Question

60% of all Americans live in cities with population greater than 100,000 people. If 36 Americans are randomly selected, find the probability that a. Exactly 21 of them live in cities with population greater than 100,000 people. b. At most 23 of them live in cities with population greater than 100,000 people. c. At least 20 of them live in cities with population greater than 100,000 people. d. Between 19 and 25 (including 19 and 25) of them live in cities with population greater than 100,000 people.

Answer 1

Normal approximation for binomial distribution

P(X < A) = P(Z < (A - mean)/standard deviation)

Given, sample size, n = 36

P(a person lives in a population greater than 100,000), p = 0.60

q = 1 - p = 0.40

Mean = np

= 36x0.6

= 21.6

Standard deviation =

=

= 2.94

a) P(X = 21) = P(20.5 < X < 21.5)

= P(X < 21.5) - P(X < 20.5)

= P(Z < (21.5 - 21.6)/2.94) - P(Z < (20.5 - 21.6)/2.94)

= P(Z < -0.034) - P(Z < -0.374)

= 0.4864 - 0.3542

= 0.1322

b) P(at most 23) = P(X < 23.5) (continuity correction is applied)

= P(Z < (23.5 - 21.6)/2.94)

= P(Z < 0.646)

= 0.7409

c) P(at least 20) = 1 - P(X < 19.5)

= 1 - P(Z < (19.5 - 21.6)/2.94)

= 1 - P(Z < -0.714)

= 1 - 0.2376

= 0.7624

d) P(19 X 25) = P(X < 25.5) - P(X < 18.5)

= P(Z < (25.5 - 21.6)/2.94) - P(Z < (18.5 - 21.6)/2.94)

= P(Z < 1.327) - P(Z < -1.054)

= 0.9077 - 0.1459

= 0.7618