Variance in linear combinations
I am having a problem understanding why the below is calculating the variance when the exercise is asking about the standard deviation.
The standard deviation of the time you take for each statistics homework problem is 1.5 minutes, and it is 2 minutes for each chemistry problem. What is the standard deviation of the time you expect to spend on statistics and physics homework for the week if you have 5 statistics and 4 chemistry homework problems assigned?
Variability of a linear combination of two independent random variables:
V(aX + bY) = $a^2$ ×V(X) + $b^2$ ×V(Y)
The standard deviation of the linear combination is the square root of the variance.
So we will use the above formula to calculate the exercise:
V(5S + 4C) = $5^2$ × V(S) + $4^2$ × V(C)
= 25 × $1.5^2$ + 16 × $2^2$
= 56.25 + 64
= 120.25
In my mind we are clearly finding the variance here, not the standard deviation, as that would be the sqrt of the 120.25, still the answer is 120.25, why?
Edit: Also, the formula says that $a^2$ and $b^2$ but still they do power of two on both 5 and 4 and V(S) and V(C).
statistics variance standard-deviation
asked Nov 28 at 20:06
StudentCoderJava
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I am having a problem understanding why the below is calculating the variance when the exercise is asking about the standard deviation.
The standard deviation of the time you take for each statistics homework problem is 1.5 minutes, and it is 2 minutes for each chemistry problem. What is the standard deviation of the time you expect to spend on statistics and physics homework for the week if you have 5 statistics and 4 chemistry homework problems assigned?
Variability of a linear combination of two independent random variables:
V(aX + bY) = $a^2$ ×V(X) + $b^2$ ×V(Y)
The standard deviation of the linear combination is the square root of the variance.
So we will use the above formula to calculate the exercise:
V(5S + 4C) = $5^2$ × V(S) + $4^2$ × V(C)
= 25 × $1.5^2$ + 16 × $2^2$
= 56.25 + 64
= 120.25
In my mind we are clearly finding the variance here, not the standard deviation, as that would be the sqrt of the 120.25, still the answer is 120.25, why?
Edit: Also, the formula says that $a^2$ and $b^2$ but still they do power of two on both 5 and 4 and V(S) and V(C).
statistics variance standard-deviation
asked Nov 28 at 20:06
StudentCoderJava
728
1
It is asked for the standard deviation and you have found the standard deviation: $sqrt{120.25}$
– callculus
Nov 28 at 20:11
1
Yes but why dont they take the square root? They just stop at 120.25
– StudentCoderJava
Nov 28 at 20:12
1
I don´t know. There is no reason not to take the square root - imho.
– callculus
Nov 28 at 20:14
1
To your question of your edit: 1.5 and 2 represent standard deviations. Thus you have to square them to get the variance.
– callculus
Nov 28 at 20:17
1
Thank you I see it now.
– StudentCoderJava
Nov 28 at 20:19
|
show 1 more comment
I am having a problem understanding why the below is calculating the variance when the exercise is asking about the standard deviation.
The standard deviation of the time you take for each statistics homework problem is 1.5 minutes, and it is 2 minutes for each chemistry problem. What is the standard deviation of the time you expect to spend on statistics and physics homework for the week if you have 5 statistics and 4 chemistry homework problems assigned?
Variability of a linear combination of two independent random variables:
V(aX + bY) = $a^2$ ×V(X) + $b^2$ ×V(Y)
The standard deviation of the linear combination is the square root of the variance.
So we will use the above formula to calculate the exercise:
V(5S + 4C) = $5^2$ × V(S) + $4^2$ × V(C)
= 25 × $1.5^2$ + 16 × $2^2$
= 56.25 + 64
= 120.25
In my mind we are clearly finding the variance here, not the standard deviation, as that would be the sqrt of the 120.25, still the answer is 120.25, why?
Edit: Also, the formula says that $a^2$ and $b^2$ but still they do power of two on both 5 and 4 and V(S) and V(C).
statistics variance standard-deviation
asked Nov 28 at 20:06
StudentCoderJava
728
I am having a problem understanding why the below is calculating the variance when the exercise is asking about the standard deviation.
The standard deviation of the time you take for each statistics homework problem is 1.5 minutes, and it is 2 minutes for each chemistry problem. What is the standard deviation of the time you expect to spend on statistics and physics homework for the week if you have 5 statistics and 4 chemistry homework problems assigned?
Variability of a linear combination of two independent random variables:
V(aX + bY) = $a^2$ ×V(X) + $b^2$ ×V(Y)
The standard deviation of the linear combination is the square root of the variance.
So we will use the above formula to calculate the exercise:
V(5S + 4C) = $5^2$ × V(S) + $4^2$ × V(C)
= 25 × $1.5^2$ + 16 × $2^2$
= 56.25 + 64
= 120.25
In my mind we are clearly finding the variance here, not the standard deviation, as that would be the sqrt of the 120.25, still the answer is 120.25, why?
Edit: Also, the formula says that $a^2$ and $b^2$ but still they do power of two on both 5 and 4 and V(S) and V(C).
statistics variance standard-deviation
statistics variance standard-deviation
asked Nov 28 at 20:06
StudentCoderJava
728
asked Nov 28 at 20:06
StudentCoderJava
728
edited Nov 28 at 20:14
asked Nov 28 at 20:06
StudentCoderJava
728
asked Nov 28 at 20:06
StudentCoderJava
728
asked Nov 28 at 20:06
StudentCoderJava
728
728
1
It is asked for the standard deviation and you have found the standard deviation: $sqrt{120.25}$
– callculus
Nov 28 at 20:11
1
Yes but why dont they take the square root? They just stop at 120.25
– StudentCoderJava
Nov 28 at 20:12
1
I don´t know. There is no reason not to take the square root - imho.
