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A survey found that about 5% of employees contacted had reported wrongdoing during the past 12 months.

Variable x be the number who have observed and reported wrongdoing in the past 12 months.

The probability of whistle-blowing is 0.05 for any federal employee over the past 12 months

X follows binomial distribution with n=25 and p=0.05

Mean of X is given by np

$$\begin{gathered} np=25脳0.05 \\ =1.25 \end{gathered}$$

Standard deviation is given by$\sqrt{\mathbf{n}\mathbf{p}\mathbf{q}}$

Where q = 1 - p .

Therefore,

$$\begin{gathered} \sqrt{npq}=\sqrt{25脳0.05脳0.95} \\ =1.0897247 \end{gathered}$$

pmf of Binomial distribution is

$$\begin{gathered} f\left(k,n,p\right)=P\left(k;n,p\right) \\ =P\left(X=k\right) \\ =\left(\begin{array}{c}n\\ k\end{array}\right)p^k{\left(1-p\right)}^{n-k} \end{gathered}$$

For k=01,2,鈥�n. where

$\left(\begin{array}{c}n\\ k\end{array}\right)=\frac{n!}{k!\left(n-k\right)!}$

Therefore,

X P(X=x)

0 0.2774

1 0.3650

2 0.0930

3........... 0.2305

4 0.0269

5 0.0060

To calculate the probability that at least 5 of the employees are whistle-blowers in terms of x it mean to calculate $\mathrm{P}(X鈮�5)$

Hence,

$$\begin{gathered} \mathrm{P}(X鈮�5) \\ =1-P(X鈮�4) \\ =0.0072 \end{gathered}$$

Hence, the probability that at least 5 of the employees are whistle-blowers in terms of x is 0.0072.

Let the 5 of the 25 contacted have been whistle-blowers over the past 12 months.

To calculating the applicability of the 5% assumption to this agency that is to calculate the$\mathrm{P}(X=5)$

Hence,

Probability for 5 whistle blowers is

role="math" localid="1663244063498" $\mathrm{P}(X=5)=0.006$from part (a)

From part (b)

$\mathrm{P}(X鈮�5)=0.0072$

Hence, it can be concluded that as the number of whistle blower (x) increase, the probability also increases. as from part (b) $\mathrm{P}\left(\right(X鈮�5)鈮�P(x=5)$