91Ó°ÊÓ

The study of 650 people is recorded.

The given sample size$n=650$ and probability of success $p=0.12$.

Now,

$$\begin{gathered} q=1-p \\ =1-0.12 \\ =0.88 \end{gathered}$$

From the given information,

$$\begin{gathered} np=650×0.12 \\ =78 \\ >5 \end{gathered}$$

$$\begin{gathered} nq=650×0.88 \\ =572 \\ >5 \end{gathered}$$

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,

$$\begin{gathered} \mathrm{μ}=np \\ =650×0.12 \\ =78 \end{gathered}$$

The standard deviation is,

$$\begin{gathered} \mathrm{σ}=\sqrt{npq} \\ =\sqrt{650×0.12×0.88} \\ =8.28 \end{gathered}$$

Let X be the random variable for the number of people with green eyes.

The probability of at least 86 people with green eyes is expressed using continuity correction formula,$P\left(xâ\copyright ¾n\right)\text{use}P\left(X>n-0.5\right)$

Here

$$\begin{gathered} P\left(Xâ\copyright ¾86\right)=P\left(X>86-0.5\right) \\ =P\left(X>85.5\right) \end{gathered}$$

Thus, the probability is expressed as $P\left(X>85.5\right)$.

The zscore for$x=85.5$using$\mathrm{μ}=78,\mathrm{σ}=8.28$ is as follows:

$$\begin{gathered} z=\frac{x-\mathrm{μ}}{\mathrm{σ}} \\ =\frac{85.5-78}{8.28} \\ =0.91 \end{gathered}$$

The z-score is 0.91.

Using the standard normal table, the cumulative area to the left of 0.91 is 0.1814.

b.

An event is significantly high if the probability of occurrence for at least that event is lesser than equal to 0.05.

If $P\left(X\text{or}\text{more}\right)â\copyright ½0.05$, then it is significantly high.

But in our case, 0.1814 > 0.05. Therefore, it is not significantly high.