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,

\(\begin{aligned}{c}q = 1 - p\\ = 1 - 0.25\\ = 0.75\end{aligned}\)

From the given information,

\(\begin{aligned}{c}np = 879 \times 0.25\\ = 219.75\\ > 5\end{aligned}\)

\(\begin{aligned}{c}nq = 879 \times 0.75\\ = 659.25\\ > 5\end{aligned}\)

Here both\({\bf{np}}\)and\({\bf{nq}}\)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,

\(\begin{aligned}{c}\mu = np\\ = 879 \times 0.25\\ = 219.75\end{aligned}\)

The standard deviation is,

\(\begin{aligned}{c}\sigma = \sqrt {npq} \\ = \sqrt {879 \times 0.25 \times 0.75} \\ = 12.84\end{aligned}\)

(a)

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

That is,

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

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:

\(\begin{aligned}{c}z = \frac{{x - \mu }}{\sigma }\\ = \frac{{230.5 - 219.75}}{{12.84}}\\ = 0.84\end{aligned}\)

The z-score is 0.84.

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

\(\begin{aligned}{c}z = \frac{{x - \mu }}{\sigma }\\ = \frac{{231.5 - 219.75}}{{12.84}}\\ = 0.92\end{aligned}\)

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,

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

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\(P\left( {X \ge n} \right)\;{\rm{use}}\;P\left( {X > n + 0.5} \right)\)

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

\(\begin{aligned}{c}P\left( {X \ge 231} \right) = P\left( {X > 231 + 0.5} \right)\\ = P\left( {X > 231.5} \right)\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;...\left( 2 \right)\end{aligned}\)

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

\(\begin{aligned}{c}z = \frac{{x - \mu }}{\sigma }\\ = \frac{{231.5 - 219.75}}{{12.84}}\\ = 0.84\end{aligned}\)

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 \(P\left( {x\;{\rm{or}}\;{\rm{more}}} \right) \le 0.05\), then it is significantly high.

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