the probability of this?

Question

What is the probability that a randomly chosen person has an IQ over 120?

(IQ scores are normally distributed with a mean of 100 and standard deviation of 15).

Answer 1

Hello Daniel K,

The formula for Z Score is

Z score = ( x – µ ) / σ

Where x is the raw score, µ is the mean and σ is the standard deviation.

... over 120? (IQ scores are normally distributed with a mean of 100 and standard deviation of 15)

So we already have the mean, standard deviation and the raw score

Substituting the same in the above Z Score formula, we get

Z Score = (120 - 100)/15

= 1.33

Next, use a Z-Table https://www.ztable.net/ and map the values for P(z>1.33)

Hope this helps,

Samantha

Answer 2

compute z = (120 - 100) / 15 = 1.33

Find P(z > 1.33) from standard normal probability table.

Answer 3

The probability that a randomly chosen person has an IQ over 120 is written as P(X>120)

P(X>120) = 1 - P(X≤120) = 1 - P((X-μ)/σ ≤ (120-100)/15) Note: Z = (X-μ)/σ

= 1 - P(Z≤1.33)

From the normal distribution calculator(https://stattrek.com/online-calculator/normal.aspx) or table 1.33 corresponds to a probability of 0.908

= 1 - 0.908

= 0.092