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

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.

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..

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

To find the probability that a randomly selected person will have an IQ score between 83 and 99, you can use the standard normal distribution table or a calculator that provides the cumulative probability function.

(a) Calculate the z-score for the lower bound of 83:
z = (83 - 96) / 12
z = -1.0833

Next, calculate the z-score for the upper bound of 99:
z = (99 - 96) / 12
z = 0.2500

Using the standard normal distribution table or a calculator, find the probability associated with each z-score. For the lower bound of -1.0833, the probability is approximately 0.1841. For the upper bound of 0.2500, the probability is approximately 0.5987.

To find the probability between these two bounds, subtract the probability of the lower bound from the probability of the upper bound:
Probability = 0.5987 - 0.1841
Probability = 0.4146

Therefore, the probability that a randomly selected person will have an IQ score between 83 and 99 is approximately 0.4146.

(b) To find the probability that a randomly selected person will have an IQ score above 93, you need to calculate the area under the curve to the right of the z-score corresponding to an IQ score of 93.

Calculate the z-score for an IQ score of 93:
z = (93 - 96) / 12
z = -0.2500

Using the standard normal distribution table or a calculator, find the probability associated with the z-score of -0.2500, which is approximately 0.4013.

This probability represents the area under the curve to the left of the z-score of -0.2500. To find the probability above 93, subtract this probability from 1:
Probability = 1 - 0.4013
Probability = 0.5987

Therefore, the probability that a randomly selected person will have an IQ score above 93 is approximately 0.5987.