Finding the sum of (x-mean)^2
$begingroup$
I am stumped on a question. It gives me the mean value as 24, standard deviation is 4 and N = 10. What is the sum of (x-mean)^2
What I've tried is as follow
$$
Standard Deviation = sqrt((sum_{}^{}(x - x')^{2}/N)
$$
$$
4 = sqrt((sum_{}^{}(x - x')^{2}/10)
$$
$$
sum_{}^{}(x - x')^{2} = 160
$$
Is that an acceptable answer?
standard-deviation
asked Oct 13 '16 at 19:12
user3276954user3276954
1156
$endgroup$
add a comment |
$begingroup$
I am stumped on a question. It gives me the mean value as 24, standard deviation is 4 and N = 10. What is the sum of (x-mean)^2
What I've tried is as follow
$$
Standard Deviation = sqrt((sum_{}^{}(x - x')^{2}/N)
$$
$$
4 = sqrt((sum_{}^{}(x - x')^{2}/10)
$$
$$
sum_{}^{}(x - x')^{2} = 160
$$
Is that an acceptable answer?
standard-deviation
asked Oct 13 '16 at 19:12
user3276954user3276954
1156
$endgroup$
add a comment |
$begingroup$
I am stumped on a question. It gives me the mean value as 24, standard deviation is 4 and N = 10. What is the sum of (x-mean)^2
What I've tried is as follow
$$
Standard Deviation = sqrt((sum_{}^{}(x - x')^{2}/N)
$$
$$
4 = sqrt((sum_{}^{}(x - x')^{2}/10)
$$
$$
sum_{}^{}(x - x')^{2} = 160
$$
Is that an acceptable answer?
standard-deviation
asked Oct 13 '16 at 19:12
user3276954user3276954
1156
$endgroup$
I am stumped on a question. It gives me the mean value as 24, standard deviation is 4 and N = 10. What is the sum of (x-mean)^2
What I've tried is as follow
$$
Standard Deviation = sqrt((sum_{}^{}(x - x')^{2}/N)
$$
$$
4 = sqrt((sum_{}^{}(x - x')^{2}/10)
$$
$$
sum_{}^{}(x - x')^{2} = 160
$$
Is that an acceptable answer?
standard-deviation
standard-deviation
asked Oct 13 '16 at 19:12
user3276954user3276954
1156
asked Oct 13 '16 at 19:12
user3276954user3276954
1156
asked Oct 13 '16 at 19:12
user3276954user3276954
1156
asked Oct 13 '16 at 19:12
user3276954user3276954
1156
asked Oct 13 '16 at 19:12
user3276954user3276954
1156
1156
add a comment |
add a comment |
1 Answer
1
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$begingroup$
It can be shown that $sigma^2_X = mu_{X^2} - mu^2_{X}$. Note that the definition of variance is the sum given. You're summing the squared deviations from the mean, which is part of computing variance. Once you know the standard deviation (which is given to you), you have the variance for free.
answered Oct 13 '16 at 19:25
Sean RobersonSean Roberson
6,42531327
$endgroup$
$begingroup$
I'm slightly confused. Using your formula, what's the answer?
$endgroup$
– user3276954
Oct 13 '16 at 19:30
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
$begingroup$
It can be shown that $sigma^2_X = mu_{X^2} - mu^2_{X}$. Note that the definition of variance is the sum given. You're summing the squared deviations from the mean, which is part of computing variance. Once you know the standard deviation (which is given to you), you have the variance for free.
answered Oct 13 '16 at 19:25
Sean RobersonSean Roberson
6,42531327
$endgroup$
$begingroup$
I'm slightly confused. Using your formula, what's the answer?
$endgroup$
– user3276954
Oct 13 '16 at 19:30
add a comment |
$begingroup$
It can be shown that $sigma^2_X = mu_{X^2} - mu^2_{X}$. Note that the definition of variance is the sum given. You're summing the squared deviations from the mean, which is part of computing variance. Once you know the standard deviation (which is given to you), you have the variance for free.
answered Oct 13 '16 at 19:25
Sean RobersonSean Roberson
6,42531327
$endgroup$
$begingroup$
I'm slightly confused. Using your formula, what's the answer?
$endgroup$
– user3276954
Oct 13 '16 at 19:30
add a comment |
$begingroup$
It can be shown that $sigma^2_X = mu_{X^2} - mu^2_{X}$. Note that the definition of variance is the sum given. You're summing the squared deviations from the mean, which is part of computing variance. Once you know the standard deviation (which is given to you), you have the variance for free.
answered Oct 13 '16 at 19:25
Sean RobersonSean Roberson
6,42531327
$endgroup$
It can be shown that $sigma^2_X = mu_{X^2} - mu^2_{X}$. Note that the definition of variance is the sum given. You're summing the squared deviations from the mean, which is part of computing variance. Once you know the standard deviation (which is given to you), you have the variance for free.
answered Oct 13 '16 at 19:25
Sean RobersonSean Roberson
6,42531327
answered Oct 13 '16 at 19:25
Sean RobersonSean Roberson
6,42531327
answered Oct 13 '16 at 19:25
Sean RobersonSean Roberson
6,42531327
answered Oct 13 '16 at 19:25
Sean RobersonSean Roberson
6,42531327
6,42531327
$begingroup$
I'm slightly confused. Using your formula, what's the answer?
$endgroup$
– user3276954
Oct 13 '16 at 19:30
add a comment |
$begingroup$
I'm slightly confused. Using your formula, what's the answer?
$endgroup$
– user3276954
Oct 13 '16 at 19:30
$begingroup$
I'm slightly confused. Using your formula, what's the answer?
$endgroup$
– user3276954
Oct 13 '16 at 19:30
$begingroup$
I'm slightly confused. Using your formula, what's the answer?
$endgroup$
– user3276954
Oct 13 '16 at 19:30
add a comment |
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