Version 
Location 
Description 
Submitted By 
Date Submitted 
Printed 
Page 6
SYP description of 2nd chart 
What's really going on might depend on something implicit in the context of the chart. For example, profit relative to a margin purchase, but showing the actual profits of the company, might easily justify a nonzero baseline. You've created the misleading impression that a nonzero baseline would be misleading.

Anonymous 
Sep 06, 2010 
PDF 
Page 7
2nd paragraph 
On page 7 where it shows the first pie chart of Manic Mango it says:
"The first thing the CEO wants to be able to do is compare the
percentage of satisfied players for each game genre. Hes started
off by plugging the data he has through some charting software,
and here are the results:"
The results of the first pie chart actually show unit sold per genre, not percentage of players satisfied per genre. Its the second pie chart on page 9 that shows the percentages.
Is this an error or have I read something wrong?

Anonymous 
Mar 12, 2013 
Printed 
Page 16
Sharpen your pencil solution 
In 2006 the Sports genre sold 15000 (according with the graph) instead of 14000, and so the strategy genre didn't sell more units than any other genre.

Hugo Juarez Corr 
Dec 19, 2010 
Printed 
Page 26
Small Diagram below the 6th paragraph. 
The scale on the number line reads "100 200 300". It *should* read "1 2 3"

Anonymous 
Feb 09, 2012 
Printed 
Page 28
Histogram 
The legend reads "Represents 1000 Days" but it *should* reads "Represents 1000 Hours"

Anonymous 
Feb 09, 2012 
Printed 
Page 33
Graph Label 
The graph "Hours Spent Gaming Per Day", has a label  "Box" represents 1000 Days. I think it should be 1000 Players. We just dont have data for Days.

Ashutosh Kar 
Feb 20, 2013 
PDF 
Page 61
Data sets on page 
The data set on page 61 for the ages of the students in the Kung Fu class don't match up with the ages listed for these students on pages 55 and 56.
This might seem pedantic, but as this is an explanation for some very basic concepts, consistency in the data is important and helps the reader/learner focus on the concept not the content.

Anonymous 
Mar 07, 2011 
Printed 
Page 80
Part b, lower right of page 
$10,000 x 1.1 should be $11,000 instead of $12,000

Anonymous 
Dec 02, 2010 
Printed 
Page 89
second data set 
In my edition of the book, the second data set is missing entirely. There is just blank space where it should appear. Using the chart to the right, I can figure out what the data should be, but I'd rather see it printed there. :)

Anonymous 
Jul 31, 2012 
Printed 
Page 96
Item 3. 
In the Exercise Solution on page 96, the answer to the question What's the interquartile range? Is described as the lower BOUND wubtreacted crom the upper BOUND. The next paragraph's formula reflects Interquartile range = upper BOUND  lower BOUND
all the references to BOUND should read as QUARTILE

Anonymous 
Dec 31, 2009 
Printed 
Page 116
Solution for player 3 
In the key to the exercise "Be the Coach Solution", the mean of 11 listed for Player 3 is inaccurate. It should be 10 (as indicated in the exercise introduction) :
2(3)+6+2(7)+3(10)+11+13+30 110
___________________________ = _____ = 10
11 11
The variance calculation should be corrected accordingly:
18+36+98+300+121+169+900
________________________ 100 = 64.2
10
Hence the corrected value for the Standard Deviation for Player 3 is 8.01 (the square root of 64.2)which is slightly higher than the 7.02 reported in the book.

Yvan Bamps 
Jul 22, 2009 
Printed 
Page 143
TANDQ 3rd ans. 
The return on an improbable investment might be qualitatively different (apart from the fact that expected value is inappropriate for extremelylowprobability events). Take, for example, the HailMary pass.

Anonymous 
Sep 06, 2010 
Printed 
Page 154
Left comment next to Venn diagram 
Key to exercise lists 50 as the total number of sport lovers, it is 60. 16 + 16+ 28 =60

Yvan Bamps 
Jul 23, 2009 
Printed 
Page 178
In the second "Vital Statistics" 
The text says "n events A1 to An", the formula uses A and A'.
The formula should use Σ (Sigma), or the text should say "two mutualy exclusive and exhaustive events A and A'".

Benjamin Poulain 
Apr 04, 2009 
Printed 
Page 181,182
Venn diagram image with think bubble 
The Venn diagram used doesn't match/represent the content and/or the think bubble. They need to be flipped. Image on page 181 should go to page 182 and vice versa.

