Problem 47 - Basic Statistics for Business and Economics with Student

Question

A recent study conducted by Penn, Shone, and Borland, on behalf of LastMinute.com, revealed that 52 percent 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: Binomial; b: Mean = 6.24, SD ≈ 1.715; c: ≈ 0.171; d: ≈ 0.403.

Step 1: Define the Probability Distribution

This problem follows a binomial distribution because there are fixed number of trials (12 travelers), two possible outcomes (planning within two weeks or not), and the probability of success is constant (52%). Let \( X \) be the number of travelers planning their trips within two weeks. Then, \( X \) follows a binomial distribution with parameters \( n = 12 \) and \( p = 0.52 \).

Step 2: Develop Probability Distribution

For \( X \), the probability of exactly \( k \) successes is given by \( P(X = k) = \binom{n}{k} p^k (1-p)^{n-k} \). Compute these probabilities for each \( k \) from 0 to 12. Example calculation for \( k = 0 \): \( P(X = 0) = \binom{12}{0} (0.52)^0 (0.48)^{12} \approx 0.0002 \). Repeat this calculation for \( k = 1, 2, ..., 12 \) and list the probabilities.

Step 3: Calculate the Mean of the Distribution

For a binomial distribution, the mean \( \mu \) is given by \( \mu = n \, p \). Thus, \( \mu = 12 \times 0.52 = 6.24 \).

Step 4: Calculate the Standard Deviation

The standard deviation \( \sigma \) for a binomial distribution is \( \sigma = \sqrt{n \, p \, (1-p)} \). \( \sigma = \sqrt{12 \times 0.52 \times 0.48} \approx 1.715 \).

Step 5: Find Probability of Exactly 5 Travelers

Use the binomial formula: \( P(X = 5) = \binom{12}{5} (0.52)^5 (0.48)^7 \approx 0.171 \).

Step 6: Find Probability of 5 or Fewer Travelers

Sum the probabilities for \( k = 0 \) through \( k = 5 \). \( P(X \leq 5) = P(X = 0) + P(X = 1) + P(X = 2) + \ldots + P(X = 5) \approx 0.403 \).

Key Concepts

Probability distribution

A probability distribution essentially outlines the likelihood of different outcomes in an experiment or process. In this particular exercise, we're discussing a binomial distribution. A binomial distribution applies when you're dealing with events that have two possible outcomes, such as success or failure—for this exercise, planning a trip within two weeks (success) versus planning it later (failure).

Key characteristics of a binomial distribution include:

• A fixed number of trials: In our case, it's 12 frequent business travelers surveyed.
• Two outcomes per trial: A "yes" or "no" scenario where travelers plan their trips within two weeks or not.
• A consistent probability of success: Here, there's a 52% chance that any given traveler plans their trip within two weeks.

For a better comprehension of this concept, visualize it as flipping a coin 12 times and counting how many times it lands heads up. Similarly, you're counting how many travelers out of the 12 planned early.

Mean calculation

The mean, often referred to as the expected value in probability, gives us an idea of the average outcome if an experiment were to be repeated many times. For binomial distributions, the mean is determined by the formula:\[ \mu = n \times p \]
where \( n \) is the number of trials, and \( p \) is the probability of success on each trial.

In this exercise, calculating the mean of our probability distribution helps quantify what we expect on average from our trial of 12 travelers, each having a 52% chance of planning their trip within two weeks. Plugging in our values:\[ \mu = 12 \times 0.52 = 6.24 \]

This tells us that from our 12 surveyed travelers, we'd expect approximately 6.24 travelers to plan their trips within two weeks. Although you can't have a fraction of a traveler, this gives a sense of tendency or central value around which data points may cluster.

Standard deviation

Standard deviation is a measure that quantifies the amount of variation or dispersion in a set of values. In other words, it tells you how much the actual outcomes can deviate from the mean you calculated.
For a binomial distribution, the standard deviation is computed using:\[ \sigma = \sqrt{n \times p \times (1-p)} \]
where, just like before, \( n \) is the number of trials, \( p \) is the probability of success, and \( 1-p \) is the probability of failure.

Applying this to our exercise provides:\[ \sigma = \sqrt{12 \times 0.52 \times 0.48} \approx 1.715 \]

The computed standard deviation of approximately 1.715 tells us the typical distance a random variable from the probability distribution is away from the mean, 6.24. This value helps assess the spread of your data, giving further insights into how varying the number of travelers (planning within two weeks) might be across different samples of 12.