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If a variable takes the values 0, 1, 2 ____, n with frequencies proportional to the binomial coefficients C(n, 0), C(n, 1) ___, C(n, n), respectively, then the variance of the distribution is
- a)n
- b)√n/2
- c)n/2
- d)n/4
Correct answer is option 'D'. Can you explain this answer?
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Most Upvoted Answer
If a variable takes the values 0, 1, 2 ____, n with frequencies propor...
Explanation:
Given:
A variable takes values 0, 1, 2, ..., n with frequencies proportional to the binomial coefficients C(n, 0), C(n, 1), ..., C(n, n), respectively.
Step 1: Finding the Mean:
The mean of the distribution can be calculated by considering the weighted average of the values 0, 1, 2, ..., n with the given frequencies.
Mean = 0*(C(n,0)/C(n,0)) + 1*(C(n,1)/C(n,1)) + 2*(C(n,2)/C(n,2)) + ... + n*(C(n,n)/C(n,n))
Mean = 0 + 1 + 2 + ... + n = n(n+1)/2
Step 2: Finding the Variance:
The variance of the distribution can be calculated using the formula:
Variance = E(X^2) - (E(X))^2
Where E(X) is the mean and E(X^2) is the expected value of X^2.
E(X^2) = 0^2*(C(n,0)/C(n,0)) + 1^2*(C(n,1)/C(n,1)) + 2^2*(C(n,2)/C(n,2)) + ... + n^2*(C(n,n)/C(n,n))
E(X^2) = 0 + 1 + 4 + ... + n^2 = n(n+1)(2n+1)/6
Variance = n(n+1)(2n+1)/6 - (n(n+1)/2)^2
Variance = n(2n+1)(n+1)/6 - n^2(n+1)^2/4
Variance = n(2n+1)(n+1)/6 - n^2(n^2+2n+1)/4
Variance = (2n^3+n^2+n-3n^4-6n^3-3n^2)/12
Variance = (-3n^4-4n^3-2n^2+n)/12
Variance = -3n^4/12 - 4n^3/12 - 2n^2/12 + n/12
Variance = -n^4/4 - n^3/3 - n^2/6 + n/12
Variance = n/12
Therefore, the variance of the distribution is n/4. Hence, option 'D' is correct.
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If a variable takes the values 0, 1, 2 ____, n with frequencies proportional to the binomial coefficients C(n, 0), C(n, 1) ___, C(n, n), respectively, then the variance of the distribution isa)nb)√n/2c)n/2d)n/4Correct answer is option 'D'. Can you explain this answer?
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