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

sure i'm up for it

step-by-step explanation: