Question

Let $a, b, c \in \mathbb{N}$ and $a<b<c$. Let the mean, the mean deviation about the mean and the variance of the 5 observations $9, 25, a, b, c$ be $18, 4$ and $\frac{136}{5}$, respectively. Then $2a + b - c$ is equal to _______.

Answer 1

Given:
$\text{Mean} = \frac{9 + 25 + a + b + c}{5} = 18.$
Solving for $a + b + c$:
$a + b + c = 56.$
The mean deviation about the mean is given by:
$\text{Mean deviation} = \frac{\sum |x_i - \bar{x}|}{n} = 4.$
Substituting values:
$|9 - 18| + |25 - 18| + |a - 18| + |b - 18| + |c - 18| = 20.$
$|18 - a| + |18 - b| + |18 - c| = 4.$
The variance is given by:
$\text{Variance} = \frac{\sum (x_i - \bar{x})^2}{n} = \frac{136}{5}.$
Calculating:
$\frac{(9 - 18)^2 + (25 - 18)^2 + (a - 18)^2 + (b - 18)^2 + (c - 18)^2}{5} = \frac{136}{5}.$
Multiplying both sides by 5:
$81 + 49 + (18 - a)^2 + (18 - b)^2 + (18 - c)^2 = 136.$
Simplifying:
$(18 - a)^2 + (18 - b)^2 + (18 - c)^2 = 6.$
Possible values:
$(18 - a)^2 = 1, \quad (18 - b)^2 = 1, \quad (18 - c)^2 = 4.$
This gives:
$18 - a = 1 \implies a = 17, \quad 18 - b = -1 \implies b = 19, \quad 18 - c = -2 \implies c = 20.$
Substituting:
$a + b + c = 17 + 19 + 20 = 56.$
Calculating $2a + b - c$:
$2a + b - c = 2 \times 17 + 19 - 20 = 34 + 19 - 20 = 33.$
Answer: 33.