Consider 10 observations x_1, x_2, \ldots, x_10 such that \sum\limits_i=1^10(x_i-\alpha)=2 and \sum\limits_i=1^10(x_i-\beta)^2=40 , where \alpha, \beta are positive integers. Let the mean and the variance of the observations be \frac65 and \frac8425 respectively. Then \frac\beta\alpha is equal to :

Statistics PYQ — JEE Main 2024 · 1 Feb · Shift 2 · Q40 with Solution

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Correct answer: A

There are $10$ observations $x_1,x_2,\dots,x_{10}$ .

We are given: $\sum_{i=1}^{10}(x_i-\alpha)=2$ and $\sum_{i=1}^{10}(x_i-\beta)^2=40.$

Also, mean $\bar{x}=\frac{6}{5}$ and variance $\sigma^2=\frac{84}{25}$ .

Since there are $10$ observations, $\sum_{i=1}^{10} x_i = 10\cdot \frac{6}{5}=12.$

Now, $\sum_{i=1}^{10}(x_i-\alpha)=\sum x_i-10\alpha=2.$ So, $12-10\alpha=2$ $10\alpha=10$ $\alpha=1.$

Variance of $10$ observations is $\sigma^2=\frac{1}{10}\sum_{i=1}^{10}(x_i-\bar{x})^2.$

Thus, $\sum_{i=1}^{10}(x_i-\bar{x})^2=10\cdot \frac{84}{25}=\frac{168}{5}.$

Now use $\sum (x_i-\bar{x})^2=\sum x_i^2-10\bar{x}^2.$ So,

=\frac{168}{5}+10\left(\frac{6}{5}\right)^2.$$ Compute: $$10\left(\frac{6}{5}\right)^2=10\cdot \frac{36}{25}=\frac{72}{5}.$$ Hence, $$\sum x_i^2=\frac{168}{5}+\frac{72}{5}=\frac{240}{5}=48.$$ --- 4. **Use the second condition to find $\beta$** Expand: $$\sum_{i=1}^{10}(x_i-\beta)^2=\sum x_i^2-2\beta\sum x_i+10\beta^2=40.$$ Substitute $\sum x_i^2=48$ and $\sum x_i=12$: $$48-2\beta(12)+10\beta^2=40.$$ So, $$48-24\beta+10\beta^2=40$$ $$10\beta^2-24\beta+8=0$$ $$5\beta^2-12\beta+4=0.$$ Solve: $$5\beta^2-12\beta+4=(5\beta-2)(\beta-2)=0.$$ Thus, $$\beta=\frac{2}{5} \quad \text{or} \quad \beta=2.$$ But $\beta$ is a positive integer, so $$\beta=2.$$ --- 5. **Find $\frac{\beta}{\alpha}$** Since $\alpha=1$ and $\beta=2$, $$\frac{\beta}{\alpha}=\frac{2}{1}=2.$$ --- 6. **Check options** The correct option is: $$\boxed{\text{A: }2}$$