Question

A survey showed that 78% of adults need correction (eyeglasses, contacts, surgery, etc.) for their eyesight. If 9 adults are randomly selected, find the probability that at leastat least 8 of them need correction for their eyesight. Is 8 a significantly highhigh number of adults requiring eyesight correction?

Answer 1

Solution:

Given that,

P = 0.78

1 - P = 0.22

n = 9

Here, BIN ( n , P ) that is , BIN (9 , 0.78)

then,

n*p = 9*0.78 = 7.02 5

n(1- P) = 9*0.22 = 1.98 5

According to normal approximation binomial,

X Normal

Mean = = n*P = 9*0.78 = 7.02

Standard deviation = =n*p*(1-p) = 9*0.78*0.22 = 1.54

We using countinuity correction factor

P(X a ) = P(X > a - 0.5)

P(x > 7.5) = 1 - P(x < 7.5)

= 1 - P((x - ) / < (7.5 - 7.02) / 1.54)

= 1 - P(z < 0.24)

= 1 - 0.5948   

= 0.4052

Probability = 0.4052