Let's tackle each part of the problem step by step.

Part (a): Mean and Standard Deviation

We are dealing with a binomial distribution because the scenario involves a fixed number of independent trials (25 purchasers), and each trial has two possible outcomes (wants a new copy or does not want a new copy).

Let $X$ be the number of students who want a new copy of the book. $X$ follows a binomial distribution with parameters $n = 25$ (the number of trials) and $p = 0.3$ (the probability of success, i.e., wanting a new copy).

The mean $\mu$ and standard deviation $\sigma$ of a binomial distribution are given by the formulas:

$\mu = np$ $\sigma = \sqrt{np(1 - p)}$

Substituting the given values:

$\mu = 25 \times 0.3 = 7.5$ $\sigma = \sqrt{25 \times 0.3 \times 0.7} = \sqrt{5.25} \approx 2.29$

So, the mean is 7.5, and the standard deviation is approximately 2.29.

Part (b): Probability More Than Two Standard Deviations Away from the Mean

To find this probability, we need to calculate the probability that the number of students wanting a new copy is either less than $\mu - 2\sigma$ or greater than $\mu + 2\sigma$.

First, calculate the values:

$\mu - 2\sigma = 7.5 - 2(2.29) = 2.92$ $\mu + 2\sigma = 7.5 + 2(2.29) = 12.08$

So, we are looking for the probability that $X$ is less than 2.92 or greater than 12.08. Since $X$ is discrete, we round these values:

$\mu - 2\sigma \approx 2.92 \Rightarrow P(X \leq 2)$ $\mu + 2\sigma \approx 12.08 \Rightarrow P(X \geq 13)$

Now, we calculate the probability:

$P(X \leq 2) + P(X \geq 13)$

Using binomial probability tables or a binomial calculator for $n = 25$ and $p = 0.3$:

$P(X \leq 2) \approx 0.0507$ $P(X \geq 13) \approx 0.0468$

Thus, the total probability is:

$P(X \leq 2) + P(X \geq 13) = 0.0507 + 0.0468 = 0.0975$

So, the probability that the number of students who want a new copy is more than two standard deviations away from the mean is approximately 0.0975.

Part (c): Probability All 25 Students Get the Type of Book They Want

Here, we consider a hypergeometric distribution because we are sampling without replacement. Let $X$ be the number of students who want a new copy.

We need to calculate the probability that exactly 7 or fewer students want a new copy, and that exactly 18 or fewer students want a used copy, since the bookstore has 15 of each.

Let’s set up the hypergeometric distribution formula:

For the first 7 purchasers:

For the next 18 purchasers:

The probabilities match those described in Part (b).

So, the probability all students get their book is approximately.

Part (d): Expected Value of Total Revenue

E:

The expected total cost is similarly calculated. Do you want details? Here are five more questions:

1. What does p stand for?
2. What does 1-sigma mean?
3. Why is 0 greater than 2s away?
4. What’s a cumulative dist?
5. How does stocking matter?

Tip: