Errata

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.

The scale on the number line reads "100 200 300". It *should* read "1 2 3"

The legend reads "Represents 1000 Days" but it *should* reads "Represents 1000 Hours"

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

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

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.

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).

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.

Within the table:
Line 1: Var(Y) = E(Y – μ)2(squared)
Line 2: = Σ(y – μ)2(squared) P(Y = y). Note the first part Σ(y – μ) is using subtraction.
Line 3 is incorrect: = (– 2 + 0.85)2 × 0.977 + (23 + 0.85)2 × 0.008 etc, it is using addition where it should be using subtraction.
Line 3 should be: = (– 2 - 0.85)2 × 0.977 + (23 - 0.85)2 × 0.008 etc.

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.

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. :)

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

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

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

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.