What does the author mean in this description about “mean”?
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I am reading the book "Intuitive Introductory Statistics" and it has the below paragraph about the mean which I don't understand. For example if you take a series 1,2,3,100 and calculate the mean, it would be 26.5. How are values less than 26.5 equally balancing values greater than 26.5?
If we interpret the visual center of a data collection to be the balance point where data values larger than the center are equally balanced by those that are smaller than the center, the numerical average or mean is a natural statistic for identifying and measuring the center. This is the case, for example, when our interpretation of the ‘visual center’ corresponds to a value for which the numerical contribution from data points that are greater than the ‘center’ is equally balanced by the numerical contribution from those that are less than it. In such settings, the appropriate statistic to measure this ‘visual center’ is naturally the average, or mean, of the collected observations.
mean
asked Dec 22 '18 at 22:20
user10697426user10697426
82
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add a comment |
$begingroup$
I am reading the book "Intuitive Introductory Statistics" and it has the below paragraph about the mean which I don't understand. For example if you take a series 1,2,3,100 and calculate the mean, it would be 26.5. How are values less than 26.5 equally balancing values greater than 26.5?
If we interpret the visual center of a data collection to be the balance point where data values larger than the center are equally balanced by those that are smaller than the center, the numerical average or mean is a natural statistic for identifying and measuring the center. This is the case, for example, when our interpretation of the ‘visual center’ corresponds to a value for which the numerical contribution from data points that are greater than the ‘center’ is equally balanced by the numerical contribution from those that are less than it. In such settings, the appropriate statistic to measure this ‘visual center’ is naturally the average, or mean, of the collected observations.
mean
asked Dec 22 '18 at 22:20
user10697426user10697426
82
$endgroup$
add a comment |
$begingroup$
I am reading the book "Intuitive Introductory Statistics" and it has the below paragraph about the mean which I don't understand. For example if you take a series 1,2,3,100 and calculate the mean, it would be 26.5. How are values less than 26.5 equally balancing values greater than 26.5?
If we interpret the visual center of a data collection to be the balance point where data values larger than the center are equally balanced by those that are smaller than the center, the numerical average or mean is a natural statistic for identifying and measuring the center. This is the case, for example, when our interpretation of the ‘visual center’ corresponds to a value for which the numerical contribution from data points that are greater than the ‘center’ is equally balanced by the numerical contribution from those that are less than it. In such settings, the appropriate statistic to measure this ‘visual center’ is naturally the average, or mean, of the collected observations.
mean
asked Dec 22 '18 at 22:20
user10697426user10697426
82
$endgroup$
I am reading the book "Intuitive Introductory Statistics" and it has the below paragraph about the mean which I don't understand. For example if you take a series 1,2,3,100 and calculate the mean, it would be 26.5. How are values less than 26.5 equally balancing values greater than 26.5?
If we interpret the visual center of a data collection to be the balance point where data values larger than the center are equally balanced by those that are smaller than the center, the numerical average or mean is a natural statistic for identifying and measuring the center. This is the case, for example, when our interpretation of the ‘visual center’ corresponds to a value for which the numerical contribution from data points that are greater than the ‘center’ is equally balanced by the numerical contribution from those that are less than it. In such settings, the appropriate statistic to measure this ‘visual center’ is naturally the average, or mean, of the collected observations.
mean
mean
asked Dec 22 '18 at 22:20
user10697426user10697426
82
asked Dec 22 '18 at 22:20
user10697426user10697426
82
asked Dec 22 '18 at 22:20
user10697426user10697426
82
asked Dec 22 '18 at 22:20
user10697426user10697426
82
asked Dec 22 '18 at 22:20
user10697426user10697426
82
82
add a comment |
add a comment |
1 Answer
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Take a thin rod of length 100 (whatever units). Now take four identical solid spheres, drill a hole through them, and attach them to the rod at positions 1, 2, 3, and 100 units along its length. Now try to balance the rod on your finger. You'll find that there is exactly one point where the rod balances, and it is 26.5 units along the rod.
answered Dec 22 '18 at 22:24
Matthew DruryMatthew Drury
26.8k267107
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$begingroup$
Ahh, the balancing point, that makes sense now.
$endgroup$
– user10697426
Dec 22 '18 at 23:25
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Take a thin rod of length 100 (whatever units). Now take four identical solid spheres, drill a hole through them, and attach them to the rod at positions 1, 2, 3, and 100 units along its length. Now try to balance the rod on your finger. You'll find that there is exactly one point where the rod balances, and it is 26.5 units along the rod.
answered Dec 22 '18 at 22:24
Matthew DruryMatthew Drury
26.8k267107
$endgroup$
$begingroup$
Ahh, the balancing point, that makes sense now.
$endgroup$
– user10697426
Dec 22 '18 at 23:25
add a comment |
$begingroup$
Take a thin rod of length 100 (whatever units). Now take four identical solid spheres, drill a hole through them, and attach them to the rod at positions 1, 2, 3, and 100 units along its length. Now try to balance the rod on your finger. You'll find that there is exactly one point where the rod balances, and it is 26.5 units along the rod.
answered Dec 22 '18 at 22:24
Matthew DruryMatthew Drury
26.8k267107
$endgroup$
$begingroup$
Ahh, the balancing point, that makes sense now.
$endgroup$
– user10697426
Dec 22 '18 at 23:25
add a comment |
$begingroup$
Take a thin rod of length 100 (whatever units). Now take four identical solid spheres, drill a hole through them, and attach them to the rod at positions 1, 2, 3, and 100 units along its length. Now try to balance the rod on your finger. You'll find that there is exactly one point where the rod balances, and it is 26.5 units along the rod.
answered Dec 22 '18 at 22:24
Matthew DruryMatthew Drury
26.8k267107
$endgroup$
Take a thin rod of length 100 (whatever units). Now take four identical solid spheres, drill a hole through them, and attach them to the rod at positions 1, 2, 3, and 100 units along its length. Now try to balance the rod on your finger. You'll find that there is exactly one point where the rod balances, and it is 26.5 units along the rod.
answered Dec 22 '18 at 22:24
Matthew DruryMatthew Drury
26.8k267107
answered Dec 22 '18 at 22:24
Matthew DruryMatthew Drury
26.8k267107
answered Dec 22 '18 at 22:24
Matthew DruryMatthew Drury
26.8k267107
answered Dec 22 '18 at 22:24
Matthew DruryMatthew Drury
26.8k267107
26.8k267107
$begingroup$
Ahh, the balancing point, that makes sense now.
$endgroup$
– user10697426
Dec 22 '18 at 23:25
add a comment |
$begingroup$
Ahh, the balancing point, that makes sense now.
$endgroup$
– user10697426
Dec 22 '18 at 23:25
$begingroup$
Ahh, the balancing point, that makes sense now.
$endgroup$
– user10697426
Dec 22 '18 at 23:25
$begingroup$
Ahh, the balancing point, that makes sense now.
$endgroup$
– user10697426
Dec 22 '18 at 23:25
add a comment |
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