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For two variables x and y, it is known that cov (x, y) = 80, variance of x is 16 and sum of squares of deviation of y from its mean is 250. The number of observations for this bivariate data is
- a)7
- b)8
- c)9
- d)10
Correct answer is option 'D'. Can you explain this answer?
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Verified Answer
For two variables x and y, it is known that cov (x, y) = 80, variance ...
we will try to solve the problem by exploiting the fact that correlations is always less than or equal to 1
correlation = 80/((standard deviation of x)*(stadard deviation of y)) standard deviation of x = squareroot(16) = 4 standard deviation of y = squareroot (250/N) = 5(10/N)^0.5 so we get 80/(4*5*(10/N)^0.5) <1 implies 4 < (10/n)^0.5 implies 16 <10/N implies N<10/16 which is impossible. the figures do not make sense once again. even if they do we cannot find out N just with the help of covariance.
correlation = 80/((standard deviation of x)*(stadard deviation of y)) standard deviation of x = squareroot(16) = 4 standard deviation of y = squareroot (250/N) = 5(10/N)^0.5 so we get 80/(4*5*(10/N)^0.5) <1 implies 4 < (10/n)^0.5 implies 16 <10/N implies N<10/16 which is impossible. the figures do not make sense once again. even if they do we cannot find out N just with the help of covariance.
Most Upvoted Answer
For two variables x and y, it is known that cov (x, y) = 80, variance ...
Solution:
Given, cov(x,y) = 80, variance of x is 16 and sum of squares of deviation of y from its mean is 250.
We know that,
Covariance of x and y, cov(x,y) = E[(x-μx)(y-μy)]
Variance of x, Var(x) = E[(x-μx)^2]
Sum of squares of deviation of y from its mean, ∑(yi-μy)^2
where,
μx, μy = mean of x and y respectively
Let the number of observations be n.
Calculating μx:
We know that, μx = E[x] and variance of x, Var(x) = E[(x-μx)^2]
Given, Var(x) = 16
So, E[(x-μx)^2] = 16
E[x^2 - 2μxx + μx^2] = 16
E[x^2] - 2μxE[x] + μx^2 = 16
We know that, E[x^2] = Var(x) + E[x]^2
So, Var(x) + E[x]^2 - 2μxE[x] + μx^2 = 16
Substituting Var(x) = 16, we get
16 + E[x]^2 - 2μxE[x] + μx^2 = 16
E[x]^2 - 2μxE[x] + μx^2 = 0
(E[x] - μx)^2 = 0
E[x] = μx
So, μx = E[x] = E[(x-μx)^2] = Var(x) = 16
Calculating μy:
We know that, ∑(yi-μy)^2 = E[(y-μy)^2]
Given, ∑(yi-μy)^2 = 250
So, E[(y-μy)^2] = 250
We know that, E[(y-μy)^2] = Var(y)
So, Var(y) = 250
Also, we know that, cov(x,y) = E[(x-μx)(y-μy)]
Substituting μx = 16, cov(x,y) = 80, and E[(x-μx)(y-μy)] = cov(x,y), we get
E[(x-16)(y-μy)] = 80
Expanding the above equation, we get
E[xy-16y-μyx+16μy] = 80
E[xy] - 16E[y] - 16μx + 16μy = 80
We know that, E[x] = μx = 16
Substituting the above values, we get
E[xy] - 16E[y] - 256 + 16μy = 80
E[xy] - 16E[y] + 16μy = 336
E[xy] - 16(E[y] - μy) = 336
We know that, Var(y) = E[(y-μy)^2]
Expanding the above equation, we get
Var(y) = E[y^2] - 2μy
Given, cov(x,y) = 80, variance of x is 16 and sum of squares of deviation of y from its mean is 250.
We know that,
Covariance of x and y, cov(x,y) = E[(x-μx)(y-μy)]
Variance of x, Var(x) = E[(x-μx)^2]
Sum of squares of deviation of y from its mean, ∑(yi-μy)^2
where,
μx, μy = mean of x and y respectively
Let the number of observations be n.
Calculating μx:
We know that, μx = E[x] and variance of x, Var(x) = E[(x-μx)^2]
Given, Var(x) = 16
So, E[(x-μx)^2] = 16
E[x^2 - 2μxx + μx^2] = 16
E[x^2] - 2μxE[x] + μx^2 = 16
We know that, E[x^2] = Var(x) + E[x]^2
So, Var(x) + E[x]^2 - 2μxE[x] + μx^2 = 16
Substituting Var(x) = 16, we get
16 + E[x]^2 - 2μxE[x] + μx^2 = 16
E[x]^2 - 2μxE[x] + μx^2 = 0
(E[x] - μx)^2 = 0
E[x] = μx
So, μx = E[x] = E[(x-μx)^2] = Var(x) = 16
Calculating μy:
We know that, ∑(yi-μy)^2 = E[(y-μy)^2]
Given, ∑(yi-μy)^2 = 250
So, E[(y-μy)^2] = 250
We know that, E[(y-μy)^2] = Var(y)
So, Var(y) = 250
Also, we know that, cov(x,y) = E[(x-μx)(y-μy)]
Substituting μx = 16, cov(x,y) = 80, and E[(x-μx)(y-μy)] = cov(x,y), we get
E[(x-16)(y-μy)] = 80
Expanding the above equation, we get
E[xy-16y-μyx+16μy] = 80
E[xy] - 16E[y] - 16μx + 16μy = 80
We know that, E[x] = μx = 16
Substituting the above values, we get
E[xy] - 16E[y] - 256 + 16μy = 80
E[xy] - 16E[y] + 16μy = 336
E[xy] - 16(E[y] - μy) = 336
We know that, Var(y) = E[(y-μy)^2]
Expanding the above equation, we get
Var(y) = E[y^2] - 2μy
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For two variables x and y, it is known that cov (x, y) = 80, variance of x is 16 and sum of squares of deviation of y from its mean is 250. The number of observations for this bivariate data isa)7b)8c)9d)10Correct answer is option 'D'. Can you explain this answer?
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