# Mathematical Statistics With Applications (7th Edition) by Wackerly, Mendenhall, and Scheaffer

Errata in the Exercises and the Solutions Manual

## What Is This About?

This started as a list of errata that either I or my students had noticed in the exercises and the solutions manual for the 7th edition of Wackerly, Mendenhall, and Scheaffer's math stat text.

This list was posted as a plain ASCII text file on my old web pages and since I took those pages down I have been surprised by the number of students and teachers who either wanted a copy of the list, or who had questions or comments about other exercises or solutions in the text. So I decided to repost them here as a public service. I hope that I have not introduced any errors in the process of converting them to org-mode. If you prefer the original text files you can grab them here (exercises) and here (solutions).

NB: I last taught from this text in 2008, not because I have any problem with the text itself, but because in the intervening time I have not been assigned to teach my department's undergrad math stat sequence. I have not checked whether any of these errata have been corrected in later printings.

## Some Errata Found in the Exercises

This is just a list of a few errata that I happen to have noticed. It is not intended to be a comprehensive list.

### Exercise 2.79

"$$P(B) < 1$$" should be "$$P(B) > 0$$".

### Exercise 2.80

Change "$$P(B) > 0$$" to "$$P(B) < 1$$".

Since $$A$$ is a subset of $$B$$, the assumption that $$P(A) > 0$$ already implies that $$P(B) > 0$$, so it is redundant to assume that $$P(B) > 0$$. If, on the other hand, $$P(B) = 1$$, then $$A$$ and $$B$$ are necessarily independent, and the objective of the exercise is to show that $$A$$ and $$B$$ must be dependent except in the uninteresting case that one or the other of the events has probability zero or one.

### Exercise 2.106

I would change this to ask "what is the smallest possible value for $$P(A)$$" or something similar. "What is $$P(A)$$?" gives the impression that one should be able to divine the exact value of $$P(A)$$ from the information given, which isn't true.

### Exercise 3.38

This is the same as the triangular taste test example that I give in class, except that here it is not carefully worded. The statement

Suppose that the two formulas are equally attractive.

should be changed to

Suppose that the two formulas are indistinguishable.

Why?

If the two formulations are equally attractive on average but distinguishable, say by everyone, then on the average half of all tasters would choose the glass containing formula B. (The other half would choose one of the two glasses containing formula A.) So, if the two formulas are "equally attractive" but distinguishable by everyone, then the distribution of $$Y$$ would be binomial on $$n = 4$$ trials with success probability $$p = 1/2$$.

If the two formulas are "equally attractive" and can be distinguished by some but not all tasters, then we would need more information to do the problem.

With the change I suggest above ("indistinguishable" rather than "equally attractive" formulas), each taster is effectively picking at random among the three glasses and $$Y$$ is binomial on $$n = 4$$ trials with success probability $$p = 1/3$$. This is what was actually intended and is the solution given in the solutions manual.

### Exercise 5.85

This exercise should probably be starred, since it refers to exercise 5.65, which is starred.

### Exercise 5.94

The word "uncorrelated" should be deleted from this problem. Nothing in the problem changes if $$Y_1$$ and $$Y_2$$ are correlated.

## Some Errata Found in the Solutions Manual

### 2.5 (b)

This is completely wrong. It should be something like

\begin{align} B \cup (A \cap \bar B) &= (B \cup A) \cap (B \cup \bar B) \\ &= (B \cup A) \cap S \\ &= B \cup A \\ &= A \end{align}

### 2.42

Should be $$P^{10}_3 = 10!/7! = 10 \times 9 \times 8 = 720$$.

### 2.44 (b)

Change $$3$$ to $$3!$$ and $$45$$ to $$90$$.

Also, the solution to 2.44 (a) is correct but expressed in an odd way. It would be better as $$\binom{8}{3} \times \binom{5}{5}.$$

### 2.45

Change the 7 to a 5. Answer should be $$\binom{17}{2\ \ 5\ \ 10} = 408,408$$.

### 2.59 (b)

There are actually 10 different kinds of straights (starting with any of A, 2, 3, …, 10).

### 2.65

Should be $$5!(2/6)(1/6)^4 = 5/162$$.

### 2.114

The answers are listed as parts a,b,b,d (two b's and no c), and there are a couple of misprints in the answer to b. It should read

\begin{align} & \text{a. } P(LL) = (.95)(.10) = 0.095 && \text{b. } P(LT) = (.95)(.9) = 0.855 \\ & \text{c. } P(TL) = (.05)(.10) = 0.005 && \text{d. } 1 ­ (.05)(.90) = 0.955 \end{align}

### 3.48(a)

For $$P(Y>=1)$$, the solution should be

\begin{align} P(Y >= 1) &= 1 - P(Y = 0) \\ &= 1- \binom{5}{0}(0.9)^0(0.1)^5 \\ &= 1 - (0.1)^5 \\ &= 0.99999 \\ \end{align}

### 3.113

There is a 3 that should be changed to 1:

$$\frac{\binom{8}{1}\binom{12}{5}}{\binom{20}{6}} + \frac{\binom{8}{0}\binom{12}{6}}{\binom{20}{6}} = 0.187.$$

### 4.57

The density of $$D$$ is wrong here. It is not equal to $$1/4$$ on the interval $$(.01, .05)$$, but to $$1/(.05 - .01) = 25$$. Also, the choice of $$d$$ as the dummy variable of integration is unfortunate here, because it leads to

$$\frac{\pi}{6} \int_{.01}^{.05} 25\, d^3 \,dd$$

and the $$dd$$ is confusing. Note that this is meant to be the same as

$$\frac{\pi}{6} \int_{.01}^{.05} 25\, y^3 \,dy.$$

Anyway, the numerical answer for the mean volume ($$.0000065\times\pi$$) is correct, but the answer for the variance of the volume is not (it should be $$0.00000000003525\,\pi^2$$, i.e., $$(3.525397 \times 10^{-11})\,\pi^2$$.

### 4.74 (d)

This part has been left unfinished. Having found that the lower quartile of test scores is $$73.98$$, we are supposed to find what proportion of test takers score $$73.98 + 5 = 78.98$$ or more. Using the normal table, the final answer is $$.4364$$ (because of rounding, this is slightly different from the answer you would get if if you did the whole exercise using a computer, $$.4369$$).

### 4.92

In the calculation of $$E(Y^3)$$ and $$E(Y^4)$$, the 100's in the integrals should be 10's, and the results should be $$E(Y^3) = \Gamma(4)\times 10^4 = 3! \times 10^4 = 6000$$ and \begin{equation*} E(Y^4) = Γ(5)× 10^5 = 4! × 10^5 = 240,000. \end{equation*}

### 5.9 (b)

The integral is set up incorrectly, but the numerical answer $$31/64$$ is correct.

### 6.14

Change $$P(Y_1 > u/Y_2)$$ to $$1 - P(Y_1 > u/Y_2)$$.

##### Brett Presnell
###### Associate Professor of Statistics

My research interests include nonparametric and computationally intensive statistics, model misspecification, statistical computing, and the analysis of directional data.