Let x, M and ^2 be respectively the mean, mode and variance of n observations x_1,x_2,,x_n…

Question

Let $\bar x$, M and $\sigma^2$ be respectively the mean, mode and variance of n observations $x_1,x_2,\ldots,x_n$ and $d_i=-x_i-a$, $i=1,2,\ldots,n$, where a is any number.

Statement I: Variance of $d_1,d_2,\ldots,d_n$ is $\sigma^2$

Statement II: Mean and mode of $d_1,d_2,\ldots,d_n$ are $-\bar x-a$ and $-M-a$, respectively.

Accepted Answer

Mean of $d_i$: $\dfrac{\sum(-x_i-a)}{n} = -\bar x-a$. Since negating/shifting each observation shifts the mode identically, mode of $d_i$ is $-M-a$. So Statement II is true.

Variance: $\dfrac1n\sum[d_i-(-\bar x-a)]^2 = \dfrac1n\sum(x-x_i)^2 = \sigma^2$ (the shift and sign flip cancel in the squared deviation). So Statement I is true.

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