Question

In: Statistics and Probability

Smartphone adoption among American younger adults has increased substantially and mobile access to the internet is...

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

Solutions

Expert Solution

n = 10

p = 0.17

It is a binomial distribution.

P(X = x) = nCx * px * (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%.


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