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Intuition for the mean for elementary school kids
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I was teaching elementary school kids (aged 10) about the mean. The intuition I gave them is roughly as follows:
You are trying to find a value such that the sum of all the distances from the mean for the values above the mean is the same as the sum of all the distances from the mean for the values below the mean.
They were happy with this and we supplemented it by plotting sets of points and drawing the line representing the mean.
But then I told them how to compute the mean. That is add up all the values and divide by the number of values.
The method of calculation seemed completely disconnected from the intuition for what the mean is.
What is a good way to give intuition for the standard method of computing the mean?
primary-education
asked 6 hours ago
Anush
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up vote
6
down vote
favorite
I was teaching elementary school kids (aged 10) about the mean. The intuition I gave them is roughly as follows:
You are trying to find a value such that the sum of all the distances from the mean for the values above the mean is the same as the sum of all the distances from the mean for the values below the mean.
They were happy with this and we supplemented it by plotting sets of points and drawing the line representing the mean.
But then I told them how to compute the mean. That is add up all the values and divide by the number of values.
The method of calculation seemed completely disconnected from the intuition for what the mean is.
What is a good way to give intuition for the standard method of computing the mean?
primary-education
asked 6 hours ago
Anush
1312
New contributor
Anush is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |Â
up vote
6
down vote
favorite
up vote
6
down vote
favorite
I was teaching elementary school kids (aged 10) about the mean. The intuition I gave them is roughly as follows:
You are trying to find a value such that the sum of all the distances from the mean for the values above the mean is the same as the sum of all the distances from the mean for the values below the mean.
They were happy with this and we supplemented it by plotting sets of points and drawing the line representing the mean.
But then I told them how to compute the mean. That is add up all the values and divide by the number of values.
The method of calculation seemed completely disconnected from the intuition for what the mean is.
What is a good way to give intuition for the standard method of computing the mean?
primary-education
asked 6 hours ago
Anush
1312
New contributor
Anush is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
I was teaching elementary school kids (aged 10) about the mean. The intuition I gave them is roughly as follows:
You are trying to find a value such that the sum of all the distances from the mean for the values above the mean is the same as the sum of all the distances from the mean for the values below the mean.
They were happy with this and we supplemented it by plotting sets of points and drawing the line representing the mean.
But then I told them how to compute the mean. That is add up all the values and divide by the number of values.
The method of calculation seemed completely disconnected from the intuition for what the mean is.
What is a good way to give intuition for the standard method of computing the mean?
primary-education
primary-education
asked 6 hours ago
Anush
1312
New contributor
Anush is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 6 hours ago
Anush
1312
New contributor
Anush is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 5 hours ago
asked 6 hours ago
Anush
1312
New contributor
Anush is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 6 hours ago
Anush
1312
asked 6 hours ago
Anush
1312
1312
New contributor
Anush is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Anush is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Anush is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
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3 Answers
3
active
oldest
votes
up vote
0
down vote
Take 3 piles of toothpicks (different amounts in each). Shove them together and split them into 3 equal piles. Make the kids do it. Do it with marbles. Do it with money. Do it a few times: "lather, rinse, repeat". ;)
answered 2 hours ago
guest
31114
add a comment |Â
up vote
0
down vote
Let's first consider the underlying algebra before turning to real-world models.
As an example, suppose we have four reals $a_1 < a_2 < a_3 <a_4$ with mean $a$ between $a_2$ and $a_3.,$ Here the equivalence connecting the two views of the mean is as follows
$$overbracea+a+a+a = a_1 + a_2 + a_3 + a_4^largetextdefinition of mean aiff overbracea!-a_1, +, a!-a_2, =, a_3 - a, +, a_4 -a^large a_i textare balanced around the mean a $$
The direction $(Rightarrow)$ arises as follows: in the definition, we can cancel a LHS $a$ from each $a_i ge a$ on the RHS, and we can cancel each RHS $a_i < a$ from an $a$ on the LHS, yielding the equivalent balanced form on the right. This method can be used more generally to check an equality of sums by rewriting it more simply, e.g.
$qquadqquadqquadbeginalign
&color#c00222+color#0a0200+1000+4119\
= &color#c00100+color#0a0328+2113+3000endalign$
$ iff beginalign
&color#c00122 + 1119\
= &color#0a0128 + 1113
endalign$
$iff beginalign
&6\
= &6
endalign$
Above we cancelled $color#c00100$ from both sides, leaving the summand $color#c00122$ on the new LHS.