– callculus
Nov 28 at 20:14
1
To your question of your edit: 1.5 and 2 represent standard deviations. Thus you have to square them to get the variance.
– callculus
Nov 28 at 20:17
1
Thank you I see it now.
– StudentCoderJava
Nov 28 at 20:19
|
show 1 more comment
1
It is asked for the standard deviation and you have found the standard deviation: $sqrt{120.25}$
– callculus
Nov 28 at 20:11
1
Yes but why dont they take the square root? They just stop at 120.25
– StudentCoderJava
Nov 28 at 20:12
1
I don´t know. There is no reason not to take the square root - imho.
– callculus
Nov 28 at 20:14
1
To your question of your edit: 1.5 and 2 represent standard deviations. Thus you have to square them to get the variance.
– callculus
Nov 28 at 20:17
1
Thank you I see it now.
– StudentCoderJava
Nov 28 at 20:19
1
1
It is asked for the standard deviation and you have found the standard deviation: $sqrt{120.25}$
– callculus
Nov 28 at 20:11
It is asked for the standard deviation and you have found the standard deviation: $sqrt{120.25}$
– callculus
Nov 28 at 20:11
1
1
Yes but why dont they take the square root? They just stop at 120.25
– StudentCoderJava
Nov 28 at 20:12
Yes but why dont they take the square root? They just stop at 120.25
– StudentCoderJava
Nov 28 at 20:12
1
1
I don´t know. There is no reason not to take the square root - imho.
– callculus
Nov 28 at 20:14
I don´t know. There is no reason not to take the square root - imho.
– callculus
Nov 28 at 20:14
1
1
To your question of your edit: 1.5 and 2 represent standard deviations. Thus you have to square them to get the variance.
– callculus
Nov 28 at 20:17
To your question of your edit: 1.5 and 2 represent standard deviations. Thus you have to square them to get the variance.
– callculus
Nov 28 at 20:17
1
1
Thank you I see it now.
– StudentCoderJava
Nov 28 at 20:19
Thank you I see it now.
– StudentCoderJava
Nov 28 at 20:19
|
show 1 more comment
1 Answer
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votes
As you have correctly determined, the answer to the question as written is the square root of the variance you have calculated. Your reasoning is correct.
It would seem that you are right and the answer book is wrong.
$$
sigma_S = 1.5,
sigma_C = 2
$$
Therefore
$$
V(S) = (1.5)^2,
V(C) = 2^2
$$
Scaling a random variable by $a$ scales the variance by $a^2$, so $V(aS) = a^2V(S)$
answered Nov 28 at 20:13
ip6
54839
add a comment |
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As you have correctly determined, the answer to the question as written is the square root of the variance you have calculated. Your reasoning is correct.
It would seem that you are right and the answer book is wrong.
$$
sigma_S = 1.5,
sigma_C = 2
$$
Therefore
$$
V(S) = (1.5)^2,
V(C) = 2^2
$$
Scaling a random variable by $a$ scales the variance by $a^2$, so $V(aS) = a^2V(S)$
answered Nov 28 at 20:13
ip6
54839
add a comment |
As you have correctly determined, the answer to the question as written is the square root of the variance you have calculated. Your reasoning is correct.
It would seem that you are right and the answer book is wrong.
$$
sigma_S = 1.5,
sigma_C = 2
$$
Therefore
$$
V(S) = (1.5)^2,
V(C) = 2^2
$$
Scaling a random variable by $a$ scales the variance by $a^2$, so $V(aS) = a^2V(S)$
answered Nov 28 at 20:13
ip6
54839
add a comment |
As you have correctly determined, the answer to the question as written is the square root of the variance you have calculated. Your reasoning is correct.
It would seem that you are right and the answer book is wrong.
$$
sigma_S = 1.5,
sigma_C = 2
$$
Therefore
$$
V(S) = (1.5)^2,
V(C) = 2^2
$$
Scaling a random variable by $a$ scales the variance by $a^2$, so $V(aS) = a^2V(S)$
answered Nov 28 at 20:13
ip6
54839
As you have correctly determined, the answer to the question as written is the square root of the variance you have calculated. Your reasoning is correct.
It would seem that you are right and the answer book is wrong.
$$
sigma_S = 1.5,
sigma_C = 2
$$
Therefore
$$
V(S) = (1.5)^2,
V(C) = 2^2
$$
Scaling a random variable by $a$ scales the variance by $a^2$, so $V(aS) = a^2V(S)$
answered Nov 28 at 20:13
ip6
54839
edited Nov 28 at 20:23
answered Nov 28 at 20:13
ip6
54839
answered Nov 28 at 20:13
ip6
54839
answered Nov 28 at 20:13
ip6
54839
54839
add a comment |
add a comment |
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1
It is asked for the standard deviation and you have found the standard deviation: $sqrt{120.25}$
– callculus
Nov 28 at 20:11
1
Yes but why dont they take the square root? They just stop at 120.25
– StudentCoderJava
Nov 28 at 20:12
1
I don´t know. There is no reason not to take the square root - imho.
– callculus
Nov 28 at 20:14
1
To your question of your edit: 1.5 and 2 represent standard deviations. Thus you have to square them to get the variance.
– callculus
Nov 28 at 20:17
1
Thank you I see it now.
– StudentCoderJava
Nov 28 at 20:19