Gowtham Raj 
Feb 03, 2020 
Printed 
Page 184
TANDQ 3rd answer 
It is a mischaracterization of independence that two spins of a wheel do not influence one another. You'd have to qualify the wheel as fair and fairly used; two spins, or even only one of them, might be subject to an unfair, perhaps common, influence. It would be possible to toss a fair coin unfairly, so as to cause it to always land heads, and if a coin lands heads the first time, since it is possible for a coin to be unfair, it is actually more likely (difficult to quantify, but there must be research in this area) for it to land heads the next time. Also, a baseball player's batting average tends to conceal streakiness; a hit one time makes a hit the next time more likely, but not necessarily because of any influence of one atbat on the next.
BTW, as of that point, you do not raise the question of whether or not three independent events every pair of which is independent might altogether be mutually independent. That would be thoughtprovoking.

Anonymous 
Sep 06, 2010 
Printed 
Page 185
Five Minute Mystery 
The problem states that there are 96 people questioned, and that 32 go to yoga (Y) and 72 go swimming (S). 24 people are exceptionally eager and go to both (B).
First, it is unclear (to me) if the 32 who go to yoga *only* go to yoga (Possibility 1), or if it is possible that they could also go to swimming (Possibility 2).
Suppose Possibility 1that those who report that they go to yoga only do yoga, those who report swimming only do swimming, and that there is a third group that does both. I would expect the following:
Y + S + B = 96
However, this is not the case.
Suppose Possibility 2that those who report that they go to yoga may do both, etc. Then I would expect the following:
(Y + S)  B = 96
However, this is also not true.
I feel like I'm missing something: my question is how are the circumstances of this problem possible?

Anonymous 
Sep 04, 2011 
Printed 
Page 201
Middle of page 
The sentence ending with "...at a glance what out gain will be." should read "...at a glance what the gain will be."

Anonymous 
Dec 25, 2010 
PDF 
Page 221
last table at bottom 
P(W = 4) has to be 0.09 instead of 0.18

Anonymous 
Jul 11, 2012 
Printed 
Page 236
In the table 
In the table, the expression for E(X^2) and the second expression for Var(X) are both correct, but I cannot see how to get them from anything in the text up to that point. I think you should add an explanation of why those 2 are correct.

Anonymous 
Aug 29, 2012 
Safari Books Online 
244
After the question 
Omits to assume that you get your initial bet back, which makes people calculate assuming x=3000

jonathanlim 
Aug 10, 2012 
Printed 
Page 256
Answer to exercise question 3 
The correct number of arrangements is:
5! / (3! 2!) = 10
As we just care about the species 5 camels finish the race consecutively:
hzhzhccccc
hzhhzccccc
hzzhhccccc
hhzhzccccc
hhzzhccccc
hhhzzccccc
zhzhhccccc
zhhzhccccc
zhhhzccccc
zzhhhccccc
Making the final answer:
10 / 2520 = 1 / 252

Anonymous 
May 17, 2019 
Printed 
Page 256
Answer to exercise question 3 
I believe the "error" posted by Anonymous for this page is incorrect. Anonymous stated:
"The correct number of arrangements is:
5! / (3! 2!) = 10
As we just care about the species 5 camels finish the race consecutively:
hzhzhccccc
hzhhzccccc..."
etc.
But the problem in the book did not state that the 5 camels must finish in positions 15, only that they had to "finish the race consecutively". There are alternative ways this could happen other than the outcomes Anonymous posted, where the camels finish in positions 26, 37, 48, etc.
There are 6 different finish orders possible where all 5 camels finish consecutively. Anonymous posted 10 different possible Horse and Zebra sequences for having the Camels finish in positions 15.
Multiplying 6 sequences by 10 horse/zebra orderings each equals the 60 given in the book.

Richard Mechaber 
May 25, 2020 
Printed 
Page 259
topright 
replace:
"If we multiply it by 17!/17!, this..."
with:
"If we multiply it by 20!/17!, this..."

Jonathan Schneider 
Mar 09, 2010 
Printed 
Page 264
bottom half 
Woah!
I tried this in lateafternoon when my head gets a little fuzzy, but even so, this exercise two with the 3 basketball players is a *TOUGHIE* !
You ought to either put something a bit simpler on this page, or else offer at least one hint if not more.
That problem 2 on the three expert players seems qualitatively different than all the other exercises and quizzes so far presented in the chapter.
Give the reader more of a helping hand???

