Head First Statistics

Errata for Head First Statistics

Submit your own errata for this product.


The errata list is a list of errors and their corrections that were found after the product was released.

The following errata were submitted by our customers and have not yet been approved or disproved by the author or editor. They solely represent the opinion of the customer.


Color Key: Serious Technical Mistake Minor Technical Mistake Language or formatting error Typo Question Note Update



Version Location Description Submitted By Date Submitted
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 
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 Z-score of X-bar should be 5 instead of 0.5 was incorrect. They were confusing the distribution of X-bar with that of X. The reason Var(X-bar) = 0.25 rather than 25 is that Var(X-bar) = Var(X)/n. So 25/100 = 0.25, and its positive square root is 0.5, which is the standard deviation of X-bar.

James  Jul 05, 2021 
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 482
Under the Binomial Distribution section

The section on the binomial distribution gives the incorrect distribution of X-bar. 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 X-bar. 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 X-bar, and presents it as the general case. Here' is the correct general formula for the distribution of X-bar, when X is binomially distributed, i.e. when X ~ B(n, p): X-bar ~ 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 "X-bar ~ 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 X-bar, 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 X-bar (i.e. "sample size for X-bar" meaning the number of independent observations of X that are averaged to get X-bar). Treating them as the same quantity gives an incorrect general formula for the distribution of X-bar. Let k be the sample size (i.e. the number of independent observations of the binomially-distributed random variable X). The central limit theorem tells us that if we have a big sample size we can model any X-bar (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 binomially-distributed X, and we plug the formula for mean and variance of the binomial distribution, we get: X-bar ~ N(np, npq/k). The book, however, says "X-bar ~ 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 X-bar for binomially-distributed X is that binomially-distributed 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 X-bar, since we are talking about the mean of a sample of binomially-distributed X, this is not one of those situations. If you consider binomially-distributed X to model the number of red gumballs in n-sized samples, then X-bar, 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 
Printed Page 590
3rd table

The 5th row of the 3rd table on the page specifies: 49 43.32 (49-43.32)²/43.32 = 5.68/43.32 = 0.131 49-43.23 equals 5.68, but the square was not applied. It should read: 49 43.32 (49-43.32)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 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 1-5, 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 2-6, 3-7, 4-8, 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 1-5. Multiplying 6 sequences by 10 horse/zebra orderings each equals the 60 given in the book.

Richard Mechaber  May 25, 2020 
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 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 646
Second paragraph, first bold word

"indpendent" should probably be "independent".

Anonymous  Jan 11, 2018 
Printed Page 539
First sentence after Second "A:"

"have0" is probably supposed to simply be "have".

Anonymous  Jan 10, 2018 
Printed Page 435
First paragraph, second sentence

I have not heard of "whilte" chocolate. Could it be white chocolate, instead?

Anonymous  Dec 27, 2017 
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(1-q^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 550
exercise solution

Z=(80-90)/3 should be changed to z=(80.5-90)/3

杜汉昌  May 28, 2016 
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 
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 
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 
Printed Page 594
step 4

prior errata for page 590 (failure to square numerator of one line of table) results in a chi-square 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 
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^(r-1) Pmax if r=1 P(X=1) = p * q^0 = p * 1 = p Kind regards, Fred

Anonymous  Apr 16, 2014 
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 379
Top of page

"...of an sdult is..." -> "...of an adult is..."

William Reardon  Aug 04, 2013 
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 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 
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 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 
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 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 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 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 re-written to lend a bit more helping hand to the reader.

Anonymous  Feb 24, 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 
Printed Page 264
bottom half

Woah! I tried this in late-afternoon 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 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 
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 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 1--that 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 2--that 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 
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 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 590
Bottom chart of solution - (O-E)^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 (O-E)^2/E is wrong. O-E = 49 - 43.32 = 5.68 (O-E)^2 = 5.68 * 5.68 = 32.262 (O-E)^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 
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 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 
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 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 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 at-bat 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 thought-provoking.

Anonymous  Sep 06, 2010 
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 extremely-low-probability events). Take, for example, the Hail-Mary pass.

Anonymous  Sep 06, 2010 
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 non-zero baseline. You've created the misleading impression that a non-zero baseline would be misleading.

Anonymous  Sep 06, 2010 
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 259
top-right

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 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 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 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 513
Using t-distribution 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 t-distribution 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 two-tailed 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 
Printed Page 293
bottom of the page

3!/(3-1)*0.25*0.5625 OK 6/2 * 0.0625*0.75 ????

Anonymous  Sep 15, 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 t-distribution is that the population distribution is normal _or_ non-normal, but in the answer to the first question on p. 518, it doesn't mention being able to use it when the population is non-normal. I'm not sure which is right.

Kathy Wolf  Jul 28, 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 
Safari Books Online 409
1st column, 1st and 2nd rows

Under both "X+Y" and "X-Y", the "N" for normal distribution is missing in Y ~ N ( mu_y , sigma^2_y )

Kathy Wolf  Jul 25, 2009 
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 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 
Safari Books Online 638-639
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=19.31-1.85x 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 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 
Safari Books Online 590
Last table

In the row corresponding to wins from Croupier B, the (O-E) part wasn't squared. So it should read, (49-43.32)^2 / 43.32 = 32.2624 / 43.32 = .745

Kathy Wolf  Jun 29, 2009 
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 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 
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 
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 178
In the second "Vital Statistics"

The text says "n events A1 to An", the formula uses A and A'. The formula should use &#931; (Sigma), or the text should say "two mutualy exclusive and exhaustive events A and A'".

Benjamin Poulain  Apr 04, 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