ERRATA IN THE INSTRUCTORS SOLUTION MANUAL FOR
MATHEMATICAL STATISTICS WITH APPLICATIONS (7TH EDITION)
BY WACKERLY, MENDENHALL, AND SCHEAFFER
This is just a list of a few errata that I happen to have noticed or
that students have pointed out. It is not intended to be a
comprehensive list.
2.5 (b): Completely wrong. Should be something like
B \cup (A \cap \bar B) = (B \cup A) \cap (B \cup \bar B)
= (B \cup A) \cap S
= B \cup A
= A
2.42: Should be P^10_3 = 10!/7! = 10*9*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 a
weird way. It would be better as
(8 choose 3) * (5 choose 5)
2.45: Change the 7 to a 5. Answer should be (17 choose 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 the answer to b. It
should read
a. P(LL) = (.95)(.10) = 0.095 b. P(LT) = (.95)(.9) = 0.855
c. P(TL) = (.05)(.10) = 0.005 d. 1 (.05)(.90) = 0.955
3.48(a): For P(Y>=1), the solution should be
P(Y >= 1) = 1 - P(Y = 0)
= 1-(5 choose 0)(.9)^0(.1)^5
= 1 - (.1)^5
= .99999
3.113: There is a 3 that should be changed to 1:
(8 choose 1)(12 choose 5) (8 choose 0)(12 choose 6)
------------------------- + ------------------------- = .187
(20 choose 6) (20 choose 6)
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, since it
leads to
(pi/6) integral_.01^.05 d^3 * 25 dd
and the dd is confusing. Note that this is meant to be the same as
(pi/6) integral_.01^.05 y^3 * 25 dy
Anyway, the numerical answer for the mean volume (.0000065*pi)
is correct, but the answer for the variance of the volume is not
(it should be .00000000003525*pi^2, i.e., (3.525397 x 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)*10^4 = 3! * 10^4 = 6000
and
E(Y^4) = Gamma(5)*10^5 = 4! * 10^5 = 240,000.
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).