91Ó°ÊÓ

It is given that 79% of the people need eyesight correction.

A sample of 20 adults is selected.

Let X denote the number of people who require eyesight correction.

Let success be defined as getting a person who needs eyesight correction

The number of trials (n) is given to be equal to 20.

The probability of success is given as follows:

$$\begin{gathered} p=79\% \\ =\frac{79}{100} \\ =0.79 \end{gathered}$$

The probability of failure is given as follows:

$$\begin{gathered} q=1-p \\ =1-0.79 \\ =0.21 \end{gathered}$$

The number of successes required in 20 trials should be at least 19.

The binomial probability formula is as follows:

$P\left(X=x\right)={}^n{C}_x{\left(p\right)}^x{\left(q\right)}^{n-x}$

Using the binomial probability formula, the probability that at least 19 of the 20 people require eyesight correction is computed below:

$$\begin{gathered} P\left(Xâ\copyright ¾19\right)=P\left(X=19\right)+P\left(X=20\right) \\ ={}^{20}{C}_{19}{\left(0.79\right)}^{19}{\left(0.21\right)}^1{+}^{20}{C}_{20}{\left(0.79\right)}^{20}{\left(0.21\right)}^0 \\ =0.04766+0.00896 \\ =0.05662 \end{gathered}$$

Therefore, the probability that at least 19 people need eyesight correction is equal to 0.057.

The number of successes (x) of a binomial probability value is said to be significantly high if $P\left(x\text{or}\text{more}\right)â\copyright ½0.05$.

Here, the number of successes (people who need eyesight correction) is equal to 19.

$$\begin{gathered} P\left(19\text{or}\text{more}\right)=0.057 \\ >0.05 \end{gathered}$$

Thus, it can be said that19 is not a significantly high number of adults who require eyesight correction.