Question

A survey that 83% of aldilts need correction ( eyeglasses, contacts, surgery, etc.) for their eyesight. if 15 adults are randomly selected, find the probability that at least 14 of them need correction for their eyesight. is 14 a significantly high number of adults requiring eyesight correction?

Answer 1

Solution:

We are given

n = 15, p = 0.83,

q = 1 – p = 1 – 0.83 = 0.17

We have to find P(X≥14)

np = 15*0.83 = 12.45

nq = 15*0.17 = 2.55

np > 5 but nq < 5

So, we cannot use normal approximation.

So, we have to use binomial distribution.

P(X≥14) = P(X=14) + P(X=15)

P(X=x) = nCx*p^x*q^(n – x)

P(X=14) = 15C14*0.83^14*0.17^(15 – 14)

P(X=14) = 15*0.83^14*0.17^1

P(X=14) = 0.187773142

P(X=15) = 15C15*0.83^15*0.17^(15 – 15)

P(X=15) = 1*0.83^15*0.17^0

P(X=15) = 0.061118317

P(X≥14) = P(X=14) + P(X=15)

P(X≥14) = 0.187773142 + 0.061118317

P(X≥14) = 0.248891459

Required probability = 0.248891459

14 is a significantly high number of adults requiring eyesight correction because probability for x=14 is greater than 0.05.