91Ó°ÊÓ

Suppose that \(30 \%\) of all students who have to buy a text for a particular course want a new copy (the successes!), whereas the other \(70 \%\) want a used copy. Consider randomly selecting 25 purchasers. a. What are the mean value and standard deviation of the number who want a new copy of the book? b. What is the probability that the number who want new copies is more than two standard deviations away from the mean value? c. The bookstore has 15 new copies and 15 used copies in stock. If 25 people come in one by one to purchase this text, what is the probability that all 25 will get the type of book they want from current stock? [Hint: Let \(X=\) the number who want a new copy. For what values of \(X\) will all 15 get what they want?] d. Suppose that new copies cost \(\$ 100\) and used copies cost \(\$ 70\). Assume the bookstore currently has 50 new copies and 50 used copies. What is the expected value of total revenue from the sale of the next 25 copies purchased? Be sure to indicate what rule of expected value you are using.

Step by step solution

01

Calculate Mean and Standard Deviation

Given that the probability of a student wanting a new copy is \( p = 0.3 \), and we are considering \( n = 25 \) purchasers, the number of students wanting a new copy follows a binomial distribution. The mean value \( \mu \) is calculated by \( \mu = n \times p \). Therefore, \( \mu = 25 \times 0.3 = 7.5 \). The standard deviation \( \sigma \) is given by \( \sigma = \sqrt{n \times p \times (1-p)} = \sqrt{25 \times 0.3 \times 0.7} = 2.29 \).

02

Determine Probability of Being More Than Two Standard Deviations Away

We need to find the range of \( X \) where the number of new copies is more than two standard deviations away from the mean. So, calculate the interval: \([\mu - 2\sigma, \mu + 2\sigma] = [7.5 - 2\times2.29, 7.5 + 2\times2.29] = [2.92, 12.08]\). Therefore, \( X \) will be more than two standard deviations away if \( X < 3 \) or \( X > 12 \). Use the binomial distribution to find \( P(X < 3) + P(X > 12) \). Calculating these probabilities involves summing the probabilities for \( X = 0, 1, 2 \) and \( X = 13, 14, \, \dots, 25 \).

03

Probability That All 25 Get the Type of Book They Want

The bookstore has enough new copies if \( X \leq 15 \), as there are 15 new books available and enough used ones for the remainder. Calculate the probability \( P(X \leq 15) \) from the binomial distribution. This can be found by \( P(X = 0), P(X = 1), \ldots, P(X = 15) \) and summing these probabilities.

04

Calculate Expected Revenue

Define \( X \) as the number of new copies sold. The expected revenue \( E(R) \) is given by the sum of the expected revenue from new and used copies. Using \( E(X) = 7.5 \) from Step 1, and \( Y = 25 - X \) for used copies: \[ E(R) = E(100X + 70Y) = 100E(X) + 70E(25-X) = 100 \times 7.5 + 70 \times 17.5 \]. Calculate to find \( E(R) = 750 + 1225 = 1975 \). This result uses the linearity of expectation.

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Key Concepts

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

Mean and Standard Deviation

To grasp the concepts of the mean and standard deviation in a binomial distribution, think of these terms as your initial stepping stones in the journey of probability. In this scenario, we are dealing with 25 students trying to buy new copies of a textbook, where the probability of each choosing a new copy is 0.3. The situation aligns perfectly with a binomial distribution.
The **mean** or average value in such a distribution is a simple calculation: you multiply the number of trials ( ext{ number of students}) by the probability of success ( ext{ wanting a new copy}). Hence, in this case: \( \mu = n \times p = 25 \times 0.3 = 7.5 \).
Moving on to the **standard deviation**, it reflects the variation or spread of the distribution from the mean. It is computed using the formula: \( \sigma = \sqrt{n \times p \times (1-p)} \), resulting in \( \sigma = \sqrt{25 \times 0.3 \times 0.7} = 2.29 \). The standard deviation helps us understand the extent of deviation expected from the mean in our binomial experiment.

Probability Calculation

Probability calculations in a binomial distribution can help affirm expectations or uncover surprising outcomes. Let’s explore the concept by examining how likely it is for the number of students wanting a new copy to differ significantly from the average, specifically by being more than two standard deviations away.
We established that the mean number of students wanting a new copy is 7.5, with a standard deviation of 2.29. We can examine the range \([\mu - 2\sigma, \mu + 2\sigma]\) to find values between which we expect most outcomes to lie: \(([7.5 - 2 \times 2.29, 7.5 + 2 \times 2.29]) = [2.92, 12.08]\).Here, the probability query targets outcomes outside this range, countably extreme events that happen less frequently.

• Calculate P(X < 3) and P(X > 12) using the binomial distribution.
• These computations require summing up values for each relevant outcome through equation inputs: \(P(X = 0, 1, 2) \) and \(P(X = 13, 14, \ldots, 25)\).

These calculations offer insights into how likely deviations beyond normal expectations occur, reinforcing the importance of probability in assessing potential risks or regularities in real-world scenarios.

Expected Value

The concept of expected value provides the forecast of average outcomes in probability and statistics. Here, it pertains to the bookstore’s revenue expectations from textbook sales. Imagine the bookstore planning financially for the demand of new and used copies.
The expected value of the revenue, denoted as E(R), accounts for the probable outcomes of selling new and used books, given certain costs (\(100 for a new copy, \)70 for a used copy).
How does it work? By combining the expected revenues from both book types:

• Let X represent the number of new copies sold, calculated as E(X) = 7.5 from earlier mean calculations.
• Complement this with the computation for used copies: Y being 25 - X (total purchases minus new copy sales) reflecting the binomial constraint.

The formula derived from the linearity of expectations aids here: \[ E(R) = 100E(X) + 70E(25-X) = 100 \times 7.5 + 70 \times 17.5\],predicting that the bookstore will average $1975 in revenue from selling these 25 textbooks. Thus, expected value elegantly predicts likely monetary outcomes from known probabilities.

Linearity of Expectation

Linearity of expectation is a fundamental principle that allows us to simplify complex probability problems, especially when dealing with sums of random variables. The power of this principle lies in its simplicity: the expected value of a sum is the sum of the expected values.
In our bookstore scenario, where new and used book costs vary, the linearity of expectation enabled us to separate and easily compute expected revenues from each type independently, and just sum them up.

• The expected revenue from new copies was calculated as \(100 \times E(X)\).
• Similarly, the expected revenue from used copies was \(70 \times (25 - E(X))\).

Together: \(E(R) = 100E(X) + 70E(25-X)\).Without requiring complex multivariable integrations or interrelations, linearity makes it straightforward.
This simplicity is a boon in numerous scenarios, establishing predictability across numerous independent factors. The bookstore effectively anticipates total revenues separately and conjointly, cheering for efficient financial planning.