Next, $, $ we cancelled $color#0a0200$ from both sides, leaving the summand $color#0a0128$ on the new RHS, etc. The OP is just the special case when all summands are equal on one side, say $a$. This replaces each $a_i$ by its distance from $a$, doing it on the equation side that keeps all the summands nonnegative.
Conversely, direction $(Leftarrow)$ follows by inverting the prior, i.e. by adding terms to both sides of the equation in order to move all negated terms to the opposite side of the equation, i.e. by eliminating all subtractions.
A simple real-world model is an old-fashioned balance scale. Then the prior additions and subtractions to both sides amount to adding or removing equal weights from both sides of the scale - which preserves balance. One can illustrate the above equivalence by explicitly performing the steps in the above sketched proof using weight manipulations. Of course one should use much smaller numbers than I did above (I chose those larger numbers only to highlight that the method can yield nonntrivial simplifications).
answered 54 mins ago
Number
8561710
This seems like massive overkill for 10-year-olds.
– Ben Crowell
18 mins ago
@Ben I've taught it at that level with much success. Note the that above exposition in written for the teacher. It should be simplified as need be for the students.
– Number
14 mins ago
@Ben After your answer I see that we have interpreted the question very differently. You interpreted it as how to motivate the mean independent of the (initial) balanced definition, whereas I interpreted it as how to explain the relationship between the two definitions.
– Number
3 mins ago
add a comment |Â
up vote
0
down vote
For basic intuition with kids this age, I would simply draw a number line and locate 2 and 4 as dots. Ask them what number is half-way in between. They say 3. Then demonstrate that $(2+4)/2=3$. Of course this isn't a proof, but the goal is to build intuition, not to prove anything.
To amplify on this simple example, describe a situation where Alice has 2 cookies and Betty has 4, but they want to be fair and share equally. How many should each have? Describe the fact that they gain and lose equal amounts (what Betty loses equals what Alice gains), and also the fact that each one gets half the cookies.
answered 9 mins ago
Ben Crowell
6,4831747
add a comment |Â
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
Take 3 piles of toothpicks (different amounts in each). Shove them together and split them into 3 equal piles. Make the kids do it. Do it with marbles. Do it with money. Do it a few times: "lather, rinse, repeat". ;)
answered 2 hours ago
guest
31114
add a comment |Â
up vote
0
down vote
Take 3 piles of toothpicks (different amounts in each). Shove them together and split them into 3 equal piles. Make the kids do it. Do it with marbles. Do it with money. Do it a few times: "lather, rinse, repeat". ;)
answered 2 hours ago
guest
31114
add a comment |Â
up vote
0
down vote
up vote
0
down vote
Take 3 piles of toothpicks (different amounts in each). Shove them together and split them into 3 equal piles. Make the kids do it. Do it with marbles. Do it with money. Do it a few times: "lather, rinse, repeat". ;)
answered 2 hours ago
guest
31114
Take 3 piles of toothpicks (different amounts in each). Shove them together and split them into 3 equal piles. Make the kids do it. Do it with marbles. Do it with money. Do it a few times: "lather, rinse, repeat". ;)
answered 2 hours ago
guest
31114
answered 2 hours ago
guest
31114
answered 2 hours ago
guest
31114
answered 2 hours ago
guest
31114
31114
add a comment |Â
add a comment |Â
up vote
0
down vote
Let's first consider the underlying algebra before turning to real-world models.
As an example, suppose we have four reals $a_1 < a_2 < a_3 <a_4$ with mean $a$ between $a_2$ and $a_3.,$ Here the equivalence connecting the two views of the mean is as follows
$$overbracea+a+a+a = a_1 + a_2 + a_3 + a_4^largetextdefinition of mean aiff overbracea!-a_1, +, a!-a_2, =, a_3 - a, +, a_4 -a^large a_i textare balanced around the mean a $$
The direction $(Rightarrow)$ arises as follows: in the definition, we can cancel a LHS $a$ from each $a_i ge a$ on the RHS, and we can cancel each RHS $a_i < a$ from an $a$ on the LHS, yielding the equivalent balanced form on the right. This method can be used more generally to check an equality of sums by rewriting it more simply, e.g.
$qquadqquadqquadbeginalign
&color#c00222+color#0a0200+1000+4119\
= &color#c00100+color#0a0328+2113+3000endalign$
$ iff beginalign
&color#c00122 + 1119\
= &color#0a0128 + 1113
endalign$
$iff beginalign
&6\
= &6
endalign$
Above we cancelled $color#c00100$ from both sides, leaving the summand $color#c00122$ on the new LHS.
