Problem 46 - Basic Statistics for Business and Economics with Student

Question

Dr. Richmond, a psychologist, is studying the daytime television viewing habits of college students. She believes 45 percent of college students watch soap operas during the afternoon. To further investigate, she selects a sample of \(10 .\) a. Develop a probability distribution for the number of students in the sample who watch soap operas. b. Find the mean and the standard deviation of this distribution. c. What is the probability of finding exactly four watch soap operas? d. What is the probability less than half of the students selected watch soap operas?

Step-by-Step Solution

Answer

The mean is 4.5, and the standard deviation is approximately 1.573. The probability of exactly four students watching soap operas is about 0.200, and the probability of fewer than five students watching is the sum of probabilities for 0, 1, 2, 3, and 4 students.

Step 1: Identify the Distribution

Since each student either watches soap operas or doesn't, and each has the same probability of doing so, this is a Binomial distribution problem. The parameters are \( n = 10 \) (trials) and \( p = 0.45 \) (probability of success).

Step 2: Define the Probability Mass Function

The probability of exactly \( k \) students watching soap operas is given by the binomial probability formula: \[ P(X = k) = \binom{n}{k} p^k (1-p)^{n-k} \]For this problem, \( n = 10 \) and \( p = 0.45 \). We apply this formula to each \( k \) from 0 to 10 to create the distribution.

Step 3: Calculate the Mean and Standard Deviation

For a binomial distribution with parameters \( n \) and \( p \), the mean is given by \( \mu = np \) and the standard deviation is \( \sigma = \sqrt{np(1-p)} \). Thus, \[ \mu = 10 \times 0.45 = 4.5 \] \[ \sigma = \sqrt{10 \times 0.45 \times 0.55} \approx 1.573 \]

Step 4: Compute Probability of Exactly Four Watching

Using the binomial formula for \( k = 4 \), \[ P(X = 4) = \binom{10}{4} (0.45)^4 (0.55)^{6} \]Calculate \( \binom{10}{4} = 210 \), and then the probability, resulting in \[ P(X = 4) \approx 0.200 \]

Step 5: Calculate the Probability Less Than Half Watching

Half of 10 is 5, so we need the probability of fewer than 5 students watching. This means calculating \( P(X < 5) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) \), using the binomial formula for each value and summing them.

Key Concepts

Probability Mass Function

In a binomial distribution, the probability mass function (PMF) allows us to find the probability of achieving exactly a certain number of successes in a series of independent trials. Each trial can only result in either a success or a failure, much like flipping a coin.The PMF for a binomial distribution is given by:\[ P(X = k) = \binom{n}{k} p^k (1-p)^{n-k} \]where:- \( n \) is the number of trials,- \( k \) is the number of successful trials,- \( p \) is the probability of success on a single trial, and- \( \binom{n}{k} \) represents the binomial coefficient, which is the number of ways to choose \( k \) successes from \( n \) trials.In the context of Dr. Richmond's study on soap opera watching among college students, each student's choice to watch or not watch is a trial. By using this function, she can compute, for example, the probability that exactly 4 out of 10 students watch soap operas.

Mean and Standard Deviation

The mean and standard deviation are crucial statistical measures that provide insights into the distribution's characteristics.- **Mean (\( \mu \))**: It indicates the expected number of successes over multiple trials. For a binomial distribution, the mean is calculated as \( \mu = np \), where \( n \) is the number of trials and \( p \) is the probability of success.- **Standard Deviation (\( \sigma \))**: It measures the distribution’s spread, indicating how much variation exists from the mean. The formula for the standard deviation in a binomial distribution is \( \sigma = \sqrt{np(1-p)} \).In the exercise, with \( n = 10 \) and \( p = 0.45 \), the mean comes out to be 4.5, suggesting that on average, 4.5 students are expected to watch soap operas. The standard deviation, approximately 1.573, shows the typical deviation from this average. This helps Dr. Richmond understand both expected value and variability.

Binomial Probability Formula

The binomial probability formula is at the heart of calculating probabilities in a binomial distribution. It enables us to determine the probability of getting exactly a defined number of successes in a fixed number of trials. This formula is:\[ P(X = k) = \binom{n}{k} p^k (1-p)^{n-k} \]Let's break down its components:- The binomial coefficient \( \binom{n}{k} \) calculates the number of possible combinations of \( k \) successes in \( n \) trials.- \( p^k \) represents the likelihood that \( k \) successes occur.- \( (1-p)^{n-k} \) calculates the probability that the remaining trials are failures.In practical application, such as Dr. Richmond's study, if you wanted to find the probability that exactly 4 students watch soap operas, you would use this formula with \( n = 10 \), \( k = 4 \), and \( p = 0.45 \). This results in a probability of approximately 0.200, or a 20% chance. This formula is a powerful tool in statistics when dealing with binary events.

Probability Distribution

A probability distribution provides a comprehensive snapshot of all potential outcomes and their probabilities in a statistical experiment. For binomial distributions, it's a table or graph that shows the probability of each possible number of successes from zero up to the number of trials. To create a probability distribution for the number of students watching soap operas in Dr. Richmond's exercise, you'd calculate the probability for each possible number of students (0 to 10) using the binomial probability formula. This distribution helps identify the likelihood of various outcomes:

• The outcome where no students watch could be very unlikely.
• Somewhere around 4 to 5 students might be the most probable scenarios, given the mean.
• It also allows for assessing less likely extremes, such as all 10 watching or none at all.

Such a probability distribution becomes an essential tool in drawing meaningful conclusions from data, allowing researchers like Dr. Richmond to make informed predictions and decisions about a population based on the sample studied.