Question

The IQ score of people is a normal random variable X with a mean of 100 and a standard deviation of 10.

Randomly choose one. Find P(85 ≤ X ≤ 90)?

Let X be the sample mean of a random sample of 100. Find P(X > 101).

Find the 90th percentile of X. (That is, find a such that P(X < a) = 0.9)

Randomly choose 10 people’ IQ scores. What is the probability that exactly 3 of the IQ scores will exceed 110?

Answer 1

Given, X follows N(mean=100,standard deviation=10) ie.,

i) To find P(85 ≤ X ≤ 90)

Using Central Limit Theorem, which states that,

Here, P(85 ≤ X ≤ 90)= P(X ≤ 90) - P(X ≤ 85)

Hence, P85 ≤ X ≤ 90) = 0.091853

ii) To find ,  

Here =100 , =10 and given n=100

We know that,

Hence, is calculated as,

  

Therefore,

iii)To find a such that P(X < a) = 0.9 (90th  Percentile)

Here, X: is a normal random variable with a mean of 100 and a standard deviation of 10

From standard normal probability table we find the value of a corresponding to 0.9

Hence, 90th percentile is 112.9

iv) Here, we are choosing 10 students randomly

Hence mean of the sample remains unchanged which is =100

For n=3, we have standard deviation  

Hence, P(X>110) is calculated as,

  

  

  

  

Hence, the required probability is = 0.42393