Question

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

Short answer

(a). The probability of the number of overturned calls is exactly 231 is 0.0003.

(b). The probability of the number of overturned calls is 231 or more is 0.0013.

The result is significantly high.

Step-by-step solution

Given information

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,

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

Check the requirement

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.

Mean and standard deviation for normal distribution

The mean value is,

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

The standard deviation is,

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

Compute the probability of exactly 231 overturned calls

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 $P\left(230.5<x<231.5\right) ...\left(1\right)$

Obtain the corresponding z-scores

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),

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

Compute the probability for 231 or more calls

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 $P\left(X>230.5\right)...\left(1\right)$

Obtain the z-score

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.