by Jim Bohan with Beth Chance
Amsco School Publications, Inc
 

Please email your corrections to mathmanjfb@aol.com or bchance@calpoly.edu

(page numbers refer to location of error)

p. 19, second paragraph

The formula for the weighted mean should read 

p. 82, problem 2

(B) P(A|B) = P(A)
(C) P(A|B)=P(B)

                 Last line should read (30/36)(30/36)(6/36) = 25/126

p. 137, 3rd line from end of example 5

Should read P(X=4 and Y=0) =^? P(X=4)^.P(Y=0)

The AMC-12 has a total of 150 points but to qualify for the AIME test, you must score 100 or more.

Probability of finding a special prize in a box of popcorn should be .20.

Change failure rate to 20% to match answer in back.

 In item II, replace "may" by "must" to read "The standard deviations of two different samples from the same population must be the same."

Change the problem to read "Find the minimum sample size so that the probability that a sample mean is less than 55 (for this distribution with mean 57 and standard devation 5.6) equals .01."  See corrections to solution as well.

Should read:

"The probability that samples of these sizes have a difference in means of 2.8 or more..."

The mean overhead cost for the pizza stores should be $764.50, not $763.50.

margin of error is 2.1, not 2.1%; use n=30 weeks as the sample of the population to compute the margin of error.

formula approximating degrees of freedom should be df=min(n₁,n₂) -1

change (C) to read "The 99% confidence interval does not contain the mean specified by the null hypothesis"

 In the confidence interval formula, t₈ for 90% confidence should equal 1.860 instead of 1.397.

the units on s_mpg should be mi/gal

Answer A should read "For a fixed p, the variance increases as n increases".

Change (D) to read "We expect the 95% confidence interval of the sample result would contain the expected population value."

Change the question to read "In a test of significance to determine the validity of the company's claim, why is it inappropriate to use the normal approximation to the binomial? Explain your answer."

Change the question to read "Estimate the approximate life of the car in the sample whose purchase price was $25,000."

Part (e) was not included in the scoring guide.

Answer C should read "Consecutive values of a discrete variable may or may not differ by a fixed(constant) amount."

Change answer A to read "Honesty of the respondent".

Start question as "In a test of the equality of the proportions..."

Answer C should read "bimodal".

The equation should read  = 10.2 + 4.15x

(A) Long distance service is predicted to cost 15 cents per minute.
(B) The monthly fixed cost of long distance service is, on average, $4.70

Add "The restaurant further claims that you have a chance to win on every visit."

Note, that if the z-score is not rounded in the middle of the calculation, then the student must increase his score by 9.02 points.  To ensure improvement (not just staying the same) then this could be rounded up, resulting in a different answer.

Change explanation to read that the percentile of a z-score of 1 is approximately 84%

The minimum is at 32.

Using height to predict weight, the regression equation is: weight = -91.9 + 3.25 height

The y-intercept should refer to the value of -91.9 not -58.

The correct prediction is -91.9+3.24(65)=119.2

Answer should be D, as the only option that is not a correct conclusion.

Answer should be E, none of these lines of fit would produce a residual plot similar to the given residual plot.

Choice should read D, explanation is correct.

P(A|B) = P(A) if A and B are independent
P(A|B) = P(B) if P(A and B) = [P(B)]²

Answer should read D.  One explanation: P(A|B) = P(A and B)/P(B) = .5/.3 > 1, not a legitimate value.  In fact, P(A and B) must always be less than (or equal to) P(B) for this fraction to be < 1.  Similarly, need P(A and B) to be less than or equal to P(A).

P(sum=8 and both even) = 3/36, should answer is 5/36 + 9/36 - 3/36 = 11/36

4/52(4/51) = .006 not .06.

z = -1.28 = (38-50)/s  or z = 1.28 = (62-50)/s

Note: to obtain the probabilities in (a), we use the fact that the random variables are independent as stated in the problem.

E(2L+M) = 5(.01)+6(.01)+7(.04)+8(.08)+9(.07)+10(.15)+11(.10)+12(.22)+13(.08)+14(.24) = 11.3 = 2E(L)+E(M)
V(2L+M) = (5-11.3)²(.01) + (6-11.3)²(.01) + (7-11.3)²(.04) + (8-11.3)²(.08) + (9-11.3)²(.07) + (10-11.3)²(.15) + (11-11.3)²(.10) + (12-11.3)²(.22) + (13-11.3)²(.08) + (14-11.3)²(.24)  = 5.01
= 4V(L)+V(M) since the two random variables are independent
SD(2L+M)=sqrt(5.01) = 2.238

The z value given is incorrect, should be -.253. For z=-.253, any value of n will work.  If the problem statement is changed as suggested above (value of n so probability that sample mean is less than 55 is .01) the answer is n>42.56 or 43.

Explanation should being P(₁-₂> 2.8)...

final statement should read P(>2.8| m=2.4)

P(boys’ sample times would be at least 4.9 seconds faster than girls’ times if the...)

The correct answer should be "The 99% confidence interval does not contain the mean specified by the null hypothesis"

The demoninator of the z statistic should be square root of 1/6(5/6)/100

Ha: (words) There is a difference by gender in science placement exam performance in 1988

z-statistic for a two-sided test would have to exceed 1.96. So n> (1.96)²(.25)/(.03)² = 1068

p-value=P(>65 or < 59 if m=62)

2(.25) = .5 (end of p-value calculation)

3. Check of Assumptions: For sample sizes, the smallest is 50(.40) = 20 > 5
7. Test statistic: =[.4(100)+.44(50)]/150 = .4133, so then z = -.469
8. Sketch should shade -.469 and below
9. p-value = .319

Technically if you allow the two points to fall at the same x value, then r is undefined and if they have the same y value, then r=0.

Answers should be A. Explanation:  Since there is no replacement of the kings in the deck, the maximum number of rotten fruit is 4
for every shipment of 24 fruit.  Since the cards are dealt without replacement, the probability of a rotten fruit is not constant.

Answer should be D (explanation is correct)

Answer should be B.  Answer D is correct since the two-sided p-value would be .032. We would fail to reject the hypothesized value at the .05 level, indicating that the hypothesized value is contained in a 95% confidence interval. Note, this is exactly true when analyzing means, but you could see a discrepancy between the CI and test with proportions since a different formula is used for the standard deviation.

Since 14.5 is the claimed value for the population standard deviation, we could recommended using the t test (with s=12.5 hours) instead of the z test. However, in order to use either the z-test or the t-test here, we would need to know/assume/believe that the population distribution is normal. There is no indication of this in the problem statement.  This leaves E as the correct answer.

Part (e) was not included in the scoring guide.  However, answers will vary. Most students will obtain an interval that contains m which should not surprise them - 95% of all intervals will contain m. An interval that does not contain m would be surprising.

Add to the explantion "On the other hand, the numbers of houses in a set of housing developments may be any set of integers, consecutive values of which do not have to differ by a fixed amount."

                    Answer is D

p. 382, problem 15

"... gives no insight into the form of the alternative..."

Add to the answer: Since these are discrete data, could instead interpret in terms of the "average deviation from the expected payment."  This would also help in comparing these results to another benefit plan payout scheme.
The data are somewhat symmetric, so applying the empirical rule is not too aggregious.