Question

A hundred squash balls are tested by dropping from a height of 100 inches and measuring the height of the bounce. A ball is “fast” if it rises above 32 inches. The average height of bounce was 30 inches and the standard deviation was ¾ inches. What is the chance of getting a “fast” standard ball?

Answer 1

A hundred squash balls are tested by dropping from a height of 100 inches and measuring the height of the bounce. A ball is "fast" if it rises above 32 inches. The average height of bounce was 30 inches and the standard deviation was $\frac{3}{4}$ inches. What is the chance of getting a "fast" standard ball?

We have normal distribution with mean 30 inches and standard deviation $-\frac{3}{4}$.

$$m = 30$$ $$\sigma = \frac{3}{4}$$

And we need to know:

$$P(x > 32)$$ $$P(x > 32) = \int_{32}^{\infty} f(x) dx$$

$f(x)$ - probability density function

The normal distribution has probability density:

$$f(x) = \frac{1}{\sqrt{2\pi}\sigma} e^{-\frac{(x - m)^2}{2\sigma^2}}$$

Therefore:

$$P(x > 32) = \int_{32}^{\infty} \frac{1}{\sqrt{2\pi}\sigma} e^{-\frac{(x - m)^2}{2\sigma^2}} dx$$

Calculating this integral:

$$P(x > 32) = 0.00383 = 0.383\%$$

Answer: $0.383\%$