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Chapter 4: Problem 16

Betting on dinner, Part I. Suppose a restaurant is running a promotion where prices of menu items are random following some underlying distribution. If you're lucky, you can get a basket of fries for \(\$ 3\), or if you're not so lucky you might end up having to pay \(\$ 10\) for the same menu item. The price of basket. of fries is drawn from a normal distribution with mean \(\$ 6\) and standard deviation of \(\$ 2 .\) The price of a fountain drink is drawn from a normal distribution with mean \(\$ 3\) and standard deviation of \(\$ 1 .\) What is the probability that you pay more than \(\$ 10\) for a dinner consisting of a basket of fries and a fountain drink?

Short Answer

Expert verified
The probability of paying more than $10 is approximately 0.327.

Step by step solution

01

Understand the Problem

To find the probability that you pay more than $10 for a meal consisting of fries and a drink, we need to consider the sum of their prices. Each item's price follows a normal distribution.
02

Define the Distributions

The price of the basket of fries, \( F \), follows a normal distribution with mean \( \mu_F = 6 \) and standard deviation \( \sigma_F = 2 \). The price of the fountain drink, \( D \), follows a normal distribution with mean \( \mu_D = 3 \) and standard deviation \( \sigma_D = 1 \).
03

Determine the Distribution of the Total Cost

The total cost, \( T \), is the sum of the prices of fries and the drink: \( T = F + D \). Since both \( F \) and \( D \) are normally distributed, their sum \( T \) is also normally distributed with mean \( \mu_T = \mu_F + \mu_D = 6 + 3 = 9 \) and variance \( \sigma_T^2 = \sigma_F^2 + \sigma_D^2 = 2^2 + 1^2 = 5 \). So, the standard deviation of the total cost is \( \sigma_T = \sqrt{5} \).
04

Calculate the Probability

Using the properties of the normal distribution, the probability that the total cost \( T \) is more than \$10 is given by \( P(T > 10) \). We standardize this to a standard normal distribution Z: \( Z = \frac{T - \mu_T}{\sigma_T} \). Thus, \( P(T > 10) = P\left(Z > \frac{10 - 9}{\sqrt{5}}\right) = P\left(Z > \frac{1}{\sqrt{5}}\right) \).
05

Use Z-tables or Software to Find Probability

Determine the probability \( P(Z > \frac{1}{\sqrt{5}}) \). Using standard normal distribution tables or software, \( Z \approx 0.4472 \); hence, \( P(Z > 0.4472) \approx 0.327 \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Probability
Probability is a measure of the likelihood of a certain event occurring. It's a fundamental concept in statistics and helps us understand how likely various outcomes are. In this exercise, we're looking at the probability of paying more than a certain amount for a meal with fries and a drink.
Key points about probability include:
  • Expressed as a number between 0 and 1. A probability of 0 means the event is impossible, while a probability of 1 means it is certain.
  • In the context of a normal distribution, probability is often calculated by standardizing to a standard normal distribution.
To find the probability that the total cost exceeds $10, we standardize the total cost to a normal distribution. We then use tools like Z-tables or software to find the corresponding probability value.
Standard Deviation
Standard deviation is a measure that quantifies the amount of variation or dispersion in a set of values. When prices of menu items vary due to their random nature, the standard deviation helps us understand how spread out these prices are from the mean.
  • A smaller standard deviation means values tend to be closer to the mean, indicating less variability.
  • A larger standard deviation means values can be far from the mean, indicating more variability.
In our example, the basket of fries has a standard deviation of $2, while the fountain drink has a standard deviation of $1. These figures tell us how much the price of these items can differ from their average price. To find the variability in the total cost, we use the formula for combining standard deviations, since the total cost is the sum of two normally distributed items.
Mean
The mean, often referred to as the average, is the central value of a set of numbers. It is calculated by summing all the individual values and dividing by their count. In relation to our dining scenario, it gives us the average expected cost of each menu item.
  • The mean price of the fries is set at $6.
  • The mean price of the drink is $3.
  • The total mean price for the meal is the sum of these means, calculated as $6 + $3 = $9.
This tells us that, on average, you would expect to pay $9 for the fries and drink combined. Knowing the mean allows us to anticipate what a typical price would be, which is crucial in gauging how likely it is to encounter prices as high as $10.
Random Variables
Random variables are quantities whose values result from the outcome of a random phenomenon. They are used to model situations where outcomes are uncertain, like the price of our combined meal consisting of fries and a drink.
  • In our scenario, both the price of the fries and the drink are considered random variables.
  • These variables follow a normal distribution, indicating their occurrences cluster around a central mean.
  • When you add two independent normally distributed random variables, the result is also normally distributed.
The total cost of the meal is a new random variable formed by summing the individual costs. Knowing these variables follow a normal distribution allows us to calculate probabilities and other statistical measures, which inform us about potential spending outcomes.

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Most popular questions from this chapter

CAPM. The Capital Asset Pricing Model (CAPM) is a financial model that assumes returns on a portfolio are normally distributed. Suppose a portfolio has an average annual return of \(14.7 \%\) (i.e. an average gain of \(14.7 \%\) ) with a standard deviation of \(33 \%\). A return of \(0 \%\) means the value of the portfolio doesn't change, a negative return means that the portfolio loses money, and a positive return means that the portfolio gains money. (a) What percent of years does this portfolio lose money, i.e. have a return less than \(0 \% ?\) (b) What is the cutoff for the highest \(15 \%\) of annual returns with this portfolio?

Stats final scores. Each year about 1500 students take the introductory statistics course at a large university. This year scores on the final exam are distributed with a median of 74 points, a mean of 70 points, and a standard deviation of 10 points. There are no students who scored above 100 (the maximum score attainable on the final) but a few students scored below 20 points. (a) Is the distribution of scores on this final exam symmetric, right skewed, or left skewed? (b) Would you expect most students to have scored above or below 70 points? (c) Can we calculate the probability that a randomly chosen student scored abave 75 using the normal distribution? (d) What is the probability that the average score for a random sample of 40 students is above \(75 ?\) (e) How would eutting the sample size in half affect the standard deviation of the mean?

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Chickenpox, Part II. We learned in Exercise 3.32 that about \(90 \%\) of American adults had chickenpox before adulthood. We now consider a random sample of 120 American adults. (a) How many people in this sample would you expect to have had chickenpox in their childhood? And with what standard deviation? (b) Would you be surprised if there were 105 people who have had chickenpox in their childhood? (c) What is the probability that 105 or fewer people in this sample have had chickenpax in their childhood? How does this probability relate to your answer to part (b)?

Betting on dinner, Part II. Exercise 4.16 introduces a promotion at a restaurant where prices of menu items are determined randomly following some underlying distribution. We are told that the price of basket of fries is drawn from a normal distribution with mean 6 and standard deviation of 2. You want to get 5 baskets of fries but you only have \(\$ 28\) in your pocket. What is the probability that you would have enough money to pay for all five baskets of fries?
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