Next, $, $ we cancelled $color#0a0200$ from both sides, leaving the summand $color#0a0128$ on the new RHS, etc. The OP is just the special case when all summands are equal on one side, say $a$. This replaces each $a_i$ by its distance from $a$, doing it on the equation side that keeps all the summands nonnegative.
Conversely, direction $(Leftarrow)$ follows by inverting the prior, i.e. by adding terms to both sides of the equation in order to move all negated terms to the opposite side of the equation, i.e. by eliminating all subtractions.
A simple real-world model is an old-fashioned balance scale. Then the prior additions and subtractions to both sides amount to adding or removing equal weights from both sides of the scale - which preserves balance. One can illustrate the above equivalence by explicitly performing the steps in the above sketched proof using weight manipulations. Of course one should use much smaller numbers than I did above (I chose those larger numbers only to highlight that the method can yield nonntrivial simplifications).
answered 54 mins ago
Number
8561710
This seems like massive overkill for 10-year-olds.
– Ben Crowell
18 mins ago
@Ben I've taught it at that level with much success. Note the that above exposition in written for the teacher. It should be simplified as need be for the students.
– Number
14 mins ago
@Ben After your answer I see that we have interpreted the question very differently. You interpreted it as how to motivate the mean independent of the (initial) balanced definition, whereas I interpreted it as how to explain the relationship between the two definitions.
– Number
3 mins ago
add a comment |Â
up vote
0
down vote
Let's first consider the underlying algebra before turning to real-world models.
As an example, suppose we have four reals $a_1 < a_2 < a_3 <a_4$ with mean $a$ between $a_2$ and $a_3.,$ Here the equivalence connecting the two views of the mean is as follows
$$overbracea+a+a+a = a_1 + a_2 + a_3 + a_4^largetextdefinition of mean aiff overbracea!-a_1, +, a!-a_2, =, a_3 - a, +, a_4 -a^large a_i textare balanced around the mean a $$
The direction $(Rightarrow)$ arises as follows: in the definition, we can cancel a LHS $a$ from each $a_i ge a$ on the RHS, and we can cancel each RHS $a_i < a$ from an $a$ on the LHS, yielding the equivalent balanced form on the right. This method can be used more generally to check an equality of sums by rewriting it more simply, e.g.
$qquadqquadqquadbeginalign
&color#c00222+color#0a0200+1000+4119\
= &color#c00100+color#0a0328+2113+3000endalign$
$ iff beginalign
&color#c00122 + 1119\
= &color#0a0128 + 1113
endalign$
$iff beginalign
&6\
= &6
endalign$
Above we cancelled $color#c00100$ from both sides, leaving the summand $color#c00122$ on the new LHS.
Next, $, $ we cancelled $color#0a0200$ from both sides, leaving the summand $color#0a0128$ on the new RHS, etc. The OP is just the special case when all summands are equal on one side, say $a$. This replaces each $a_i$ by its distance from $a$, doing it on the equation side that keeps all the summands nonnegative.
Conversely, direction $(Leftarrow)$ follows by inverting the prior, i.e. by adding terms to both sides of the equation in order to move all negated terms to the opposite side of the equation, i.e. by eliminating all subtractions.
A simple real-world model is an old-fashioned balance scale. Then the prior additions and subtractions to both sides amount to adding or removing equal weights from both sides of the scale - which preserves balance. One can illustrate the above equivalence by explicitly performing the steps in the above sketched proof using weight manipulations. Of course one should use much smaller numbers than I did above (I chose those larger numbers only to highlight that the method can yield nonntrivial simplifications).
answered 54 mins ago
Number
8561710
This seems like massive overkill for 10-year-olds.
– Ben Crowell
18 mins ago
@Ben I've taught it at that level with much success. Note the that above exposition in written for the teacher. It should be simplified as need be for the students.
– Number
14 mins ago
@Ben After your answer I see that we have interpreted the question very differently. You interpreted it as how to motivate the mean independent of the (initial) balanced definition, whereas I interpreted it as how to explain the relationship between the two definitions.
– Number
3 mins ago
add a comment |Â
up vote
0
down vote
up vote
0
down vote
Let's first consider the underlying algebra before turning to real-world models.
