Question

At the end of 2013, it was reported that 92% of U.S. adults owned a cell-phone. If a sample of 10 U.S. adults is selected, what is the probability that:

8 own a cell phone?

All ten own a cell phone?

At least 8 own a cell phone?

Between 6 and 8 own a cell phone (including 6 and 8)?

Answer 1

n = 10

p = 0.92

It is a binomial distribution.

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

a) P(X = 8) = 10C8 * (0.92)^8 * (0.08)^2 = 0.1478

b) P(X = 10) = 10C10 * (0.92)^10 * (0.08)^0 = 0.4344

c) P(X > 8) = P(X = 8) + P(X = 9) + P(X = 10)

                   = 10C8 * (0.92)^8 * (0.08)^2 + 10C9 * (0.92)^9 * (0.08)^1 + 10C10 * (0.92)^10 * (0.08)^0

                   = 0.9599

d) P(6 < X < 8) = P(X = 6) + P(X = 7) + P(X = 8)

                         = 10C6 * (0.92)^6 * (0.08)^4 + 10C7 * (0.92)^7 * (0.08)^3 + 10C8 * (0.92)^8 * (0.08)^2

                         = 0.1873