Question

this is a normal distribution question :
Let us suppose that a particular IQ test gives results that are normally distributed with a mean score of 100 and a standard deviation of 15
(a) If a randomly chosen person has a score on this test that is greater than 125 what is the probability their score is greater than 135 ?
(b) If five people were randomly selected what is the probability that at least three would have a scroe greater than 120 ?

Answer 1

Please tell me the formula that are need to be known , before explaining that particular question

Answer 2

familiar with zscores ?

Answer 3

A z-score is calculated as \[ Z = \frac{X-\mu}{\sigma}\], where \(\mu \) is the mean, and \(\sigma\) is the standard deviation.

So you will need to convert your normal distribution to a standard normal by the formula above.

For a), this is a conditional probability. And you are asked to find: \[{P(X>135|X>125)} \] Recall from conditional probability that: \[ P(A|B)=\frac{P(A\cap B)}{P(B)}\], so apply the same to get
\[P(X>135|X>125)=\frac{P[(X>135) \cap (X>125)]}{P(X>125)} \]

For b) This is a problem mixing up the binomial and normal distribution. How to analyze this:
They select 5 ppl, think of this as 5 trials, and for each trial, there are 2 outcomes:
i) A person gets a score over 120
ii) a person does not get a score over 120

and the probability of success is the probability of getting a score over 120.

So let \(Y\) be the number of people scoring over 120, then you are asked to find \[P(Y\ge3)=\\P(Y=3)+P(Y=4)+P(Y=5)={ 5\choose 3}p^3(1-p)^2+...+{ 5\choose 5}p^5(1-p)^0 \]

The probability of success \(p\) is the probability of getting a score over 120. This is the same as the previous problem where \(X\) is normally distributed. So

\(p = P(X>120)\), and you do this by standardizing \(X\) as in a)