As an example, suppose we have four reals $a_1 < a_2 < a_3 <a_4$ with mean $a$ between $a_2$ and $a_3.,$ Here the equivalence connecting the two views of the mean is as follows
$$overbracea+a+a+a = a_1 + a_2 + a_3 + a_4^largetextdefinition of mean aiff overbracea!-a_1, +, a!-a_2, =, a_3 - a, +, a_4 -a^large a_i textare balanced around the mean a $$
The direction $(Rightarrow)$ arises as follows: in the definition, we can cancel a LHS $a$ from each $a_i ge a$ on the RHS, and we can cancel each RHS $a_i < a$ from an $a$ on the LHS, yielding the equivalent balanced form on the right. This method can be used more generally to check an equality of sums by rewriting it more simply, e.g.
$qquadqquadqquadbeginalign
&color#c00222+color#0a0200+1000+4119\
= &color#c00100+color#0a0328+2113+3000endalign$
$ iff beginalign
&color#c00122 + 1119\
= &color#0a0128 + 1113
endalign$
$iff beginalign
&6\
= &6
endalign$
Above we cancelled $color#c00100$ from both sides, leaving the summand $color#c00122$ on the new LHS.
Next, $, $ we cancelled $color#0a0200$ from both sides, leaving the summand $color#0a0128$ on the new RHS, etc. The OP is just the special case when all summands are equal on one side, say $a$. This replaces each $a_i$ by its distance from $a$, doing it on the equation side that keeps all the summands nonnegative.
Conversely, direction $(Leftarrow)$ follows by inverting the prior, i.e. by adding terms to both sides of the equation in order to move all negated terms to the opposite side of the equation, i.e. by eliminating all subtractions.
A simple real-world model is an old-fashioned balance scale. Then the prior additions and subtractions to both sides amount to adding or removing equal weights from both sides of the scale - which preserves balance. One can illustrate the above equivalence by explicitly performing the steps in the above sketched proof using weight manipulations. Of course one should use much smaller numbers than I did above (I chose those larger numbers only to highlight that the method can yield nonntrivial simplifications).
answered 54 mins ago
Number
8561710
Let's first consider the underlying algebra before turning to real-world models.
As an example, suppose we have four reals $a_1 < a_2 < a_3 <a_4$ with mean $a$ between $a_2$ and $a_3.,$ Here the equivalence connecting the two views of the mean is as follows
$$overbracea+a+a+a = a_1 + a_2 + a_3 + a_4^largetextdefinition of mean aiff overbracea!-a_1, +, a!-a_2, =, a_3 - a, +, a_4 -a^large a_i textare balanced around the mean a $$
The direction $(Rightarrow)$ arises as follows: in the definition, we can cancel a LHS $a$ from each $a_i ge a$ on the RHS, and we can cancel each RHS $a_i < a$ from an $a$ on the LHS, yielding the equivalent balanced form on the right. This method can be used more generally to check an equality of sums by rewriting it more simply, e.g.
$qquadqquadqquadbeginalign
&color#c00222+color#0a0200+1000+4119\
= &color#c00100+color#0a0328+2113+3000endalign$
$ iff beginalign
&color#c00122 + 1119\
= &color#0a0128 + 1113
endalign$
$iff beginalign
&6\
= &6
endalign$
Above we cancelled $color#c00100$ from both sides, leaving the summand $color#c00122$ on the new LHS.
Next, $, $ we cancelled $color#0a0200$ from both sides, leaving the summand $color#0a0128$ on the new RHS, etc. The OP is just the special case when all summands are equal on one side, say $a$. This replaces each $a_i$ by its distance from $a$, doing it on the equation side that keeps all the summands nonnegative.
Conversely, direction $(Leftarrow)$ follows by inverting the prior, i.e. by adding terms to both sides of the equation in order to move all negated terms to the opposite side of the equation, i.e. by eliminating all subtractions.
A simple real-world model is an old-fashioned balance scale. Then the prior additions and subtractions to both sides amount to adding or removing equal weights from both sides of the scale - which preserves balance. One can illustrate the above equivalence by explicitly performing the steps in the above sketched proof using weight manipulations. Of course one should use much smaller numbers than I did above (I chose those larger numbers only to highlight that the method can yield nonntrivial simplifications).
answered 54 mins ago
Number
8561710
edited 18 mins ago
answered 54 mins ago
Number
8561710
answered 54 mins ago
Number
8561710
answered 54 mins ago
Number
8561710
8561710
This seems like massive overkill for 10-year-olds.
– Ben Crowell
18 mins ago
@Ben I've taught it at that level with much success. Note the that above exposition in written for the teacher. It should be simplified as need be for the students.