Anonymous 
Feb 16, 2012 
Printed 
Page 265
all of it 
This page also (just as was true for 264) is a bit too tough.
Please come up with something easier, or else offer more hints.
Thanks. : )

Anonymous 
Feb 16, 2012 
Safari Books Online 
272
very last calculation on page 
It says "We can add these probabilities because they're independent." I think it should say "We can add these probabilities because they're mutually exclusive."

Kathy W. 
May 11, 2009 
Printed 
Page 272
top of page and also bottom calcs 
trouble number #1: It is confusing at best and misleading at worst to set things up this way. First of all, the whole example seems to be against the grain of what was taught in previous expectation value sections, e.g., expectations of winnings at Fat Dan's Casino.
It seems to me that expectations are always concerned with values and probabilites. So this problem comes across as very strange and unnatural, that we are not setting things up in the format of (say) X = zero = fail (with prob = .8), X = 1 = succeed (with prob=.2), and then proceed to get expectation from there.
trouble number #2: It may be true but it is extremely confusing and unmotivated to refer to the different skiing trials as "independent." After all, consider two rolls of the roulette wheel. In roll number two, we DO NOT CARE about what happened in roll number one. But this is NOT the case here: if there is success in skiing trial one, then we do NOT even execute trial two.
It is confusing to the reader why the two trials are still independent variables under such circumstances.
trouble number #3: At the bottom of the page, you refer to an intersection of two sets: failure in trial one and success in trial 2. Again, this seems very confusing. In the past, when such "trees" were employed, showing the various possible outcomes, things were quite different. See, e.g. page 159, entitled "Trees also help you calculate conditional probabilities." On that page it is pointed out that the far branches of a tree refer to conditional probabilties.
So, where does the intersection of sets come from at bottom of p 272?

Anonymous 
Feb 16, 2012 
PDF 
Page 278
2nd from the buttom up 
In my opinion, the sentence printed in bold letters
"the mode of any geometric distribution is always 1,..."
should be
"the mode of any geometric distribution is always p,..."
explanation:
P(X=r) = p * q^(r1)
Pmax if r=1
P(X=1) = p * q^0 = p * 1 = p
Kind regards,
Fred

Anonymous 
Apr 16, 2014 
Printed 
Page 280
4th and 5th paragraphs 
I believe that the notations and words are so close to one another that confusion is hard to avoid.
And "Head First" Books are usually eminently good at avoiding ALL confusions!
specifically, the notation xP(X=x) is VERY VERY close to xP(X<=x)
the reader can all too easily slip into confusion in your paragraph which begins "The values of xP(X=x) start off slow and then..."
I am not saying that your overall strategy is poor, just that this part might be rewritten to lend a bit more helping hand to the reader.

Anonymous 
Feb 24, 2012 
Printed, PDF 
Page 280ff
The tables shown 
The tables claim that x(P(X) ≤ x) is equal to the sum of x(P(X) = x)  which it isn't.
If you Plot out x(P(X) ≤ x) (defined as x(1q^r)), it does not converge. It diverges into infinity.
In the converging graph, it says that the sum of x(P(X) ≤ x) is converging to five, but it's as well wrong (the sum of x(P(X) = x) is converging to five).
The same error is repeated on variance calculation.

Anonymous 
May 04, 2017 
Printed 
Page 281
top two paragra 
The following may be perfectly correct, but the book is making essentially the same error that other books make: you're not making the concepts clear and obvious enough
(Which "Head First" books are usually so fantastically good at doing...)
What is confusing is this: why would we consider many many, or possibly infinitely many trial runs, and then find information regarding what is true after the FIFTH run?
Doesn't that seem like something that might befuddle the new reader?

Anonymous 
Feb 24, 2012 
Printed 
Page 293
bottom of the page 
3!/(31)*0.25*0.5625 OK
6/2 * 0.0625*0.75 ????

Anonymous 
Sep 15, 2009 
Printed 
Page 313
Combine Poisson Variables section 
In order for the following relations to be true:
P(X+Y)=P(X)+P(Y)
E(X+Y)=E(X)+E(Y)
X+Y~Po(lambda_x+lambda_y)
shouldn't be required that the two variables be 'mutally exclusive' too?

