Question

A survey showed that 77% of us need correction (eyeglasses, contacts, surgery, etc.) for their eyesight. 13 adults are randomly selected, find the probability that at least 12 of them need correction for their eyesight is 12 a significantly high
number of adults requiring eyesight correction?
The probability that at least 12 of the 13 adults require eyesight correction is I
(Round to three decimal places as needed)
Is 12 a significantly high number of adults requiring eyesight correction? Note that a small probability is one that is less than 0.05.
this purring is not small

Answer 1

Using the binomial distribution, it is found that the probability that at least 12 of the 13 adults require eyesight correction is of 0.163 = 16.3%. Since this probability is greater than 5%, it is found that 12 is not a significantly high number of adults requiring eyesight correction.

For each person, there are only two possible outcomes, either they need correction for their eyesight, or they do not. The probability of a person needing correction is independent of any other person, hence, the binomial distribution is used to solve this question.

What is the binomial distribution formula?

The formula is:

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$C_{n,x} = \frac{n!}{x!(n-x)!}$

The parameters are:

• x is the number of successes.

• n is the number of trials.

• p is the probability of a success on a single trial.

In this problem:

• A survey showed that 77% of us need correction, hence p = 0.77.

• 13 adults are randomly selected, hence n = 13.

The probability that at least 12 of them need correction for their eyesight is given by:

$P(X \geq 12) = P(X = 12) + P(X = 13)$

In which:

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 12) = C_{13,12}.(0.77)^{12}.(0.23)^{1} = 0.1299$

$P(X = 13) = C_{13,13}.(0.77)^{13}.(0.23)^{0} = 0.0334$

Then:

$P(X \geq 12) = P(X = 12) + P(X = 13) = 0.1299 + 0.0334 = 0.163$

The probability that at least 12 of the 13 adults require eyesight correction is of 0.163 = 16.3%. Since this probability is greater than 5%, it is found that 12 is not a significantly high number of adults requiring eyesight correction.

More can be learned about the binomial distribution at brainly.com/question/24863377