Standard Deviation gives you a sense of how spread out the data is from the mean. In other words, it shows you the variation of the sample which you are analyzing.

StdDev informs us the amount of variation from the mean

Most of the data points will fall within 1.044 away from the mean value of 2.9 tickets.

this would depend on the distribution

What is the difference between variance and standard deviation? Why is standard deviation the square root of the variance? Thanks!

Variance is spread. If you see the way we calculate the variance basically we are changing the unit of variable to variable square that is the area. While Standard deviation basically tells you by how much value the variable is deviated from expected value.

It seems like the standard deviation is similar to the mean difference from the mean--mean average deviation--but since you square everything and then put it through a square root, it comes out slightly differently. What is the purpose behind that? Why is it better to use standard deviation with the mean as opposed to MAD?

Taking the square root of the sum of squares for the standard deviation results in the following behavior; a few points that are very far from the mean will increase the standard deviation more than many points deviating slightly from the mean. Example: The mean average deviations for both of the sets {2, 2, 6, 6} and {0, 8, 4, 4} equal 2. However, the standard deviation for the first set is 2 and the standard deviation for the second set is 2.828. In the first set, all of the points deviate slightly from the mean. In the second set, a couple of points deviate largely from the mean. This results in the second standard deviation being larger than the first, even though they share the same mean average deviation.

Number of tickets bought by the most of customers (about 68% of total customers) will fall between <mean>-<st dev> and <mean>+<st dev>. Or 2.9-1.044 <= number of tickets by customer <= 2.9+1.044

Remember that 68% would be if the distribution is normal, which it isn't. I'm not sure if my method is correct, but I think that if we say that a person buying 𝑛 tickets is actually buying at least (𝑛 − 0.5) tickets and less than (𝑛 + 0.5) tickets, then we can view the distribution as if it were continuous. The fraction of customers that are within 1 standard deviation of the mean is equal to the area under the distribution curve between 𝑛 = 1.856 and 𝑛 = 3.944, which would be (2.5 − 1.854)⋅0.3 + 0.2 + (3.944 − 3.5)⋅0.4 = 0.5714 or about 57% Then again, this would only make sense if customers could actually buy fractions of tickets.

Main content

Course: AP®︎/College Statistics > Unit 8

Lesson 2: Mean and standard deviation of random variables

Mean and standard deviation of a discrete random variable

Q: Number of tickets bought by the most of customers (about 68% of total customers) will fall between <mean>-<st dev> and <mean>+<st dev>. Or 2.9-1.044 <= number of tickets by customer <= 2.9+1.044

Remember that 68% would be if the distribution is normal, which it isn't. I'm not sure if my method is correct, but I think that if we say that a person buying 𝑛 tickets is actually buying at least (𝑛 − 0.5) tickets and less than (𝑛 + 0.5) tickets, then we can view the distribution as if it were continuous. The fraction of customers that are within 1 standard deviation of the mean is equal to the area under the distribution curve between 𝑛 = 1.856 and 𝑛 = 3.944, which would be (2.5 − 1.854)⋅0.3 + 0.2 + (3.944 − 3.5)⋅0.4 = 0.5714 or about 57% Then again, this would only make sense if customers could actually buy fractions of tickets.

Google Classroom

Practice calculating and interpreting the mean and standard deviation of a discrete random variable.

Example: Ticket sales for a concert

Organizers of a concert are limiting tickets sales to a maximum of

4

tickets per customer. Let

T

be the number of tickets purchased by a random customer. Here is the probability distribution of

T

$T = # of tickets$ ‍	$1$ ‍	$2$ ‍	$3$ ‍	$4$ ‍
$P (T)$ ‍	$0.1$ ‍	$0.3$ ‍	$0.2$ ‍	$0.4$ ‍

Question 1

Calculate the expected value of $T$ ‍.

μ_{T} =

tickets

Question 2

Choose the best interpretation of the mean (or expected value) that you found in the previous question.

Question 3

Calculate the standard deviation of $T$ ‍.
Round your answer to three decimal places.

σ_{T} =

tickets

For reference, here is the distribution again:

$T = # of tickets$ ‍	$1$ ‍	$2$ ‍	$3$ ‍	$4$ ‍
$P (T)$ ‍	$0.1$ ‍	$0.3$ ‍	$0.2$ ‍	$0.4$ ‍

What do you think is the best interpretation of the standard deviation that you found in the previous question?

Feel free to discuss this in the comments! Keep in mind that standard deviation describes how far away from the mean data points tend to fall.

