In: Statistics and Probability
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? ANSWER: (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? ANSWER: (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? ANSWER: (Report your answer to 3 decimal places, using conventional rounding rules)
p = 0.7
n = 11
P(X = x) = 11Cx * 0.7x * (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