1841

Therefore 0.1841 or (18.41%) of students had a higher score than molly