Errata

In the equation of PDF, the square root of (2*pi) should be in the denominator.

Note from the Author or Editor:
Yes, this needs to be fixed. Correct formula:
www.gstatic.com/education/formulas2/553212783/en/normal_distribution.svg

I looked up the formula and it should be (including some HTML for the formatting:

<math display="block">
f(x;\mu,\sigma^2) = \frac{1}{\sigma\sqrt{2\pi}} e^{ -\frac{1}{2}\left(\frac{x-\mu}{\sigma}\right)^2 }.
</math>

This formula shows that the 2-Pi should be in the denominator.

This section is about finding the 95% confidence level for the sample mean of 64.408 being in line with the population mean. The mean should be the center of the plot of the normal distribution, in Figure 3.15. However, the figure shows the mean as 18. I don't know where this number came from.

Note from the Author or Editor:
Yes, this needs to be fixed. Please use this image:
imgur.com/a/4JPvv1k

Squared root is missing from the "Standard error of estimate" formula

Note from the Author or Editor:
Yes, this needs to be fixed. Math latex is:
S_e = \sqrt{\frac{\sum{(y-\hat{y})^2}}{n-2}}

A line is missing in the book due to which error "x is not defined appears". I added the line
x = symbols('x') and it worked.

I am talking about
Book: Essential math for data science by Thomas Neild
Page # 31
Example: 1 - 24

Note from the Author or Editor:
Yes, this needs to be fixed. Code is:

from sympy import *

x = symbols('x')

z = (x**2 + 1)**3 - 2
dz_dx = diff(z, x)
print(dz_dx)

# 6*x*(x**2 + 1)**2

Both in PDF and ePub.
This is more of a serious mathematical error/omission in hypothesis testing.
p98 should be "H0: population mean =18 ".
More importantly, in all the following Python codes and the examples, the standard deviation should have been divided by the square root of the sample size,
e.g p98 Example 3-19 should be corrected to
std_dev=1.5/sqrt(40).
The same applies to Examples 3-20,3-21,3-22.
The probabilities will be different and the conclusions of the tests as well.
Otherwise, really love the book and the examples provided.
thank you.

Note from the Author or Editor:
Initially this felt a little pedantic but I understand where they are coming from, considering I said "Past studies have shown that the mean recovery time for a cold is 18 days, with a standard deviation of 1.5 days, and follows a normal distribution.""I think we might be able to fix this by saying "The population of people with a cold has a mean recovery time of 18 days, with a standard deviation of 1.5 days, and follows a normal distribution.""

I think that will make examples work from that point forward.

stackoverflow.com/questions/48554161/why-divide-sample-standard-deviation-by-sqrtsample-size-when-calculating-z-sco

www.youtube.com/watch?v=JQc3yx0-Q9E

I might be wrong about it. From what I understood, Central Limit Theorem is saying that standard deviation of sample means (sampling standard deviation) is equal to the population standard deviation, divided by square root of sample size.
Books is saying - "sample standard deviation" instead of "sampling standard deviation".

Note from the Author or Editor:
Haha, man so much semantics. Let's just change that "sample standard deviation" formula to this latex:

\sigma_{\bar{x}} = \frac{\sigma}{\sqrt{n}}

i_hat and j_hat values are should be the other way around in the right hand example for it to work as a visualisation of example 4-9.

Note from the Author or Editor:
Corrected image here:

/i.imgur.com/njmVoze.png

The inverse matrix needs to have -4/3 in the right center, not -4/3.
Also applies to repeated depictions of this matrix on pages

Note from the Author or Editor:
That 4/3 in the two instances on page 137 needs to be -4/3.

Math latex for the first matrix should be:
\left[\begin{matrix}- \frac{1}{2} & 0 & \frac{1}{3}\\\frac{11}{2} & -2 & - \frac{4}{3}\\-2 & 1 & \frac{1}{3}\end{matrix}\right]

That also needs to be carried over into the second equation:
\left[\begin{matrix}- \frac{1}{2} & 0 & \frac{1}{3}\\\frac{11}{2} & -2 & - \frac{4}{3}\\-2 & 1 & \frac{1}{3}\end{matrix}\right]\left[\begin{matrix}4 & 2 & 4\\5 & 3 & 7\\9 & 3 & 6\end{matrix}\right]= \left[\begin{matrix}1 & 0 & 0\\0 & 1 & 0\\0 & 0 & 1\end{matrix}\right]

'x_0 + X_mean' part should be 'x_0 - X_mean'

Note from the Author or Editor:
change + to -

the second multiplier misses '1 - '

Note from the Author or Editor:
That 1.0 - needs to be put in the second multiplier just like on page 202

log is missing in the formula

Note from the Author or Editor:
A log() needs to be wrapped around each of the two multiplied expressions.

The parentheses around the subtraction of log likelihood fit and log likelihood is missing in the formula of the Chi Square value.
Should be: chi_2 = 2 * ( (log likelihood fit) - (log likelihood) )

Note from the Author or Editor:
Wrap parantheses around the whole expression following the "2"

In the third paragraph, the author mentioned that using 'stratify' option in scikit-learn can duplicate samples in the minority class until it is equally represented in the dataset. However, that is not the function of stratified split. Using 'stratify' will retain the same class distribution in the train set and test set as in the original dataset, not duplicating the samples.

Note from the Author or Editor:
Yup, I misunderstood this parameter as I don't use it often. Change "You can do this in scikit-learn as shown in Example 6-19 when doing your train-test splits. Pass the stratify option with the column containing the class values, and it will attempt to equally represent the data of each class." to "Another technique is shown in Example 6-19 when doing your train-test splits. Pass the stratify option with the column containing the class values, and it will keep the class distribution consistent between the train-test split."

Figure shows Logistic function as an Activation Function of the Output Layer, while Neural Network suppose to solve classification problem of 10 classes (numbers 0 - 9).

Note from the Author or Editor:
Replace that "s-shaped" curve chart on the top-right with this figure: imgur.com/3LuEjJy

The union probability answer that is given in the book is:

(1 - 0.3) + 0.4 - (.03 x 0.4) = 0.98

There are two serious mistakes here:

1) In the second part of the calculation, the subtraction of the joint probability, the probability of rain is 30%, or 0.3, but the author is multiplying against .03, or 3%. The answer with the calculation provided ends up being 1.088, which is above 1.0, so it's incorrect.
2) Even if you correct for the percentage, the calculation is still incorrect. The calculation for the Union probability is P(A OR B) = P(A) + P(B) - P(A AND B). In this exercise, A stands for (NOT RAIN), which the author calculates correctly for the P(A) part, but in the joint probability he reverts to using P(RAIN) instead of P(NOT RAIN).
The correct calculation should be: (1 - 0.3) + 0.4 - ((1 - 0.3) * 0.4) = 0.82

Note from the Author or Editor:
Change as advised. That line should be:
(1 - 0.3) + 0.4 - ((1 - 0.3) * 0.4) = 0.82

I don't know what I did there. It must have been late.