It was said that a sample variance is calculated by dividing by n-1. Why is the sample standard deviation here not dividing by n-1?

It should be. That being said, in this case, it doesn't matter so much, as Sal is not asking about how the SD will behave when changing a number. So, we're really just interested in whether the numerator is increasing or decreasing, dividing by `n` or by `n-1` is only scaling that numerator. So, yes, the formula should have an `n-1` in the denominator, but having just `n` there will not change the answer

when using the formula of standard deviation that statitians use how do you know what the sign (negative or positive the answer to the devation is going to be?

This question is thoroughly answered if you click [Explain] below the last question in the article.

So the formula for MAD is (∑|x-mean|)/n ? Why not x_i ? I mean, the ith value in the data set.

There are two ways how we can define this sum. First, we can use indexing as you suggest. Second, we can use ∈ operator to show this is the sum of all elements x such that x ∈ S where S is the set of given data. Here the second way is used with hiding x ∈ S under the sigma symbol.

Main content

Course: Statistics and probability > Unit 3

Lesson 4: Variance and standard deviation of a population

Concept check: Standard deviation

Google Classroom

Six questions that will help you understand standard deviation more deeply.

Introduction

The questions below are designed to help you think deeply about standard deviation and its formula.

Unlike most questions on Khan Academy, some of these questions aren't graded by a computer. You'll learn the most if you try answering each question yourself before clicking "explain".

The formula (for reference)

The formula for standard deviation (SD) is

$SD = \sqrt{\frac{\sum_{}^{} | x - \bar{x} |^{2}}{n}}$ ‍

where

\sum

means "sum of",

x

is a value in the data set,

\bar{x}

is the mean of the data set, and

n

is the number of values in the data set.

Part 1

Consider the simple data set

{1, 4, 7, 2, 6}

How does the standard deviation change when $7$ ‍ is replaced with $12$ ‍?

How can we see this in the formula for standard deviation?

How does the standard deviation change?

The standard deviation increases because the data becomes more spread out:

How can we see this in the formula?

We can see this in the formula

$SD = \sqrt{\frac{\sum_{}^{} | x - \bar{x} |^{2}}{n}}$ ‍

because

\sum_{}^{} | x - \bar{x} |^{2}

is the sum of the squares of the distances from each data point to the mean. As the data, gets more spread out, the value of

\sum_{}^{} | x - \bar{x} |^{2}

increases.

I'm curious, what are the actual standard deviations of the data sets?

The standard deviation of

{1, 2, 4, 6, 7}

is approximately

2.28

The standard deviation of

{1, 2, 4, 6, 12}

is approximately

3.90

Part 2

Is it possible to create a data set with $4$ ‍ data points that has a standard deviation of $0$ ‍?

If it is possible, do it! Can you create two different data sets? How about three?

Yes, it's possible!

In fact, there are an infinite number of possible data sets.

Here's one:

$5, 5, 5, 5$ ‍

Here's another:

$8, 8, 8, 8$ ‍

Any data set where all of the data points are the same has a standard deviation of

0

because the distance from each data point to the mean is

0

Show me the calculation for ${5, 5, 5, 5}$ ‍.

Step 1: Find the mean

\bar{x} = \frac{5 + 5 + 5 + 5}{4} = \frac{20}{4} = 5

Step 2: Find the square of the distances from each of the data points to the mean

$x$ ‍		$\| x - \bar{x} \|^{2}$ ‍
$5$ ‍		$\| 5 - 5 \|^{2} = 0^{2} = 0$ ‍
$5$ ‍		$\| 5 - 5 \|^{2} = 0^{2} = 0$ ‍
$5$ ‍		$\| 5 - 5 \|^{2} = 0^{2} = 0$ ‍
$5$ ‍		$\| 5 - 5 \|^{2} = 0^{2} = 0$ ‍

Step 3: Apply the formula

\begin{aligned} SD & = \sqrt{\frac{\sum_{}^{} | x - \bar{x} |^{2}}{n}} \\ = \sqrt{\frac{0 + 0 + 0 + 0}{4}} \\ = \sqrt{\frac{0}{4}} \\ = \sqrt{0} \\ = 0 \end{aligned}

Part 3

Can standard deviation be negative?

