## Related questions with answers

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 25 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. [Hint: Let h(X) = the revenue when X of the 25 purchasers want new copies. Express this as a linear function.]

Solution

VerifiedLet

$X = \text{number of student who wants new copy}.$

Since $30\%$ students want a new copy, $p=0.3$ and there are $n=25$ purchasers. This means that $X \sim Bin(25,0.3)$ (Binomial Distribution). $\textcolor{#19804f}{\textbf{(a):}}$

$\text{\textcolor{#4257b2}{\textbf{Proposition:}}}$ For a binomial random variable X with parameters n, p, and $q=1-p$, the following is true

$\begin{equation*} \begin{split} &E(X) = np; \\ &V(X) = np(1-p)=npq;\\ &\sigma_X = \sqrt{npq}. \end{split} \end{equation*}$

From the proposition, the following holds

$\begin{equation*} \begin{split} \textcolor{#4257b2}{E(X)} = np = 25 \cdot 0.3 = \textcolor{#4257b2}{7.5}. \end{split} \end{equation*}$

Also, we have

$\begin{equation*} \begin{split} \textcolor{#4257b2}{V(X)} = np(1-p) = 25 \cdot 0.3 \cdot 0.7 = \textcolor{#4257b2}{5.25}, \end{split} \end{equation*}$

which means that Standard Deviation is

$\begin{equation*} \begin{split} \textcolor{#4257b2}{\sigma_X} = \sqrt{V(X)} = \sqrt{5.25} = \textcolor{#4257b2}{2.291} \end{split} \end{equation*}$

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