Question

A survey of U.S. adults who own smartphones found that 35% of the respondents check their phones at least once an hour for each hour during the waking hours. Such smartphone owners are classified as "constant checkers." Suppose you were to draw a random sample of 10 smartphone owners.(a) The number in your sample who are constant checkers has a binomial distribution. What are n and p?(b) Use the binomial formula to find the probability that exactly two of the 10 are constant checkers in your sample. (Round your answer to four decimal places.)(c) Use the binomial formula to find the probability that two or fewer are constant checkers in your sample. (Round your answer to four decimal places.)

Answer 1

Answer:

a) $n = 10, p = 0.35$

b) 0.1757

c) 0.2617

Step-by-step explanation:

For each adult who owns a smarthphone, there are only two possible outcomes. Either they check their phones at least once an hour. Or they do not. The probability of each adult checking their phone at least once an hour is independent from other adults. So we use the binomial probability distribution to solve this problem.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

In which $C_{n,x}$ is the number of different combinations of x objects from a set of n elements, given by the following formula.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

And p is the probability of X happening.

35% of the respondents check their phones at least once an hour for each hour during the waking hours.

This means that $p = 0.35$

Suppose you were to draw a random sample of 10 smartphone owners.

This means that $n = 10$

(a) The number in your sample who are constant checkers has a binomial distribution. What are n and p?

$n = 10, p = 0.35$

(b) Use the binomial formula to find the probability that exactly two of the 10 are constant checkers in your sample. (Round your answer to four decimal places.)

This is P(X = 2)

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 2) = C_{10,2}.(0.35)^{2}.(0.75)^{8} = 0.1757$

(c) Use the binomial formula to find the probability that two or fewer are constant checkers in your sample. (Round your answer to four decimal places.)

$P(X \leq 2) = P(X = 0) + P(X = 1) + P(X = 2)$

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 0) = C_{10,0}.(0.35)^{0}.(0.75)^{10} = 0.0135$

$P(X = 1) = C_{10,1}.(0.35)^{1}.(0.75)^{9} = 0.0725$

$P(X = 2) = C_{10,2}.(0.35)^{2}.(0.75)^{8} = 0.1757$

So

$P(X \leq 2) = P(X = 0) + P(X = 1) + P(X = 2) = 0.0135 + 0.0725 + 0.1757 = 0.2617$