Question:

Consider 10 observations x1,x2,,x10x_1, x_2, \ldots, x_{10} such that i=110(xiα)=2andi=110(xiβ)2=40,\sum_{i=1}^{10} (x_i - \alpha) = 2 \quad \text{and} \quad \sum_{i=1}^{10} (x_i - \beta)^2 = 40,where α,β\alpha, \beta are positive integers. Let the mean and the variance of the observations be 65\frac{6}{5} and 8425\frac{84}{25}, respectively. The value of βα\frac{\beta}{\alpha} is equal to:

Updated On: Mar 20, 2025
  • 2
  • 32\frac{3}{2}
  • 52\frac{5}{2}
  • 1
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The Correct Option is A

Solution and Explanation

We are given:

i=110Xi10A=2    i=110Xi=10A+2\sum_{i=1}^{10} X_i - 10A = 2 \implies \sum_{i=1}^{10} X_i = 10A + 2.

i=110Xi10B=40    i=110Xi=10B+40\sum_{i=1}^{10} X_i - 10B = 40 \implies \sum_{i=1}^{10} X_i = 10B + 40.

Equating both expressions for i=110Xi\sum_{i=1}^{10} X_i, we get:

10A+2=10B+40    10A10B=38    AB=3.810A + 2 = 10B + 40 \implies 10A - 10B = 38 \implies A - B = 3.8.

Since A and B are integers, A=4A = 4 and B=2B = 2.

Thus, B=2B = 2.

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