In Exercises 25–28, find the probabilities and answer the questions.

Vision Correction A survey sponsored by the Vision Council showed that 79% of adults need correction (eyeglasses, contacts, surgery, etc.) for their eyesight. If 20 adults are randomly selected, find the probability that at least 19 of them need correction for their eyesight. Is 19 a significantly high number of adults requiring eyesight correction?

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:

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

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 $P\left(x\text{or}\text{more}\right)⩽0.05$.

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.