Problem 29 - A First Course in Probability

Question

The gross weekly sales at a certain restaurant are a normal random variable with mean \(\$ 2200\) and standard deviation \(\$ 230 .\) What is the probability that (a) the total gross sales over the next 2 weeks exceeds \(\$ 5000\) (b) weekly sales exceed \(\$ 2000\) in at least 2 of the next 3 weeks? What independence assumptions have you made?

Step-by-Step Solution

Answer

The probability that the total gross sales over the next 2 weeks exceed $\$5000$ is approximately \(1 - P(Z

Step 1: Determine the combined probability distribution of the sales over 2 weeks

Since the weekly sales are normally distributed and we assume independence, the sum of two normal random variables is also a normal random variable. The new mean will be the sum of the individual means, and the new variance will be the sum of the individual variances. For our problem, we have \(\mu = 2200\), \(\sigma = 230\), so for the sum of the sales over the next two weeks, the mean will be \(\mu_{total} = 2 \times 2200 = 4400\) and the variance will be \(\sigma^2_{total} = 2 \times 230^2\), so the standard deviation is \(\sigma_{total} = \sqrt{2 \times 230^2}\).

Step 2: Calculate the z-score

Now we can convert the problem to a z-score by subtracting the mean and dividing by the standard deviation. \(z = \frac{X - \mu_{total}}{\sigma_{total}} \) where X is 5000 in our case. So, we get: \( z = \frac{5000 - 4400}{\sqrt{2 \times 230^2}} \)

Step 3: Calculate the probability

To find the probability that the total sales exceed \(\$ 5000\), we want \(P(Z > z)\). We can find this probability using a standard normal distribution table or a calculator with a normal distribution function. In our case: \( P(Z > z) = 1 - P(Z < z) \) The resulting probability is the answer to part (a). II. Weekly sales exceed \( \$ 2000\) in at least 2 of the next 3 weeks

Step 4: Calculate the probability of exceeding \$2000 in a single week

First, we will calculate the z-score for a single week with \(X = 2000\), \(\mu = 2200\), and \(\sigma = 230\). \(z = \frac{X - \mu}{\sigma} = \frac{2000 - 2200}{230}\) Now, we will calculate \(P(Z < z)\) using a standard normal distribution table or a normal distribution calculator, and this will give us the probability of a single week's sales being below \( \$ 2000\). Let's call this probability p. The probability of a single week's sales being above \(\$ 2000\) will be \(1 - p\).

Step 5: Use a binomial model for 3 weeks

Now we will apply the binomial probability model to find the probability of at least 2 successful weeks out of 3. Let success mean a week's sales exceeding \( \$ 2000\). Calculate the probabilities: \(P(X = 2) = {3\choose2}(1-p)^2p\) \(P(X = 3) = (1-p)^3\) Now, add these two probabilities to get the probability of at least 2 successful weeks: \(P(X \geq 2) = P(X = 2) + P(X = 3)\) The resulting probability is the answer to part (b).

Step 6: Independence assumptions

In this exercise, we made the assumption that the sales in each week are independent events. This is a reasonable assumption as long as there are no external factors that could cause sales to be correlated across weeks, like seasonality or special promotions affecting multiple weeks.

Key Concepts

Normal Distribution

The normal distribution is a fundamental concept in probability and statistics. It is often used to model real-world phenomena where data tends to cluster around a central value. This distribution is characterized by its bell-shaped curve, which is symmetric about the mean.

In mathematics, a normal distribution is defined by two parameters: the mean \( \mu \) and the standard deviation \( \sigma \). The mean is the center of the distribution, while the standard deviation measures the spread of the data around the mean.

• A higher standard deviation indicates more spread out data.
• The total area under the normal distribution curve is 1, representing the complete probability space.

In practical applications like the one in the exercise, given weekly sales at a restaurant, a normal distribution helps to assess the likelihood of different sales outcomes.

Random Variable

A random variable is a variable that can take on various values, each with an associated probability. In the context of probability distributions, a random variable is used to quantify uncertain outcomes.

