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

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?

Solution.

Using the Central Limit Theorem we can conclude that if $S_{n} = \sum_{i=1}^{n} X_{i}$, then $\frac{S_{n - \mu n}}{\sigma \sqrt{n}} \sim N(0,1)$, where $X_{i}$ - random variable means the height of the bounce with $E[X_{i}] = 30$ and $\sigma[X_{i}] = 0.75$. This is equivalent to $X \sim N(\mu, \sigma^{2})$, because $S_{n} = \sum_{i=1}^{n} X_{i} \sim N(\mu n, \sigma^{2} n)$. So $P(X > 32) = P(X^{*}\sigma[X] - E[X] > 32)$, where $X^{*} = \frac{X - E[X]}{\sigma[X]} \sim N(0,1)$.

$P(X > 32) = P(X^{*} * 0.75 - 30 > 32) = P(X^{*} > 2.6) = 1 - P(X^{*} < 2.66) = 1 - \Phi(2.66) = 1 - 0.9961 = 0.0039$ where $\Phi(a)$ - function of standard normal distribution.

Answer: 0.0039