Problem 21 - Basic Statistics for Business & Economics

Question

In a recent study, \(90 \%\) of the homes in the United States were found to have large-screen TVs. In a sample of nine homes, what is the probability that: a. All nine have large-screen TVs? b. Less than five have large-screen TVs? c. More than five have large-screen TVs? d. At least seven homes have large-screen TVs?

Step-by-Step Solution

Answer

a) 0.38742; b) use binomial sum; c) use complement; d) use complement.

Step 1: Identify the Type of Distribution

The problem involves independent trials with two possible outcomes: having a large-screen TV or not having one. This is a typical scenario for a binomial distribution where the probability of success (having a large-screen TV) is 0.9.

Step 2: Define the Parameters

For this binomial distribution, the number of trials (n) is 9, and the probability of success (p) for each trial is 0.9.

Step 3: Calculate Probability for All Nine Homes

For part (a), we need the probability that all nine homes have large-screen TVs. Use the binomial probability formula:\[ P(X = 9) = \binom{9}{9} (0.9)^{9} (0.1)^{0} \]Calculate: \( P(X = 9) = 1 \times (0.9)^9 = 0.38742 \).

Step 4: Calculate Probability for Less Than Five Homes

For part (b), calculate the probability that less than five homes have large-screen TVs, i.e., P(X < 5). Sum the probabilities from 0 to 4:\[ P(X < 5) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) \]Calculate each using the formula and sum them.

Step 5: Calculate Probability for More Than Five Homes

For part (c), calculate the probability that more than five homes have large-screen TVs, i.e., P(X > 5). Use the complement rule:\[ P(X > 5) = 1 - P(X \leq 5) = 1 - (P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5)) \]Calculate each probability and find the result.

Step 6: Calculate Probability for At Least Seven Homes

For part (d), calculate the probability that at least seven homes have large-screen TVs, i.e., P(X \geq 7). Use:\[ P(X \geq 7) = 1 - P(X \leq 6) = 1 - (P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5) + P(X = 6)) \]Find each probability and compute the result.

Key Concepts

Probability

Probability is at the heart of many statistical problems, including those involving binomial distributions. It measures how likely an event is to occur and ranges from 0 (impossible) to 1 (certain). In this example, the probability of a home having a large-screen TV is 0.9 or 90%. Understanding probability allows us to predict outcomes such as the number of homes that might have a large-screen TV in a sample.

• If the probability of success in one trial is given (e.g., a home having a large-screen TV), this can be extended to multiple trials using formulas suited for different distributions.
• Probability calculations like these help in making informed predictions in real-life situations.

As you encounter problems involving probability, remember to identify the type of events and use the correct approach to calculate their likelihood.

Independent Trials

In probability theory, trials are independent if the outcome of one does not affect the others. This is a key feature of the binomial distribution. When analyzing whether homes have large-screen TVs, each home is considered an independent trial.

• For example, knowing that one home has a large-screen TV does not change the likelihood of another home having one.
• This independence simplifies calculations since each trial has the same probability of success, in this case, 0.9.

Keep in mind that real-world situations might sometimes introduce dependencies, but the assumption of independence is a core concept for binomial probability calculations.

Binomial Probability Formula

The binomial probability formula is vital when working with binomial distributions. It's used to find the probability of a specific number of successes over a series of trials.To calculate the probability of exactly "k" successes in "n" trials, use this formula:\[ P(X = k) = \binom{n}{k} p^{k} (1-p)^{n-k} \]Where:

• \(n\) is the total number of trials (e.g., 9 homes).
• \(p\) is the probability of success on a single trial (e.g., 0.9).
• \(k\) is the desired number of successful trials.

This formula gives the likelihood of observing a specific number of successes, and is crucial when determining probabilities for discrete outcomes, like how many homes have large-screen TVs.

Complement Rule

The complement rule is a useful tool in probability, allowing you to find the probability of an event happening by subtracting its complement from 1. It's especially handy when computing probabilities for 'at least' or 'more than' scenarios.For instance, if you need the probability of more than five homes having large-screen TVs, calculate the probability of five or fewer homes first, then use the complement:\[ P(X > 5) = 1 - P(X \leq 5) \]

• The complement rule helps simplify complex probability questions by breaking them into more manageable parts.
• It is often used in conjunction with the binomial probability formula to give a clearer understanding of different situations.

Remembering and correctly applying the complement rule can save time and reduce errors in calculations.