Why or why not?
Hint: Think about the formula.

No, standard deviation cannot be negative!

To see why, think about the numerator and denominator inside the radical:

$SD = \sqrt{\frac{\sum_{}^{} | x - \bar{x} |^{2}}{n}}$ ‍

Notice how

n

is always positive. It's the number of data points, and we can't have a negative number of data points.

Also notice that

\sum | x - \bar{x} |^{2}

involves a quantity getting squared. Whenever we square something, we get a non-negative number.

Since both the denominator and numerator are positive, the entire expression must be positive too.

Part 4

Standard deviation is a measure of spread of a data distribution.

What do you think deviation means?

Part 5

Here are the formulas for standard deviation (SD) and the formula for mean absolute deviation (MAD), both of which are measures of spread:

$SD = \sqrt{\frac{\sum_{}^{} | x - \bar{x} |^{2}}{n}}$ ‍

$MAD = \frac{\sum_{}^{} | x - \bar{x} |}{n}$ ‍

What are the similarities between the formulas? What are the differences?

What are the similarities?

The formulas are very similar! They are both based on the distance from each data point to the mean

| x - \bar{x} |

, and they both include dividing by the number of data points

n

What are the differences?

The difference between the two formulas is that when calculating standard deviation, we square the distance from each data point to the mean, and we take the square root as the last step of the formula.

Which one is better?

Standard deviation is more complicated, but it has some nice properties that make it statisticians' preferred measure of spread.

Part 6

Here's the formula that we've been using to calculate standard deviation:

$\sqrt{\frac{\sum_{}^{} | x - \bar{x} |^{2}}{n}}$ ‍

Here's the formula that statisticians actually use:

$\sqrt{\frac{\sum_{}^{} (x - \bar{x})^{2}}{n}}$ ‍

Are the two formulas equivalent?

What's the difference between the formulas?

In the formula that we've been using, we take the absolute value of

x - \bar{x}

$\sqrt{\frac{\sum_{}^{} {| x - \bar{x} |}^{2}}{n}}$ ‍

In the formula that statisticians use, they put parentheses around

x - \bar{x}

$\sqrt{\frac{\sum_{}^{} (x - \bar{x})^{2}}{n}}$ ‍

Are the formulas equivalent?

Yes, both formulas are equivalent!

Statisticians realize that squaring will make the distance positive, so they don't bother using absolute value signs and just use parentheses instead.

For example, let's evaluate

| x - \bar{x} |^{2}

and

(x - \bar{x})^{2}

for

x = 2

and

\bar{x} = 5

$| x - \bar{x} |^{2} = | 2 - 5 |^{2} = | - 3 |^{2} = 3^{2} = 9$ ‍

$(x - \bar{x})^{2} = (2 - 5)^{2} = (- 3)^{2} = 9$ ‍

They're both positive! They're both

9

! They're equivalent!

Want to join the conversation?

Sort by:

Spencer Black
Posted 9 years ago. Direct link to Spencer Black's post “I agree with snowball1984...”
I agree with snowball1984...what makes this the preferred measure? It seems that there could also be another way to calculate deviation based on any way to calculate mean. Could we say that there is:
1) a linear mean deviation, [or mean absolute deviation,]
2) a standard deviation, [using the squared differences.]
3) a geometric deviation, [using the cubed differences,]
4) a harmonic deviation, [using a harmonic mean...
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(23 votes)
Answer
Tom
Posted 9 years ago. Direct link to Tom's post “It was said that a sample...”
It was said that a sample variance is calculated by dividing by n-1. Why is the sample standard deviation here not dividing by n-1?
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(11 votes)
Answer
- Dr C
  Posted 9 years ago. Direct link to Dr C's post “It should be. That being...”
  It should be.
  