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Nguyen Nguyen
Posted 6 years ago. Direct link to Nguyen Nguyen's post “Standard Deviation gives ...”
Standard Deviation gives you a sense of how spread out the data is from the mean. In other words, it shows you the variation of the sample which you are analyzing.
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(44 votes)
Answer
- Paul Shekoski
  Posted 6 years ago. Direct link to Paul Shekoski's post “StdDev informs us the amo...”
  StdDev informs us the amount of variation from the mean
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  (6 votes)
Naoya Okamoto
Posted 5 years ago. Direct link to Naoya Okamoto's post “Most of the data points w...”
Most of the data points will fall within 1.044 away from the mean value of 2.9 tickets.
Button navigates to signup pageComment on Naoya Okamoto's post “Most of the data points w...”
(23 votes)
Answer
- prithvi.dhyanii
  Posted a year ago. Direct link to prithvi.dhyanii's post “this would depend on the ...”
  this would depend on the distribution
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  (3 votes)
katieleonard032
Posted 5 years ago. Direct link to katieleonard032's post “What is the difference be...”
What is the difference between variance and standard deviation? Why is standard deviation the square root of the variance? Thanks!
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(5 votes)
Answer
- Kunal Kumar Gupta
  Posted 4 years ago. Direct link to Kunal Kumar Gupta's post “Variance is spread. If yo...”
  Variance is spread. If you see the way we calculate the variance basically we are changing the unit of variable to variable square that is the area. While Standard deviation basically tells you by how much value the variable is deviated from expected value.
  Button navigates to signup page
  (6 votes)
mirahbob
Posted 4 years ago. Direct link to mirahbob's post “It seems like the standar...”
It seems like the standard deviation is similar to the mean difference from the mean--mean average deviation--but since you square everything and then put it through a square root, it comes out slightly differently. What is the purpose behind that? Why is it better to use standard deviation with the mean as opposed to MAD?
Button navigates to signup pageButton navigates to signup page
(4 votes)
Answer
- KLaudano
  Posted 4 years ago. Direct link to KLaudano's post “Taking the square root of...”
  Taking the square root of the sum of squares for the standard deviation results in the following behavior; a few points that are very far from the mean will increase the standard deviation more than many points deviating slightly from the mean.
  
  Example: The mean average deviations for both of the sets {2, 2, 6, 6} and {0, 8, 4, 4} equal 2. However, the standard deviation for the first set is 2 and the standard deviation for the second set is 2.828. In the first set, all of the points deviate slightly from the mean. In the second set, a couple of points deviate largely from the mean. This results in the second standard deviation being larger than the first, even though they share the same mean average deviation.
  Comment on KLaudano's post “Taking the square root of...”
  (7 votes)
elliots
Posted a year ago. Direct link to elliots's post “If many many tickets are ...”
If many many tickets are purchased, the amount of tickets purchased will typically vary by about 1.044 tickets from the mean of 2.9 tickets.
Button navigates to signup pageComment on elliots's post “If many many tickets are ...”
(6 votes)
Answer
Sandra Malaquias
Posted 5 years ago. Direct link to Sandra Malaquias's post “If we look at a large num...”
If we look at a large number of customers, then 68% customer, will have purchased between 1.856 and 3.94 tickets (corresponding 1.044 bellow and above the mean)
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(4 votes)
Answer
Aleksandr Nokhrin
Posted 2 years ago. Direct link to Aleksandr Nokhrin's post “Number of tickets bought ...”
Number of tickets bought by the most of customers (about 68% of total customers) will fall between <mean>-<st dev> and <mean>+<st dev>.
Or 2.9-1.044 <= number of tickets by customer <= 2.9+1.044
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(2 votes)
Answer
- Jerry Nilsson
  Posted 2 years ago. Direct link to Jerry Nilsson's post “Remember that 68% would b...”
  Remember that 68% would be if the distribution is normal, which it isn't.
  
  I'm not sure if my method is correct, but I think that if we say that a person buying 𝑛 tickets is actually buying at least (𝑛 − 0.5) tickets and less than (𝑛 + 0.5) tickets, then we can view the distribution as if it were continuous.
  
  The fraction of customers that are within 1 standard deviation of the mean is equal to the area under the distribution curve between 𝑛 = 1.856 and 𝑛 = 3.944,
  which would be (2.5 − 1.854)⋅0.3 + 0.2 + (3.944 − 3.5)⋅0.4 = 0.5714
  or about 57%
  
  Then again, this would only make sense if customers could actually buy fractions of tickets.
  Comment on Jerry Nilsson's post “Remember that 68% would b...”
  (3 votes)
BeeGee
Posted 4 years ago. Direct link to BeeGee's post “I used the 1-Var Stats fu...”
I used the 1-Var Stats function of the TI-84 Plus CE, and it gave me 0.1104 for the standard deviation. My List was {0.1, 0.3, 0.2, 0.4}, and my FreqList was just {1, 2, 3, 4}.
Did I do something wrong, or am I supposed to multiply σx by 10 to get the correct answer?

EDIT:

7 months later: I found out that the FreqList option is used only for denoting multiple occurrences of whatever value it maps to in the List option. TI-84 users beware!

Also, leaving the FreqList option blank is the equivalent to having a FreqList that has the same dimensions as the List but filled with 1's.
Button navigates to signup pageComment on BeeGee's post “I used the 1-Var Stats fu...”
(2 votes)
Answer
- Peterson Torio
  Posted 3 years ago. Direct link to Peterson Torio's post “You can still use it but ...”
  You can still use it but instead of leaving it blank or writing the probability (i.e. 0.84, 0.16), you have to input the product of the probability (the numbers with decimal points to show percentages) and the number of discrete random variables given. I do that and I think that this technique is the fastest way to get the standard deviation and mean.
  
  Note: This only works for problems where the given infos are discrete random variables and their probabilities.
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  (1 vote)
Abdullah S.
Posted 2 months ago. Direct link to Abdullah S.'s post “I would say that if we ta...”
I would say that if we take many, many customers, the number of tickets purchased would vary by 1.044 from the mean of 2.9 tickets
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(2 votes)
Answer
Saugat Bhattarai
Posted 4 years ago. Direct link to Saugat Bhattarai's post “The pattern of data sprea...”
The pattern of data spread from mean value.
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(2 votes)
Answer