Question

When Jeremiah goes bowling, his scores are normally distributed with a mean of 105 and a standard deviation of 10. What is the probability that the next game Jeremiah bowls, his score will be between 93 and 103, to the nearest thousandth?

Answer 1

Let X ~ N(105, 10^2). Convert to z:

z1 = (93−105)/10 = −1.2, z2 = (103−105)/10 = −0.2.

P(93 ≤ X ≤ 103) = Φ(−0.2) − Φ(−1.2) = Φ(1.2) − Φ(0.2)
≈ 0.884930 − 0.579260 = 0.305671 ≈ 0.306 (to the nearest thousandth).