  That being said, in this case, it doesn't matter so much, as Sal is not asking about how the SD will behave when changing a number. So, we're really just interested in whether the numerator is increasing or decreasing, dividing by n or by n-1 is only scaling that numerator. So, yes, the formula should have an n-1 in the denominator, but having just n there will not change the answer
  Button navigates to signup page
  (6 votes)
Jason Farrell
Posted 8 years ago. Direct link to Jason Farrell's post “It appears this and the p...”
It appears this and the previous module are not using the correct variance equation. They should use the population variance equation and not the sample variance equation. N and mu, not n and x-bar.
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(11 votes)
Answer
pierdolonymistrzmatematyki
Posted 7 months ago. Direct link to pierdolonymistrzmatematyki's post “No way, dude. That's craz...”
No way, dude. That's crazy. 💀💀💀💀
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(4 votes)
Answer
Jasmine Corona
Posted 8 years ago. Direct link to Jasmine Corona's post “when using the formula of...”
when using the formula of standard deviation that statitians use how do you know what the sign (negative or positive the answer to the devation is going to be?
Button navigates to signup pageButton navigates to signup page
(2 votes)
Answer
- Jordan Cooper
  Posted 7 years ago. Direct link to Jordan Cooper's post “This question is thorough...”
  This question is thoroughly answered if you click [Explain] below the last question in the article.
  Comment on Jordan Cooper's post “This question is thorough...”
  (5 votes)
SkulldyvanKhan
Posted a year ago. Direct link to SkulldyvanKhan's post “Does it make sense to cal...”
Does it make sense to calculate the standard deviation of a collection of complex numbers? That final question in Part 6 makes me wonder.
Button navigates to signup pageComment on SkulldyvanKhan's post “Does it make sense to cal...”
(3 votes)
Answer
- Chuck B
  Posted 4 months ago. Direct link to Chuck B's post “I tend to think of comple...”
  I tend to think of complex numbers as vectors that lie on the plane formed by the real and imaginary axes. I'd want to think about, what am I looking to accomplish or understand through a "standard deviation" of a set of vectors?
  Button navigates to signup page
  (1 vote)
The first integral proponent
Posted 6 years ago. Direct link to The first integral proponent's post “So the formula for MAD is...”
So the formula for MAD is (∑|x-mean|)/n ? Why not x_i ? I mean, the ith value in the data set.
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(2 votes)
Answer
- Vyacheslav Shults
  Posted 5 years ago. Direct link to Vyacheslav Shults's post “There are two ways how we...”
  There are two ways how we can define this sum. First, we can use indexing as you suggest. Second, we can use ∈ operator to show this is the sum of all elements x such that x ∈ S where S is the set of given data. Here the second way is used with hiding x ∈ S under the sigma symbol.
  Button navigates to signup page
  (2 votes)
Anna Cherdakova
Posted 7 years ago. Direct link to Anna Cherdakova's post “I am a bit confused: is S...”
I am a bit confused: is SD an average of how far the point of data are from each other or from the mean? if second - what is the formula for "how far the points of data are from each other on average"? thank you!
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(2 votes)
Answer
ilya112358
Posted 5 years ago. Direct link to ilya112358's post “Part 3 Can standard devia...”
Part 3
Can standard deviation be negative?
Hint: Think about the formula.

I thought about the formula and came up with the following.

Since symbol √ always means non-negative square root then SD cannot be negative.

The content of the Explain section then is irrelevant. In fact, it explains why variance cannot be negative which is a different question after all.

Correct me if I am wrong.
Button navigates to signup pageComment on ilya112358's post “Part 3 Can standard devia...”
(2 votes)
Answer
Cullen ONeill
Posted 7 months ago. Direct link to Cullen ONeill's post “I don't think the explana...”
I don't think the explanation in the section 'No, standard deviation cannot be negative!' is useful. It explains very well why what is under the square root sign cannot be negative, but that is not the same thing as the StdDev itself not being negative (just means it is not imaginary I think). I am of the impression that the reason the StdDev will not be negative is because we select the principal (non-negative) square root. Am I missing something, or am I correct in thinking that this section is not quite showing the right thing?
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(2 votes)
Answer
- John Montgomery
  Posted 7 months ago. Direct link to John Montgomery's post “It's not about the square...”
  It's not about the square root. It's because of the squaring of the distances from the mean - the "2" at the top right of the formula. Every square is positive (note that negative times negative is always positive).
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  (1 vote)