– Number
14 mins ago
@Ben After your answer I see that we have interpreted the question very differently. You interpreted it as how to motivate the mean independent of the (initial) balanced definition, whereas I interpreted it as how to explain the relationship between the two definitions.
– Number
3 mins ago
add a comment |Â
This seems like massive overkill for 10-year-olds.
– Ben Crowell
18 mins ago
@Ben I've taught it at that level with much success. Note the that above exposition in written for the teacher. It should be simplified as need be for the students.
– Number
14 mins ago
@Ben After your answer I see that we have interpreted the question very differently. You interpreted it as how to motivate the mean independent of the (initial) balanced definition, whereas I interpreted it as how to explain the relationship between the two definitions.
– Number
3 mins ago
This seems like massive overkill for 10-year-olds.
– Ben Crowell
18 mins ago
This seems like massive overkill for 10-year-olds.
– Ben Crowell
18 mins ago
@Ben I've taught it at that level with much success. Note the that above exposition in written for the teacher. It should be simplified as need be for the students.
– Number
14 mins ago
@Ben I've taught it at that level with much success. Note the that above exposition in written for the teacher. It should be simplified as need be for the students.
– Number
14 mins ago
@Ben After your answer I see that we have interpreted the question very differently. You interpreted it as how to motivate the mean independent of the (initial) balanced definition, whereas I interpreted it as how to explain the relationship between the two definitions.
– Number
3 mins ago
@Ben After your answer I see that we have interpreted the question very differently. You interpreted it as how to motivate the mean independent of the (initial) balanced definition, whereas I interpreted it as how to explain the relationship between the two definitions.
– Number
3 mins ago
add a comment |Â
up vote
0
down vote
For basic intuition with kids this age, I would simply draw a number line and locate 2 and 4 as dots. Ask them what number is half-way in between. They say 3. Then demonstrate that $(2+4)/2=3$. Of course this isn't a proof, but the goal is to build intuition, not to prove anything.
To amplify on this simple example, describe a situation where Alice has 2 cookies and Betty has 4, but they want to be fair and share equally. How many should each have? Describe the fact that they gain and lose equal amounts (what Betty loses equals what Alice gains), and also the fact that each one gets half the cookies.
answered 9 mins ago
Ben Crowell
6,4831747
add a comment |Â
up vote
0
down vote
For basic intuition with kids this age, I would simply draw a number line and locate 2 and 4 as dots. Ask them what number is half-way in between. They say 3. Then demonstrate that $(2+4)/2=3$. Of course this isn't a proof, but the goal is to build intuition, not to prove anything.
To amplify on this simple example, describe a situation where Alice has 2 cookies and Betty has 4, but they want to be fair and share equally. How many should each have? Describe the fact that they gain and lose equal amounts (what Betty loses equals what Alice gains), and also the fact that each one gets half the cookies.
answered 9 mins ago
Ben Crowell
6,4831747
add a comment |Â
up vote
0
down vote
up vote
0
down vote
For basic intuition with kids this age, I would simply draw a number line and locate 2 and 4 as dots. Ask them what number is half-way in between. They say 3. Then demonstrate that $(2+4)/2=3$. Of course this isn't a proof, but the goal is to build intuition, not to prove anything.
To amplify on this simple example, describe a situation where Alice has 2 cookies and Betty has 4, but they want to be fair and share equally. How many should each have? Describe the fact that they gain and lose equal amounts (what Betty loses equals what Alice gains), and also the fact that each one gets half the cookies.
answered 9 mins ago
Ben Crowell
6,4831747
For basic intuition with kids this age, I would simply draw a number line and locate 2 and 4 as dots. Ask them what number is half-way in between. They say 3. Then demonstrate that $(2+4)/2=3$. Of course this isn't a proof, but the goal is to build intuition, not to prove anything.
To amplify on this simple example, describe a situation where Alice has 2 cookies and Betty has 4, but they want to be fair and share equally. How many should each have? Describe the fact that they gain and lose equal amounts (what Betty loses equals what Alice gains), and also the fact that each one gets half the cookies.
answered 9 mins ago
Ben Crowell
6,4831747
answered 9 mins ago
Ben Crowell
6,4831747
answered 9 mins ago
Ben Crowell
6,4831747
answered 9 mins ago
Ben Crowell
6,4831747
6,4831747
add a comment |Â
add a comment |Â
Anush is a new contributor. Be nice, and check out our Code of Conduct.
Anush is a new contributor. Be nice, and check out our Code of Conduct.
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