Question

A recent study conducted by Penn, Shone, and Borland, on behalf of LastMinute, revealed that 52% of business travelers plan their trips less than two weeks before departure. The study is to be replicated in the tri-state area with a sample of 12 frequent business travelers.
a. Develop a probability distribution for the number of travelers who plan their trips within two weeks of departure.

b. Find the mean and the standard deviation of this distribution.

c. What is the probability exactly 5 of the 12 selected business travelers plan their trips within two weeks of departure?

Answer 1

Answer:

a) Binomial, with $p = 0.52, n = 12$

b) The mean is 6.24 and the standard deviation is 1.73.

c) 17.68% probability exactly 5 of the 12 selected business travelers plan their trips within two weeks of departure

Step-by-step explanation:

For each traveler, there are only two possible outcomes. Either they plan their trips less than two weeks before departure, or they do not. The travelers are independent of each other. So we use the binomial probability distribution to solve this question.

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.

The expected value of the binomial distribution is:

$E(X) = np$

The standard deviation of the binomial distribution is:

$\sqrt{V(X)} = \sqrt{np(1-p)}$

52% of business travelers plan their trips less than two weeks before departure.

This means that $p = 0.52$

12 frequent business travelers.

This means that $n = 12$

a. Develop a probability distribution for the number of travelers who plan their trips within two weeks of departure.

Binomial, with $p = 0.52, n = 12$

b. Find the mean and the standard deviation of this distribution.

$E(X) = np = 12*0.52 = 6.24$

$\sqrt{V(X)} = \sqrt{np(1-p)} = \sqrt{12*0.52*0.48} = 1.73$

The mean is 6.24 and the standard deviation is 1.73.

c. What is the probability exactly 5 of the 12 selected business travelers plan their trips within two weeks of departure?

This is P(X = 5).

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

$P(X = 5) = C_{12,5}.(0.52)^{5}.(0.48)^{7} = 0.1768$

17.68% probability exactly 5 of the 12 selected business travelers plan their trips within two weeks of departure