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.

What is the probability that no more than five of the sampled individuals have a smart phone? (Use only the appropriate formula and/or statistical table in your textbook to answer this question. Report your answer to 3 decimal places, using conventional rounding rules)

What is the probability that at least 7, but less than 11, of the sampled individuals have a smart phone? (Use only the appropriate formula and/or statistical table in your textbook to answer this question. Report your answer to 3 decimal places, using conventional rounding rules)

On average, how many of the eleven sampled individuals would you expect to have a smart phone? (Report your answer to 3 decimal places, using conventional rounding rules)

Answer 1

X ~ Binomial (n,p)

Where n = 11 , p = 0.70

Binomial probability distribution is

P(X) = ⁿC_x p^x (1-p)^n-x

a)

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

= ¹¹C₀ 0.70⁰ 0.30¹¹ +¹¹C₁ 0.70¹ 0.30¹⁰ +¹¹C₂ 0.70² 0.30⁹ +¹¹C₃ 0.70³ 0.30⁸ +

¹¹C₄ 0.70⁴ 0.30⁷ +¹¹C₅ 0.70⁵ 0.30⁶

= 0.078

b)

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

= ¹¹C₇ 0.70⁷ 0.30⁴ +¹¹C₈ 0.70⁸ 0.30³ +¹¹C₉ 0.70⁹ 0.30² +¹¹C₁₀ 0.70¹⁰ 0.30¹

= 0.770

c)

E(X) = np

= 11 * 0.70

= 7.7