Anonymous 
Sep 21, 2008 
Printed 
Page 313
First equation on page 
The probability for X and Y is listed as:
P(X + Y) = P(X) + P(Y)
Shouldn't this be:
P(X + Y) = P(X) * P(Y)
I understand that lambdaX + lambdaY = 5.7
Also, P(X=0) for lambda = 3.4 is 0.033 and
P(Y=0) for lambda = 2.3 is 0.100
Thus, P(X=0) * P(Y=0) = 0.003, which matches
P(X+Y=0) when lambda is 5.7, which is also 0.003 (as on p314).

mudphone 
Jun 18, 2009 
PDF 
Page 313
First paragraph 
Combine Poisson variables
You saw in previous chapters that if X and Y are independent random
variables, then
P(X + Y) = P(X) + P(Y)
E(X + Y) = E(X) + E(Y)
>>>>> should be corrected to
Combine Poisson variables
You saw in previous chapters that if X and Y are independent random
variables, then
E(X + Y) = E(X) + E(Y)
Var(X + Y) = Var(X) + Var(Y)
>>>>> and perhaps an additional paragraph before declaring that it's also poisson
something to the effect of showing the student that the Expectation and Variance of X+Y are again equal, fulfilling one criteria of a Poisson Distribution.

Aditya Liviandi 
Jan 25, 2011 
Printed 
Page 343
top paragraph 
variance cannot have same units as the quantity itself.
the mean is 71 inches, so that variance must have units of inches squared, not inches.

Anonymous 
Feb 29, 2012 
Printed 
Page 366
Solution #4 
Solution #4 "Var(X + Y) = Var(X) + Var)(Y)" should be "Var(X  Y) = Var(X) + Var)(Y)"

Anonymous 
Jan 15, 2010 
Printed 
Page 379
Top of page 
"...of an sdult is..." > "...of an adult is..."

William Reardon 
Aug 04, 2013 
Safari Books Online 
409
1st column, 1st and 2nd rows 
Under both "X+Y" and "XY", the "N" for normal distribution is missing in Y ~ N ( mu_y , sigma^2_y )

Kathy Wolf 
Jul 25, 2009 
Safari Books Online 
411
next to last row, last column 
In the last column, under "Condition" it lists npq > 5 as the 2nd condition, when it should just be nq > 5.

Kathy Wolf 
Jul 25, 2009 
Printed 
Page 435
First paragraph, second sentence 
I have not heard of "whilte" chocolate. Could it be white chocolate, instead?

Anonymous 
Dec 27, 2017 
Safari Books Online 
466
2nd paragraph of middle column under "there are no dumb questions" 
It says "The expectation of the sampling population of proportions..." I believe it should say "The expectation of the sampling distribution of proportions..."

Anonymous 
Jun 04, 2009 
Printed 
Page 468
Entire page 
The original question is:
25% of the gumball population are red. What’s the probability that in a box of 100 gumballs, at least 40% will be red? We’ll guide you through the steps
My question is:
Couldn't we directly use the binomial distribution to solve this question instead of using the sampling distribution of proportions?
We are looking for P(X >= 40).
n = 100
p = 0.25
q = 0.75
We could use B(100, 0.25). Since np and npq are greater than 5, we could even use the normal approximation of the binomial to simplify the calculation.
Why use the sampling distribution of proportions instead of the binomial?

