Eye ColorBased on a study by Dr. P. Sorita at Indiana University, assume that 12% of us have green eyes. In a study of 650 people, it is found that 86 of them have green eyes.

a.Find the probability of at least 86 people with green eyes among 650 randomly selected people.

b.Is 86 people with green eyes significantly high?

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,

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

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

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,

$$\begin{gathered} \mu =np \\ =650\times 0.12 \\ =78 \end{gathered}$$

The standard deviation is,

$$\begin{gathered} \sigma =\sqrt{npq} \\ =\sqrt{650\times 0.12\times 0.88} \\ =8.28 \end{gathered}$$

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,$P\left(x⩾n\right)\text{use}P\left(X>n-0.5\right)$

Here

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

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

The zscore for$x=85.5$using$\mu =78,\sigma =8.28$ is as follows:

$$\begin{gathered} z=\frac{x-\mu }{\sigma } \\ =\frac{85.5-78}{8.28} \\ =0.91 \end{gathered}$$

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 $P\left(X\text{or}\text{more}\right)⩽0.05$, then it is significantly high.

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