Statistics
Variance
Grade 11
Question:
<p>If a variable \(x\) takes values \(0, 1, 2, \ldots, n\) with frequencies proportional to the binomial coefficients \({}^nC_0, {}^nC_1, {}^nC_2, \ldots, {}^nC_n\), then \(\text{var}(X)\) is</p>
<p>\(\dfrac{n^2-1}{12}\)</p>
<p>\(\dfrac{n}{2}\)</p>
<p>\(\dfrac{n}{4}\)</p>
<p>None of these</p>
Step-by-Step Solution
Key Concept: Recognize that frequencies proportional to binomial coefficients form a binomial distribution. Use the property that the mean and variance of a binomial distribution with parameters n and p=1/2 are n/2 and n/4 respectively.
<p><strong>Step 1:</strong> The frequencies are proportional to binomial coefficients ${}^nC_0, {}^nC_1, \ldots, {}^nC_n$. By the binomial theorem: $$\sum_{r=0}^{n} {}^nC_r = 2^n$$ So the probability mass function is: $$P(X=r) = \frac{{}^nC_r}{2^n}$$</p><p><strong>Step 2:</strong> This is a binomial distribution with parameters $n$ and $p = \frac{1}{2}$. This is because: $$P(X=r) = {}^nC_r \left(\frac{1}{2}\right)^r \left(\frac{1}{2}\right)^{n-r} = \frac{{}^nC_r}{2^n}$$</p><p><strong>Step 3:</strong> For a binomial distribution $B(n,p)$: $$\text{var}(X) = np(1-p)$$ With $n=n$ and $p=\frac{1}{2}$: $$\text{var}(X) = n \cdot \frac{1}{2} \cdot \frac{1}{2} = \frac{n}{4}$$</p><p>∴ Answer: C</p>
Correct Answer: C