Anonymous 
Mar 10, 2015 
Printed 
Page 482
Under the Binomial Distribution section 
The section on the binomial distribution gives the incorrect distribution of Xbar. This section confuses the concept of the number of Bernoulli trials parameterizing X ~ B(n, p) with the number of independent observations of X (AKA sample size) parameterizing our distribution of Xbar. It uses the variable n to refer to both concepts, laying out the very specific case where the number of trials parameterizing the distribution of X is equal to the sample size parameterizing the distribution of Xbar, and presents it as the general case.
Here' is the correct general formula for the distribution of Xbar, when X is binomially distributed, i.e. when X ~ B(n, p):
Xbar ~ N(np, npq/k), where k is the sample size, and n is the number of Bernoulli trials parameterizing the distribution of X ~ B(n, p).
The book says "Xbar ~ N(np, pq)" which in general is incorrect, regardless of whether you're using n to denote sample size (AKA the number of independent observations of X going into each Xbar, which I refer to as k), or to denote the number of Bernoulli trials parameterizing the distribution of X. It's only true in the very specific case when n happens to be equal to k, so the formula is incorrect generally.
Longer explanation:
The text in the binomial distribution section confuses two different concepts, and uses the variable n to refer to both. Specifically, n is being used both as the number of Bernoulli trials parameterizing the binomial distribution of X, and as the sample size for Xbar (i.e. "sample size for Xbar" meaning the number of independent observations of X that are averaged to get Xbar). Treating them as the same quantity gives an incorrect general formula for the distribution of Xbar.
Let k be the sample size (i.e. the number of independent observations of the binomiallydistributed random variable X).
The central limit theorem tells us that if we have a big sample size we can model any Xbar (regardless of the distribution of X) as N(mean_of_X, variance_of_X / k).
Let n be the number of trials parameterizing the binomial distribution of X, so that X ~ B(n, p). If we take a sample of k independent observations of this binomiallydistributed X, and we plug the formula for mean and variance of the binomial distribution, we get:
Xbar ~ N(np, npq/k).
The book, however, says "Xbar ~ N(np, pq)" which in general isn't true. As you can see from the formula, it will only be true in the specific case when n = k, or in other words, the formula given in the book is only true when the number of Bernoulli trials just so happens to be equal to the sample size (to be clear: sample size means the number of independent observations of the binomially distributed random variable X).
What makes it easy to get confused on the topic of finding the distribution of Xbar for binomiallydistributed X is that binomiallydistributed X can actually itself model a sample, because a binomially distributed random variable can be thought of a "sample" of Bernoulli trials. So, to be clear, there are some situations where you can model sample size as the number of Bernoulli trials parameterizing a binomial distribution, but in this case where we want the distribution of Xbar, since we are talking about the mean of a sample of binomiallydistributed X, this is not one of those situations. If you consider binomiallydistributed X to model the number of red gumballs in nsized samples, then Xbar, in turn, would be what you get when you take k random samples of X ("k samples of n samples of gumballs"), and calculate, for your k samples of n gumballs, on average, how many red gumballs you got out of n total gumballs.

James 
May 23, 2021 
Safari Books Online 
507
Answer to last question in first column 
The answer is about the difference between the terms confidence level and confidence interval. The first sentence should read "The confidence level is the probability that your statistic is contained within the confidence interval," instead of "the confidence interval..."

Kathy Wolf 
Jun 19, 2009 
Printed 
Page 507
3rd answer 
It should read "A: The confidence level is the probability", not "The confidence interval".

Anonymous 
Mar 21, 2010 
Printed 
Page 511
2nd paragraph 
The expectation of X bar is mu, and the standard deviation is sigma/n
it should be sigma/sqrt(n)

Anonymous 
Jul 08, 2009 
Printed 
Page 513
Using tdistribution probability tables 
The first graphic on the page shows the two tails, measuring 0.025 each. Then the next graphic shows the single tail used to find the value of t, which is great  and implies that for our data p=0.025.
In the first paragraph in the "Using tdistribution probability tables" you state that p=0.0025.
What follows in an example, but for the example you use v=7 and p=0.05. This is a little confusing. I think most people will wonder if our p=0.025 is related to this new p=0.05. Since the first graphic shows two tails, this example might mean that you multiply the p times 2.
In future editions I would explain the one and twotailed difference. I would change the example to use a p that is wildly different than 0.025 and 0.05. This will allow the reader to clearly understand that the lookup value is different from the current example (which he/she will have to look up for the following page's excercise).

Anonymous 
Oct 24, 2009 
Safari Books Online 
515 or 518
Table at bottom of 515, first question on 518 
In the table at the bottom of p. 515, it says that one of the conditions for using the tdistribution is that the population distribution is normal _or_ nonnormal, but in the answer to the first question on p. 518, it doesn't mention being able to use it when the population is nonnormal. I'm not sure which is right.

Kathy Wolf 
Jul 28, 2009 
Printed 
Page 518
Exercise Solution 
This error (if it is one) appears at several locations in the chapter.
The formula for calculating the confidence interval is x +/ c * sqrt( s / n). However in several places (including here it's show as ( c * s) / sqrt(n).
The formula is used correctly when values are plugged, but *not* in the description with variables.
Hopefully this comes across  I can't submit this errata with latex. :)

Anonymous 
Oct 14, 2013 
Printed 
Page 539
First sentence after Second "A:" 
"have0" is probably supposed to simply be "have".

Anonymous 
Jan 10, 2018 
Printed 
Page 550
exercise solution 
Z=(8090)/3 should be changed to z=(80.590)/3

杜汉昌 
May 28, 2016 
Printed 
Page 564
Solution Step 2 
The standard deviation used in step 2 of the solution is listed as sqrt(.25)=0.5. However, on the previous page the distribution of the syrup in the bottles was listed as N(355,25), meaning the standard deviation should be 5. This changes the answers in part 2 of the exercise.

