Question

IQ scores are normally distributed with a mean of 100 and a standard deviation of 15. What is the probability that a randomly chosen person's IQ score will be between 62 and 106, to the nearest thousandth?

Answer 1

Let X ~ N(100,15^2). Compute z-scores:

z1 = (62−100)/15 = −38/15 ≈ −2.5333
z2 = (106−100)/15 = 6/15 = 0.4

P(62 ≤ X ≤ 106) = Φ(0.4) − Φ(−2.5333)
= Φ(0.4) − (1 − Φ(2.5333))
≈ 0.6554 − 0.0056 ≈ 0.6498

To the nearest thousandth: 0.650.