Let X = 11, 12, 13, , 40, 41 and Y = 61, 62, 63, , 90, 91 be the two sets of observations.

Question

Let $X = \{11, 12, 13, \ldots, 40, 41\}$ and $Y = \{61, 62, 63, \ldots, 90, 91\}$ be the two sets of observations. If $\bar{x}$ and $\bar{y}$ are their respective means and $\sigma^2$ is the variance of all the observations in $X \cup Y$, then $|\bar{x}+\bar{y}-\sigma^2|$ is equal to \_\_\_\_\_\_\_.

Answer: 603

Accepted Answer

$\bar{x} = \dfrac{11+41}{2}=26$ (31 terms), $\bar{y} = \dfrac{61+91}{2}=76$ (31 terms).

Combined mean: $\mu = \dfrac{31(26)+31(76)}{62} = \dfrac{26+76}{2}=51$

$$\sigma^2 = \frac{1}{62}\left[\sum_{i=1}^{31}(x_i-\mu)^2+\sum_{i=1}^{31}(y_i-\mu)^2\right]$$

Since both $X$ and $Y$ are 31-term A.P.s with common difference 1, and by symmetry about $\mu=51$:
$$\sum(x_i-\mu)^2 = \sum(y_i-\mu)^2 = 10^2+11^2+\cdots+40^2 = \frac{40\times41\times81}{6}-\frac{9\times10\times19}{6} = 21855$$

$$\sigma^2 = \frac{2\times21855}{62} = 705$$

$$|\bar{x}+\bar{y}-\sigma^2| = |26+76-705| = 603$$

Solutions · 1