Anonymous 
Dec 01, 2009 
Printed 
Page 564
Under Step 3 
The critical value is reported as 2.32, but a closer estimate from the probability table is 2.33. The true value is between 2.32 and 2.33, but it is closer to 2.33.

James 
Jul 05, 2021 
Printed 
Page 564
Solution step 2 
This is just to clarify that despite the reported "minor technical mistake," there is no error in Step 2 on this page. The person who said the standard deviation used to calculate the Zscore of Xbar should be 5 instead of 0.5 was incorrect. They were confusing the distribution of Xbar with that of X. The reason Var(Xbar) = 0.25 rather than 25 is that Var(Xbar) = Var(X)/n. So 25/100 = 0.25, and its positive square root is 0.5, which is the standard deviation of Xbar.

James 
Jul 05, 2021 
Safari Books Online 
590
Last table 
In the row corresponding to wins from Croupier B, the (OE) part wasn't squared. So it should read, (4943.32)^2 / 43.32 = 32.2624 / 43.32 = .745

Kathy Wolf 
Jun 29, 2009 
Printed 
Page 590
Bottom chart of solution  (OE)^2/E 
In the solution shown on p 590, the line for Croupier B Wins
shows the observed frequency as 49, the expected frequency as 43.32.
The solution for the (OE)^2/E is wrong.
OE = 49  43.32 = 5.68
(OE)^2 = 5.68 * 5.68 = 32.262
(OE)^2/E = 32.262/43.32 = 0.745
Not the 0.131 that is shown in my book.
This brings the overall total at the bottom to 5.618
not the 5.004 shown.

Anonymous 
Jan 30, 2011 
Printed 
Page 590
3rd table 
The 5th row of the 3rd table on the page specifies:
49 43.32 (4943.32)²/43.32 = 5.68/43.32 = 0.131
4943.23 equals 5.68, but the square was not applied. It should read:
49 43.32 (4943.32)²/43.32 = 32.2624/43.32 = 0.745
This would also change the sum at the bottom of the page to 5.618

Robert Kearney 
Feb 02, 2021 
Printed 
Page 594
step 4 
prior errata for page 590 (failure to square numerator of one line of table) results in a chisquare value of 5.618
Therefore, the long solution at step 4 should also be changed from 5.004 to 5.618

Peter Partch 
Feb 05, 2015 
Printed 
Page 603
3rd line in the table, result in the last line of the table, result below the table and in Step 5. 
The X² result for A Black is given as s"0.005", but it should be 0.052 instead. This leads to the X² sum of 1.583, which should be 1.63 instead.

Susanne 
Sep 03, 2015 
PDF 
Page 623
line 21 
the mean of y should be 38.875,here all typo 38.75
and the first group of (y  mean of y) should be 16.875
122.53 is wrong. should be 122.84
so the result of b and a of "y= a+ bx" all wrong next pages.
use Excel I got
Coefficients
Intercept 15.7283193
X Variable 1 5.336410535

Anonymous 
Jul 14, 2011 
PDF 
Page 626
Bottom part of page 
It says "you can use it to predict the value of y, given a value b"
Shouldn't it say "you can use it to predict the value of y, given a value x"?

Anonymous 
Feb 19, 2015 
Safari Books Online 
638639
Exercise Solution 
I know this has already been mentioned in the errata, but in case, like me, you didn't notice the error until you had done the whole exercise already with the (original) data set on p. 636:
n=9
xbar=5
ybar=10.06
b=1.85
a=19.31
So y=19.311.85x
r=.92
If Swindler exposes Captain Amazing to radiation for 5 minutes, we would expect Captain Amazing to be able to lift 10.06 tons.

Kathy Wolf 
Jul 09, 2009 
Printed 
Page 646
Second paragraph, first bold word 
"indpendent" should probably be "independent".

Anonymous 
Jan 11, 2018 
Printed 
Page 651
First sentence 
It is claimed on this page that "You've seen how you can find confidence intervals for [the mean] and [variance]", but that overstates what the book covered regarding how a confidence interval for variance is computed. The only time that particular topic is specifically addressed at all is in the Q & A on page 491: "Theoretically, yes [it's possible to construct a confidence interval for the variance], but we haven’t covered the sampling distribution for the variance, and we’re not going to."

James 
Jul 17, 2021 