Question
Download Solution PDFLet X be a discrete variable taking values -3, -2, -1 and 0 with probabilities 1/3, 1/4 ,1/12 and 1/3. The variance of X
Answer (Detailed Solution Below)
Detailed Solution
Download Solution PDF| Values (Xi) | Probabilities (Pi) | \(\prod X_iP(X_i) \) |
| \((-3)\) | \(1\over3\) | \((-3) \times {1\over3} = -1 \) |
| \((-2)\) | \(1\over4\) | \((-2) \times {1\over4} = - {1\over 2} \) |
| \((-1)\) | \(1\over12\) | \((-1) \times {1\over12} = - {1\over 12} \) |
| \(0\) | \(1\over3\) | \(0 \times {1\over 3} = 0 \) |
| Total | 1 | \(\mu = \sum X_iP(X_i) = -{19 \over 12} \) |
| Values (Xi) | Probability (Pi) | \((X_i - \mu) \) | \((X_i - \mu)^2 \) | \((X_i - \mu)^2 \times P_i \) |
| \((-3)\) | \(1\over 3\) | \([(-3) - ({-{19\over 12}})] = - {17 \over 12} \) | \((- {17\over 12})^2 = {289 \over 144} \) | \({289 \over 144} \times {1\over 3}\) |
| \((-2)\) | \(1\over 4\) | \([(-2) - ({-{19\over 12}})] = - {5 \over 12} \) | \((- {5\over 12})^2 = {25 \over 144} \) | \({25 \over 144} \times {1\over 4}\) |
| \((-1)\) | \(1\over12\) | \([(-1) - ({-{19\over 12}})] = {7 \over 12} \) | \((- {7\over 12})^2 = {34 \over 144} \) | \({49 \over 144} \times {1\over 12}\) |
| \(0\) | \(1\over 3\) | \([(0) - ({-{19\over 12}})] = {19\over 12} \) | \(({19\over 12})^2 = {361 \over 144} \) | \({361 \over 144} \times {1\over 3}\) |
| Total | 1 | \(\sum_i [(X_i - \mu)^2\times P_i] = {227\over 144} \) |
Hence, the variance of X = Var(X) = \(227\over 144\)
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