Question

A survey showed that 73% of adults need correction (eyeglasses, contacts, surgery, etc.) for their eyesight. If16 adults are randomly selected, find the probability that

no more than 1 of them need correction for their eyesight. Is 1 a significantly low number of adults requiring eyesight correction?

The probability that no more than 1 of the 16 adults require eyesight correction is _______? (round to 3 decimal places)

Answer 1

Solution:

Given:

p = probability of adults need correction for their eyesight = 0.73

n = sample size =16

Find:

P( no more than 1 of the 16 adults require eyesight correction) =..............?

That is:

P(X <1) = ........

P(X <1) = P(X = 0) + P(X = 1)

Using Binomial probability formula :

P(X = x) =n Cx pul xqN-1

Where

n! nort! x (n-1)!

and q = 1 - p = 1 - 0.73 = 0.27

Thus

P(X=0) =16 Co x 0.730 x 0.2716-0

16! O! (16 X1 X 0.2716

16! P(X = 0) = V = 1x 16! x1 x 0.000000000798

P(X = 0) = 1x1 x 0.000000000798

P(X = 0) = 0.000000000798

and

P(X = 1) =16 C1 x 0.731 x 0.2716-1

P(X = 1) - P(X = 1) = 1! * (16 - 1)! 16! 1! X (16_1Y X 0.73 x 0.2715

P(X = 1) = 1 x 15!" 16! X 0.73 x 0.000000002954

P(X = 1) = 16 x 0.73 x 0.000000002954

P(X = 1) = 0.000000034506

Thus

P(X <1) = P(X = 0) + P(X = 1)

P(X < 1) = 0.000000000798 + 0.000000034506

P(X < 1) = 0.00000003530

P(X < 1) = 0.000

Yes, 1 is a significantly low number of adults requiring eyesight correction, since its probability is very small ( < 0.05)