Question

When Emily goes bowling, her scores are normally distributed with a mean of 100 and a standard deviation of 10. What percentage of the games that Emily bowls does she score between 86 and 125, to the nearest tenth?

Answer 1

Let X ~ N(100, 10^2). Compute z-scores:
z1 = (86−100)/10 = −1.4, z2 = (125−100)/10 = 2.5.

P(86 ≤ X ≤ 125) = Φ(2.5) − Φ(−1.4) ≈ 0.9938 − 0.0808 = 0.9130.

So about 91.3% of her games (to the nearest tenth).