Question

Seventy percent of all individuals living in the United States have a smart phone. Suppose you select a random sample of 11 individuals who live in the United States. Define X to be a binomial random variable representing whether or not an individual living in the United States has a smart phone.

a.) What is the probability that no more than five of the sampled individuals have a smart phone?

b.) What is the probability that at least 7, but less than 11, of the sampled individuals have a smart phone?

c.) On average, how many of the eleven sampled individuals would you expect to have a smart phone?

Answer 1

p = 0.7

n = 11

P(X = x) = 11Cx * 0.7^x * (1 - 0.7)^11-x

a) P(X < 5) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5)

                  = 11C0 * 0.7⁰ * 0.3¹¹ + 11C1 * 0.7¹ * 0.3¹⁰ + 11C2 * 0.7² * 0.3⁹ + 11C3 * 0.7³ * 0.3⁸ + 11C4 * 0.7⁴ * 0.3⁷ + 11C5 * 0.7⁵ * 0.3⁶

                  = 0.078

b) P(7 < x < 11) = P(X = 7) + P(X = 8) + P(X = 9) + P(X = 10)

                          = 11C7 * 0.7⁷ * 0.3⁴ + 11C8 * 0.7⁸ * 0.3³ + 11C9 * 0.7⁹ * 0.3² + 11C10 * 0.7¹⁰ * 0.3¹

                          = 0.770

c) Expected value = n * p = 11 * 0.7 = 7.7