Problem 47 - Basic Statistics for Business & Economics

Question

A recent study conducted by Penn, Shone, and Borland, on behalf of LastMinute .com, 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? d. What is the probability 5 or fewer of the 12 selected business travelers plan their trips within two weeks of departure?

Step-by-Step Solution

Answer

a) Use binomial distribution formula. b) Mean: 6.24, SD: 1.7191. c) Use binomial formula for P(X=5). d) Sum probabilities P(X=0) to P(X=5).

Step 1: Define the random variable

Let the random variable \( X \) represent the number of travelers who plan their trips within two weeks of departure. \( X \) follows a binomial distribution because each traveler either plans a trip within two weeks or not. The parameters are \( n = 12 \) (number of trials) and \( p = 0.52 \) (probability of success).

Step 2: Probability distribution setup

The probability mass function for a binomial distribution is given by:\[ P(X = k) = \binom{n}{k} p^k (1-p)^{n-k} \]where \( k \) is the number of successes (travelers who plan within two weeks) and \( \binom{n}{k} \) is the binomial coefficient. We need to compute this for all possible values of \( k \) from 0 to 12.

Step 3: Compute mean and standard deviation

The mean \( \mu \) of a binomial distribution is given by \( \mu = n \times p = 12 \times 0.52 = 6.24 \).The standard deviation \( \sigma \) is given by \( \sigma = \sqrt{n \times p \times (1-p)} = \sqrt{12 \times 0.52 \times 0.48} \approx 1.7191 \).

Step 4: Probability of exactly 5 travelers

Using the binomial probability formula:\[ P(X=5) = \binom{12}{5} (0.52)^5 (0.48)^{7} \]Calculate this probability using the binomial coefficient and simplify.

Step 5: Probability of 5 or fewer travelers

The probability of 5 or fewer successful outcomes is the sum of probabilities from 0 to 5:\[ P(X \leq 5) = \sum_{k=0}^{5} \binom{12}{k} (0.52)^k (0.48)^{12-k} \]Calculate and add these probabilities to find the cumulative probability.

Key Concepts

Probability Distribution

In statistics, a probability distribution assigns a probability to each possible outcome of a random experiment. For this exercise, we focus on a binomial distribution. The binomial distribution applies here because each business traveler can either plan their trip within two weeks or not. This creates two possible outcomes, making it a binomial scenario. The probability distribution is represented by a probability mass function (PMF). For a binomial distribution, the PMF is given by:\[ P(X = k) = \binom{n}{k} p^k (1-p)^{n-k} \]where:- \( \binom{n}{k} \) = the binomial coefficient- \( n \) = total number of trials (here, 12 travelers)- \( k \) = number of successful outcomes (i.e., travelers planning within two weeks)- \( p \) = probability of success on an individual trial (here, 0.52) In the exercise, you would calculate the probability for each \( k \) from 0 to 12 to see the likelihood of different numbers of travelers planning their trips early. This forms the complete binomial probability distribution for this scenario.

Random Variable

A random variable is a variable whose possible values are outcomes of a random phenomenon. In our exercise, the random variable \( X \) is defined as the number of travelers planning their trips within two weeks of departure. It quantifies the outcome of our focus, the early planners among the travelers.In the context of the binomial distribution:- \( X \) can take any integer value from 0 to 12.- Each value corresponds to a different number of travelers planning early.This random variable is crucial as it forms the basis of calculating probabilities and understanding the behavior of early trip planning among the business travelers sampled. Each outcome or value of \( X \) has an associated probability, determined using the probability mass function of the binomial distribution.

Mean and Standard Deviation

The mean and standard deviation of a binomial distribution tell us about the distribution's central tendency and variability. The mean (or expected value) gives the average number of successes in the trials. In this exercise:- The formula for the mean \( \mu \) is: \[ \mu = n \times p = 12 \times 0.52 = 6.24 \]- This means, on average, we expect about 6.24 travelers out of 12 to plan their trips early.The standard deviation tells us how much spread or variability there is in those outcomes:- The formula for the standard deviation \( \sigma \) is: \[ \sigma = \sqrt{n \times p \times (1-p)} = \sqrt{12 \times 0.52 \times 0.48} \approx 1.7191 \]- This scale of 1.7191 indicates the typical distance of a number from the mean value, reflecting how consistent the results are expected to be around the average.

Cumulative Probability

Cumulative probability is the sum of probabilities for all outcomes up to a certain point in a probability distribution. It reflects the likelihood that the random variable is less than or equal to that point. This is particularly useful when you want to understand the probability of not exceeding a certain number of successes.In the given exercise, finding the probability that 5 or fewer travelers plan early involves calculating the cumulative probability for \( X \leq 5 \):- You'll use the probability mass function to find: \[ P(X \leq 5) = \sum_{k=0}^{5} \binom{12}{k} (0.52)^k (0.48)^{12-k} \]- This calculation adds up the individual probabilities from 0 through 5 travelers planning early.The cumulative probability helps provide insight into the likelihood of observing up to a particular number of successes, in this case helping you understand how likely it is to have 5 or fewer early planners out of the 12 travelers.