Question

When Brody goes bowling, his scores are normally distributed with a mean of 125 and a standard deviation of 11. What is the probability that the next game Brody bowls, his score will be between 97 and 115, to the nearest thousandth?

Answer 1

Let X ~ N(125, 11^2). Compute z-scores:
z1 = (97 − 125)/11 = −28/11 ≈ −2.545,
z2 = (115 − 125)/11 = −10/11 ≈ −0.909.

P(97 ≤ X ≤ 115) = Φ(−0.909) − Φ(−2.545) ≈ 0.181 − 0.005 ≈ 0.176.

To the nearest thousandth: 0.176.