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# Given that x is the mean and σ² is the variation of n observations x_{1}, x_{2}, .... , x_{n} Prove that the mean and variance of the observations ax_{1}, ax_{2}, .... , ax_{n} recpectively. (a ≠ 0)

**Solution:**

The given n observations are x_{1}, x_{2}, .... , x_{n}

Mean = x

Variance = σ^{2}

Therefore,

σ² = 1/n ∑^{n}_{i = 1}y_{i }(x_{i} - x)² ....(1)

If each observation is multiplied by a and the new observations are y_{i}, then

y_{i} = ax_{i} i.e., x_{i} = 1/n y_{i}

Hence,

y = 1/n ∑^{n}_{i = 1}y_{i}

= 1/n ∑^{n}_{i = 1}ax_{i}

= a/n ∑^{n}_{i = 1}x_{i}

= a x [∵ x = 1/n ∑^{n}_{i = 1}x_{i}]

Therefore, mean of the observations, x_{1}, x_{2}, .... , x_{n} is a x

Substituting the values of x and x_{i} in (1) , we obtain

σ² = 1/n ∑^{n}_{i = 1} (1/a y_{i} - 1/a y)²

a²σ² = 1/n ∑^{n}_{i = 1} (y_{i} - y)²

Thus, the variance of the observations ax_{1}, ax_{2}, .... , ax_{n} is a a²σ²

NCERT Solutions Class 11 Maths Chapter 15 Exercise ME Question 4

## Given that x is the mean and σ² is the variation of n observations x_{1}, x_{2}, .... , x_{n} Prove that the mean and variance of the observations ax_{1}, ax_{2}, .... , ax_{n} recpectively. (a ≠ 0)

**Summary:**

Therefore, mean of the observations, x_{1}, x_{2}, .... , x_{n} is a x and thus the variance of the observations ax_{1}, ax_{2}, .... , ax_{n} is a a^{2}σ^{2}

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