Question

Tennis Replay In the year that this exercise was written, there were 879 challenges made to referee calls in professional tennis singles play. Among those challenges, 231 challenges were upheld with the call overturned. Assume that in general, 25% of the challenges are successfully upheld with the call overturned. a. If the 25% rate is correct, find the probability that among the 879 challenges, the number of overturned calls is exactly 231. b. If the 25% rate is correct, find the probability that among the 879 challenges, the number of overturned calls is 231 or more. If the 25% rate is correct, is 231 overturned calls among 879 challenges a result that is significantly high

Answer 1

Answer:

a. 0.0209 = 2.09% probability that among the 879 challenges, the number of overturned calls is exactly 231.

b. 231 is less than 2.5 standard deviations above the mean, which means that 231 overturned calls among 879 challenges is not a significantly high result.

Step-by-step explanation:

For each challenge, there are only two possible outcomes. Either it was overturned, or it was not. The probability of a challenge being overturned is independent of any other challenge. This means that the binomial probability distribution is used to solve this question.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

In which $C_{n,x}$ is the number of different combinations of x objects from a set of n elements, given by the following formula.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

And p is the probability of X happening.

Significantly high:

The expected value of the binomial distribution is:

$E(X) = np$

The standard deviation of the binomial distribution is:

$\sqrt{V(X)} = \sqrt{np(1-p)}$

If a value is more than 2.5 standard deviations above the mean, this value is considered significantly high.

25% of the challenges are successfully upheld with the call overturned.

This means that $p = 0.25$

879 challenges

This meas that $n = 879$

a. If the 25% rate is correct, find the probability that among the 879 challenges, the number of overturned calls is exactly 231.

This is P(X = 231). So

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 231) = C_{879,231}.(0.25)^{231}.(0.75)^{648} = 0.0209$

0.0209 = 2.09% probability that among the 879 challenges, the number of overturned calls is exactly 231.

b. If the 25% rate is correct, find the probability that among the 879 challenges, the number of overturned calls is 231 or more. If the 25% rate is correct, is 231 overturned calls among 879 challenges a result that is significantly high

The mean is:

$E(X) = np = 879*0.25 = 219.75$

The standard deviation is:

$\sqrt{V(X)} = \sqrt{np(1-p)} = \sqrt{879*0.25*0.75} = 12.84$

$219.75 + 2.5*12.84 = 251.85 > 231$

231 is less than 2.5 standard deviations above the mean, which means that 231 overturned calls among 879 challenges is not a significantly high result.