91Ó°ÊÓ

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,

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

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

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,

$$\begin{gathered} \mathrm{μ}=np \\ =879×0.25 \\ =219.75 \end{gathered}$$

The standard deviation is,

$$\begin{gathered} \mathrm{σ}=\sqrt{npq} \\ =\sqrt{879×0.25×0.75} \\ =12.84 \end{gathered}$$

(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 $x=230.5$, $\mathrm{μ}=219.75\mathrm{σ}=12.84$as follows:

$$\begin{gathered} z=\frac{x-\mathrm{μ}}{\mathrm{σ}} \\ =\frac{230.5-219.75}{12.84} \\ =0.84 \end{gathered}$$

The z-score is 0.84.

Find zscore using$x=231.5$, $\mathrm{μ}=219.75,\mathrm{σ}=12.84$as follows:

$$\begin{gathered} z=\frac{x-\mathrm{μ}}{\mathrm{σ}} \\ =\frac{231.5-219.75}{12.84} \\ =0.92 \end{gathered}$$

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$P\left(Xâ\copyright ¾n\right)\text{use}P\left(X>n+0.5\right)$

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

$$\begin{gathered} P\left(Xâ\copyright ¾231\right)=P\left(X>231+0.5\right) \\ =P\left(X>231.5\right)...\left(2\right) \end{gathered}$$

Find zscore using$x=231.5$,$\mathrm{μ}=219.75,\mathrm{σ}=12.84$as follows:

$$\begin{gathered} z=\frac{x-\mathrm{μ}}{\mathrm{σ}} \\ =\frac{231.5-219.75}{12.84} \\ =0.84 \end{gathered}$$

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 $P\left(x\text{or}\text{more}\right)â\copyright ½0.05$, then it is significantly high.

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