Question

A survey showed that 82​% of adults need correction​ (eyeglasses, contacts,​ surgery, etc.) for their eyesight. If 12 adults are randomly​ selected, find the probability that at least 11 of them need correction for their eyesight. Is 11 a significantly high number of adults requiring eyesight​ correction?

Answer 1

Probability of an adult who needs correction = 82% or 0.82

So, probability of an adult who does not needs correction = 1 - 0.82  = 0.18

If 12 adults are randomly​ selected, the probability that at least 11 of them need correction for their eyesight is:

12C11$(0.82)^{11}(0.18)^{12-11}$ + 12C12$(0.82)^{12}(0.18)^{12-12}$

= 12*0.11*0.18 + 1*0.09*1

= 0.238+0.09

= 0.328 or 0.33

And, yes, 11 is significantly a high number of adults requiring eyesight​ correction.

Answer 2

Answer:

Pr(X >= 11) = 0.33586800937

Step-by-step explanation:

This scenario can be modeled by a binomial distribution model.

The probability of success, p = 0.82 is constant.

The are 12 independent trials.

We let the random variable X denote the number of adults who need correction for their eyesight. We are then to determine the probability that X is at least 11;

Pr(X=11 or 12) = Pr(X=11) + Pr(X=12)

                       =  0.33586800937

11 is significantly a high number of adults requiring eyesight​ correction.