Question

A survey showed that 72​% of adults need correction​ (eyeglasses, contacts,​ surgery, etc.) for their eyesight. If

18 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?

Answer 1

Solution:

We are given n = 18, p = 0.72, q = 1 – p = 1 – 0.72 = 0.28

We have to find P(X≤1)

Here, np = 18*.72 = 12.96 > 5

And nq = 18*.28 = 5.04 > 5

So, we can use normal approximation.

Mean = np = 12.96

SD = sqrt(npq) = sqrt(18*.72*.28) = 1.904941

P(X≤1) = P(X<1.5) (by adding continuity correction factor 0.5)

Z = (X – mean)/SD

Z = (1.5 - 12.96) / 1.904941

Z = -6.01593

P(Z<-6.01593) = P(X≤1) = 0.0000

(by using z-table or excel)

Required probability = 0.0000

1 is a significantly low number of adults requiring eyesight correction, because we get corresponding probability as 0.0000 which is less than significance level 0.05.