Question

IQ scores in a certain population are normally distributed with a mean of 96 and a standard deviation of 12. (Give your answers correct to four decimal places.)
(a) Find the probability that a randomly selected person will have an IQ score between 83 and 99.
83-96/12=75 the looked at graph and got 0.2266-0.1841=.0425

(b) Find the probability that a randomly selected person will have an IQ score above 93.
93-96/12=85 =0.1977

Answer 1

z-score for 83
= (83-96)/12 = -1.0833

z-score for 99
= (99-96)/12 = .25

so from your tables you should have
.5987 - .1393
= .4594

83-96/12=75 is incorrect, the order of operation is critical, and you need brackets like I used in my above solution
you have to calculate the numerator first

Now try the second part, I am sure you will get it correct this time, since you made the same error in finding the z-score.

Answer 2

So this is how I did the last one and it came up wrong, (93-96)/12=-.25 I looked it up on graph and it came up to 0.4013 but it says it is wrong, what did I do wrong..

Answer 3

The numbers you see on the graph or find in tables always gives you the probability of below the event.

It had asked for the IQ score to be above 93
So your value you have is correct for an IQ of below 93
so for ABOVE 93 it would be 1 - .4013 = .5987