91影视

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,

\(\begin{aligned}{c}q = 1 - p\\ = 1 - 0.22\\ = 0.78\end{aligned}\)

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

From the given information,

\(\begin{aligned}{c}np = 879 \times 0.22\\ = 193.38\\ > 5\end{aligned}\)

\(\begin{aligned}{c}nq = 879 \times 0.78\\ = 685.62\\ > 5\end{aligned}\)

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

\(\begin{aligned}{c}\mu = np\\ = 879 \times 0.22\\ = 193.38\end{aligned}\)

The standard deviation is,

\(\begin{aligned}{c}\sigma = \sqrt {npq} \\ = \sqrt {879 \times 0.22 \times 0.78} \\ = 12.28\end{aligned}\)

a.

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

Here

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

Thus, the expression is \[P\left( {230.5 < x < 231.5} \right)\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\,...\left( 1 \right)\]

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

\(\begin{aligned}{c}z = \frac{{x - \mu }}{\sigma }\\ = \frac{{230.5 - 193.38}}{{12.28}}\\ = 3.02\end{aligned}\)

The z-score is 3.02

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

\(\begin{aligned}{c}z = \frac{{x - \mu }}{\sigma }\\ = \frac{{231.5 - 193.38}}{{12.28}}\\ = 3.10\end{aligned}\)

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),

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

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

\(\begin{aligned}{c}P\left( {X \ge 231} \right) = P\left( {X > 231 - 0.5} \right)\\ = P\left( {X > 230.5} \right)\end{aligned}\)

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:

\(\begin{aligned}{c}z = \frac{{x - \mu }}{\sigma }\\ = \frac{{230.5 - 193.38}}{{12.28}}\\ = 3.02\end{aligned}\)

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),

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

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

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

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