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Statistics question

2022 · 29 Jul · Shift 1 · Q41
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Statistics question

2022 · 29 Jul · Shift 1 · Q41

JEE MainMathematicsStatisticsNumerical+4 / −1
Let the mean and the variance of 20 observations x1,x2,…,x20x_{1}, x_{2}, \ldots, x_{20}x1​,x2​,…,x20​ be 15 and 9 , respectively. For α∈R\alpha \in \mathbf{R}α∈R, if the mean of (x1+α)2,(x2+α)2,…,(x20+α)2\left(x_{1}+\alpha\right)^{2},\left(x_{2}+\alpha\right)^{2}, \ldots,\left(x_{20}+\alpha\right)^{2}(x1​+α)2,(x2​+α)2,…,(x20​+α)2 is 178 , then the square of the maximum value of α\alphaα is equal to ‾\underline{\hspace{2cm}}​.
Numerical answer
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Correct answer: 4

  1. Let the mean of the 20 observations be xˉ=15\bar{x}=15xˉ=15 and the variance be 999.

  2. Using the formula

Variance=1n∑xi2−xˉ2,\text{Variance} = \frac{1}{n}\sum x_i^2 - \bar{x}^2,Variance=n1​∑xi2​−xˉ2,

we get

9=120∑i=120xi2−152.9 = \frac{1}{20}\sum_{i=1}^{20} x_i^2 - 15^2.9=201​i=1∑20​xi2​−152.

So,

120∑i=120xi2=9+225=234.\frac{1}{20}\sum_{i=1}^{20} x_i^2 = 9+225=234.201​i=1∑20​xi2​=9+225=234.

Hence,

mean of xi2=234.\text{mean of } x_i^2 = 234.mean of xi2​=234.
  1. Now consider the numbers (xi+α)2(x_i+\alpha)^2(xi​+α)2. Their mean is given as 178178178.

Expand:

(xi+α)2=xi2+2αxi+α2.(x_i+\alpha)^2 = x_i^2 + 2\alpha x_i + \alpha^2.(xi​+α)2=xi2​+2αxi​+α2.

Taking mean on both sides,

mean of (xi+α)2=mean of xi2+2α(mean of xi)+α2.\text{mean of }(x_i+\alpha)^2 = \text{mean of }x_i^2 + 2\alpha(\text{mean of }x_i) + \alpha^2.mean of (xi​+α)2=mean of xi2​+2α(mean of xi​)+α2.

Therefore,

178=234+2α(15)+α2.178 = 234 + 2\alpha(15) + \alpha^2.178=234+2α(15)+α2.

So,

178=234+30α+α2.178 = 234 + 30\alpha + \alpha^2.178=234+30α+α2.

Rearranging,

α2+30α+56=0.\alpha^2 + 30\alpha + 56 = 0.α2+30α+56=0.
  1. Solve the quadratic:
α2+30α+56=0.\alpha^2 + 30\alpha + 56 = 0.α2+30α+56=0.

Factorizing,

(α+2)(α+28)=0.(\alpha+2)(\alpha+28)=0.(α+2)(α+28)=0.

Thus,

α=−2orα=−28.\alpha=-2 \quad \text{or} \quad \alpha=-28.α=−2orα=−28.

The maximum value of α\alphaα is −2-2−2.

  1. Hence, the square of the maximum value is
(−2)2=4.(-2)^2 = 4.(−2)2=4.

Therefore, the required integer is 444.

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