Tennis ReplayIn 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?

The number of challenges made to referee calls in professional tennis singles play is recorded.

The given sample size$n=879$ and probability of success $p=0.25$.

Then,

$$\begin{gathered} q=1-p \\ =1-0.25 \\ =0.75 \end{gathered}$$

From the given information,

$$\begin{gathered} np=879\times 0.25 \\ =219.75 \\ >5 \end{gathered}$$

$$\begin{gathered} nq=879\times 0.75 \\ =659.25 \\ >5 \end{gathered}$$

Here both and are greater than 5. Thus, probabilities from a binomial

probability distribution can be approximated reasonably well by using a normal distribution.

The mean value is,

$$\begin{gathered} \mu =np \\ =879\times 0.25 \\ =219.75 \end{gathered}$$

The standard deviation is,

$$\begin{gathered} \sigma =\sqrt{npq} \\ =\sqrt{879\times 0.25\times 0.75} \\ =12.84 \end{gathered}$$

(a)

The probability of getting exactly 231 overturned calls is expressed using continuity correction,For $P\left(X=n\right)\text{use}P\left(n-0.5<X<n+0.5\right)$

That is,

$$\begin{gathered} P\left(X=231\right)=P\left(231-0.5<X<231+0.5\right) \\ =P\left(230.5<X<231.5\right) \end{gathered}$$

Thus, the expression is $P\left(230.5<X<231.5\right)...\left(1\right)$

The z-scores using $x=230.5$, $\mu =219.75\sigma =12.84$as follows:

$$\begin{gathered} z=\frac{x-\mu }{\sigma } \\ =\frac{230.5-219.75}{12.84} \\ =0.84 \end{gathered}$$

The z-score is 0.84.

Find zscore using$x=231.5$, $\mu =219.75,\sigma =12.84$as follows:

$$\begin{gathered} z=\frac{x-\mu }{\sigma } \\ =\frac{231.5-219.75}{12.84} \\ =0.92 \end{gathered}$$

The z-score is 0.92.

Using standard normal table, the z-scores $z=0.84$ and $z=0.92$, yield cumulative areas of 0.2005 and 0.1788.

The equation (1) implies,

$$\begin{gathered} P\left(0.84<Z<0.92\right)=P\left(Z<0.92\right)-P\left(Z<0.84\right) \\ =0.2005-0.1788 \\ =0.0217 \end{gathered}$$

Thus, the probability that exactly 231 calls are overturned is 0.0217.

b.

The probability that 231 or more calls are overturned is expressed using continuity correction$P\left(X⩾n\right)\text{use}P\left(X>n+0.5\right)$

Thus, the probability that 231 or more calls are overturned,

$$\begin{gathered} P\left(X⩾231\right)=P\left(X>231+0.5\right) \\ =P\left(X>231.5\right)...\left(2\right) \end{gathered}$$

Find zscore using$x=231.5$,$\mu =219.75,\sigma =12.84$as follows:

$$\begin{gathered} z=\frac{x-\mu }{\sigma } \\ =\frac{231.5-219.75}{12.84} \\ =0.84 \end{gathered}$$

The z-score is 0.84.

Using standard normal table, the z-score corresponding to 231.5 is $z=0.84$, which yield cumulative areas 0.2012.

If $P\left(x\text{or}\text{more}\right)⩽0.05$, then it is significantly high.

But In our case, 0.2012 > 0.05. Therefore, it is not significantly high.