Tennis ReplayRepeat the preceding exercise after changing the assumed rate of overturned calls from 25% to 22%.

Refer to the previous exercise for the information.

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 $p=0.22$.

Then,

$$\begin{gathered} q=1-p \\ =1-0.22 \\ =0.78 \end{gathered}$$

Let X be the random variable for the number of overturned calls in 879 challenges.

From the given information,

$$\begin{gathered} np=879\times 0.22 \\ =193.38 \\ >5 \end{gathered}$$

$$\begin{gathered} nq=879\times 0.78 \\ =685.62 \\ >5 \end{gathered}$$

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

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

The mean value is,

$$\begin{gathered} \mu =np \\ =879\times 0.22 \\ =193.38 \end{gathered}$$

The standard deviation is,

$$\begin{gathered} \sigma =\sqrt{npq} \\ =\sqrt{879\times 0.22\times 0.78} \\ =12.28 \end{gathered}$$

a.

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

Here

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

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

Find zscore using$x=230.5$, $\mu =193.38,\sigma =12.28$are as follows:

$$\begin{gathered} z=\frac{x-\mu }{\sigma } \\ =\frac{230.5-193.38}{12.28} \\ =3.02 \end{gathered}$$

The z-score is 3.02

Find zscore using$x=231.5$, $\mu =193.38,\sigma =12.28$as follows:

$$\begin{gathered} z=\frac{x-\mu }{\sigma } \\ =\frac{231.5-193.38}{12.28} \\ =3.10 \end{gathered}$$

The z-score is 3.10

Using Table A-2, the cumulative area corresponding to z-scores $z=3.02$ and $z=3.10$, are 0.9987 and 0.9990.

Substitute the values in equation (1),

$$\begin{gathered} P\left(230.5<X<231.5\right)=P\left(3.02<Z<3.10\right) \\ =P\left(Z<3.10\right)-P\left(Z<3.02\right) \\ =0.9990-0.9987 \\ =0.0003 \end{gathered}$$

b. The probability of 231 or more overturned calls is obtained by using continuity correction $P\left(x⩾n\right)\text{use}P\left(X>n+0.5\right)$as,

$$\begin{gathered} P\left(X⩾231\right)=P\left(X>231-0.5\right) \\ =P\left(X>230.5\right) \end{gathered}$$

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

Find zscore using$x=230.5$,$\mu =193.38,\sigma =12.28$ as follows:

$$\begin{gathered} z=\frac{x-\mu }{\sigma } \\ =\frac{230.5-193.38}{12.28} \\ =3.02 \end{gathered}$$

The z-score is 3.02.

Using standard normal table, the cumulative area corresponding to 3.02 is 0.9987.

Substitute the value in equation (1),

$$\begin{gathered} P\left(X>230.5\right)=P\left(Z>3.02\right) \\ =1-0.9987 \\ =0.0013 \end{gathered}$$

Thus, the probability of 231 or more calls is 0.0013.

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

In our case, 0.0013 < 0.05. Therefore, it is significantly high.