There are two types of random variables:

• Discrete random variables: These take on a countable number of distinct values, like the number of sales above a threshold in our exercise.
• Continuous random variables: These can take on any value within a certain range and are represented by smooth probability distributions like the normal distribution.

In the restaurant sales problem, weekly sales are treated as a continuous random variable because they can vary continuously from week to week. Understanding random variables is crucial for analyzing any problem involving uncertainty and probability.

Independence Assumption

The independence assumption is a key concept in probability theory. It states that two events are independent if the occurrence of one does not affect the probability of the occurrence of the other.

In the given exercise, the independence assumption allows us to treat each week's sales as unconnected to previous or subsequent weeks. Therefore:

• The probability distribution of total sales over two weeks is derived by simply adding the individual means and variances, assuming they are independent.
• This simplification allows for straightforward calculations using standard probability methods.

Recognizing when events are independent is important in statistical modeling since it impacts how probabilities are computed and combined.

Binomial Probability

The binomial probability concept applies when there are a fixed number of independent trials, each with two possible outcomes: success or failure. The probability of a specific number of successes is determined using the binomial formula.

In the exercise at hand, weekly sales exceeding a set amount is considered a 'success' in a sequence of weekly trials over 3 weeks. Here,

• Success probability: The chance of weekly sales exceeding $2000.
• Trials: The number of weeks (3 weeks in this case).
• The binomial model is applied to determine the probability that the sales exceed $2000 in at least two out of these three weeks.

This form of probability distribution is crucial for problems involving repeated trials with defined success criteria, such as predicting outcomes in weekly sales performance.

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Problem 28

Next

Problem 30

Other exercises in this chapter

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Problem 28

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Problem 31

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Key Concepts

Normal Distribution

The normal distribution is a fundamental concept in probability and statistics. It is often used to model real-world phenomena where data tends to cluster around a central value. This distribution is characterized by its bell-shaped curve, which is symmetric about the mean.

In mathematics, a normal distribution is defined by two parameters: the mean \( \mu \) and the standard deviation \( \sigma \). The mean is the center of the distribution, while the standard deviation measures the spread of the data around the mean.

• A higher standard deviation indicates more spread out data.
• The total area under the normal distribution curve is 1, representing the complete probability space.

In practical applications like the one in the exercise, given weekly sales at a restaurant, a normal distribution helps to assess the likelihood of different sales outcomes.

Random Variable

A random variable is a variable that can take on various values, each with an associated probability. In the context of probability distributions, a random variable is used to quantify uncertain outcomes.

There are two types of random variables:

• Discrete random variables: These take on a countable number of distinct values, like the number of sales above a threshold in our exercise.
• Continuous random variables: These can take on any value within a certain range and are represented by smooth probability distributions like the normal distribution.

In the restaurant sales problem, weekly sales are treated as a continuous random variable because they can vary continuously from week to week. Understanding random variables is crucial for analyzing any problem involving uncertainty and probability.

Independence Assumption

The independence assumption is a key concept in probability theory. It states that two events are independent if the occurrence of one does not affect the probability of the occurrence of the other.

In the given exercise, the independence assumption allows us to treat each week's sales as unconnected to previous or subsequent weeks. Therefore:

• The probability distribution of total sales over two weeks is derived by simply adding the individual means and variances, assuming they are independent.
• This simplification allows for straightforward calculations using standard probability methods.

Recognizing when events are independent is important in statistical modeling since it impacts how probabilities are computed and combined.

Binomial Probability

The binomial probability concept applies when there are a fixed number of independent trials, each with two possible outcomes: success or failure. The probability of a specific number of successes is determined using the binomial formula.

In the exercise at hand, weekly sales exceeding a set amount is considered a 'success' in a sequence of weekly trials over 3 weeks. Here,

• Success probability: The chance of weekly sales exceeding $2000.
• Trials: The number of weeks (3 weeks in this case).
• The binomial model is applied to determine the probability that the sales exceed $2000 in at least two out of these three weeks.

This form of probability distribution is crucial for problems involving repeated trials with defined success criteria, such as predicting outcomes in weekly sales performance.