# Statistics Question Please Help Asap!!?

A certain flight arrives on time 81 percent of the time. Suppose 197 flights are randomly selected. Use the normal approximation to the binomial to approximate the probability that

(a) exactly 166 flights are on time.

(b) at least 166 flights are on time.

(c) fewer than 169 flights are on time.

(d) between 169 and 175, inclusive are on time

### 1 Answer

- fcas80Lv 74 weeks ago
Normal approximation to binomial

n = 197

p = 0.81

xbar = np = 159.57

s = √[n * p * (1-p)] = 5.51

I am going to use the continuity correction factor. Some books and teachers use it, and some do not. If yours does not, then the following may be wrong.

x = 166

x - .5 = 165.5

x + .5 = 166.5

z for x - .5 = 165.5 is 1.08

P(z < 1.08) = 0.8599

z for x + .5 = 166.5 is 1.26

P(z < 1.26) = 0.8962

a. P(x=166) = 0.8962 - 0.8599 = 0.0363

b. P(x>=166) = P(x>=165.5) = 1 - .8962 = .1038

c. P(x < 169) = P(x < 168.5) = P(z < 1.62) = 0.9474

P(x < 175) = P(x < 174.5) = P(z < 1.62) = 0.9966

d. P(169 < x < 175) = 0.0492