Question

Smartphone adoption among American younger adults has increased substantially and mobile access to the internet is pervasive. 17% of young adults, ages 18-29, who own a smartphone are “smartphone – dependent”, meaning that they do not have home broadband service and have limited options for going online other than their mobile device.

If a sample of 10 American young adults is selected what is the probability that

a. 3 are smartphone-dependent?

b. At least 3 are smartphone-dependent?

c. At most 6 are smartphone-dependent?

d. If you selected the sample in a particular geographical area and found that none of the 10 respondents are smartphone-dependent, what conclusions might you reach about whether the percentage of smartphone-dependent young adults in this area was 17%?

PLEASE ANSWER ALL PARTS

Answer 1

n = 10

p = 0.17

It is a binomial distribution.

P(X = x) = nCx * p^x * (1 - p)^{n - x}

a) P(X = 3) = 10C3 * (0.17)^3 * (0.83)^7 = 0.159983

b) P(X > 3) = 1 - P(X < 3)

                = 1 - (P(X = 0) + P(X = 1) + P(X = 2))

                = 1 - (10C0 * (0.17)^0 * (0.83)^10 + 10C1 * (0.17)^1 * (0.83)^9 + 10C2 * (0.17)^2 * (0.83)^8)

                = 1 - 0.765869

                = 0.234131

c) P(X < 6) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5) + P(X = 6)

                = 10C0 * (0.17)^0 * (0.83)^10 + 10C1 * (0.17)^1 * (0.83)^9 + 10C2 * (0.17)^2 * (0.83)^8 + 10C3 * (0.17)^3 * (0.83)^7 + 10C4 * (0.17)^4 * (0.83)^6 + 10C5 * (0.17)^5 * (0.83)^5 + 10C6 * (0.17)^6 * (0.83)^4

                = 0.999696

d) P(X = 0) = 10C0 * (0.17)^0 * (0.83)^10 = 0.155160

Since the probability is greater than is 0.05, so we can conclude that the percentage of smartphone-dependent young adults in this area was 17%.