# Thread: Help in Normal Distribution!

1. ## Help in Normal Distribution!

A group of 9 people take an IQ test.
1. What is the probability that their average is greater than 93?
2. What is the probability that their average is less than 97?
3. What is the probability that their average is between 90 and 107?
4. What is the probability that their average is 100?

Thanks a lot!

2. Originally Posted by Vedicmaths

A group of 9 people take an IQ test.
1. What is the probability that their average is greater than 93?
2. What is the probability that their average is less than 97?
3. What is the probability that their average is between 90 and 107?
4. What is the probability that their average is 100?

Thanks a lot!
You're sampling from a normal population assumed to have mean 100 and standard deviation of either 16 (Stanford-Binet) or 15 (Wechsler Adult Intelligence Scale). You should have been given the appropriate information in the question - I don't know which one you're expected to use. I'll use 15.

Therefore the sample mean (average) also follows a normal distribution, with mean 100 and standard deviation 15/sqrt{9} = 5.

In other words:

Average ~ Normal(Mean = 100, sd = 5).

You need to find:

Pr(average > 93) = .....

Pr(average < 97) = .....

Pr(90 < average < 107) = .....

Pr(average = 100) = 0 (since the average is a continuous random variable).

And you've no doubt been taught how to find the probabilities concerning a normal random variable ......

3. I am sorry for an incomplete problem.
But this is really amazing, you actually used the correct mean = 100 and standard deviation = 5.
And thank you very much for your help. I solved the problem. I was getting confused on how to get the standard deviation value here and the last problem of 100.

Quick question though: so like in the last problem we are asked to find for 100 and since you said this is a a continuous random variable, that will be zero.
So is it always going to be zero, whenever we are asked to find the 100% of the total? What if we are asked to find for 50, will that be the same procedure we used above?

Thanks again for your help Sir!

4. Originally Posted by Vedicmaths
I am sorry for an incomplete problem.
But this is really amazing, you actually used the correct mean = 100 and standard deviation = 5.
And thank you very much for your help. I solved the problem. I was getting confused on how to get the standard deviation value here and the last problem of 100.

Quick question though: so like in the last problem we are asked to find for 100 and since you said this is a a continuous random variable, that will be zero.
So is it always going to be zero, whenever we are asked to find the 100% of the total? What if we are asked to find for 50, will that be the same procedure we used above?

Thanks again for your help Sir!
When X is a continuous random variable, Pr(X = a) = 0 for all values of a. It's to do with probability being the area under the pdf.