Applied Mathematics for Electrical Engineering (3130908)

Question-5b

BE | Semester-3 Winter-2019 | 26-11-2019

Q5) (b)

Moment about actual mean is

μ_{r} = \frac{\sum_{} {(x_{i} - \bar{x})}^{r}}{n}; r = 0, 1, 2, 3, 4, . . .

Therefore, we need to find

\bar{x}

\bar{x} = \frac{\sum_{} x_{i}}{n} = \frac{5 + 10 + 8 + 13 + 4}{5} = \frac{40}{5} = 8 .

$x_{i}$	$x_{i} - \bar{x}$	${(x_{i} - \bar{x})}^{2}$	${(x_{i} - \bar{x})}^{3}$	${(x_{i} - \bar{x})}^{4}$
$5$	$- 3$	$9$	$- 27$	$81$
$10$	$2$	$4$	$8$	$16$
$8$	$0$	$0$	$0$	$0$
$13$	$5$	$25$	$125$	$625$
$4$	$- 4$	$16$	$- 64$	$256$
	$\sum_{} (x_{i} - \bar{x}) = 0$	$\sum_{} {(x_{i} - \bar{x})}^{2} = 54$	$\sum_{} {(x_{i} - \bar{x})}^{3} = 42$	$\sum_{} {(x_{i} - \bar{x})}^{4} = 978$

μ_{r} = \frac{\sum_{} {(x_{i} - \bar{x})}^{r}}{n}; r = 0, 1, 2, 3, 4, . . .

μ_{1} = \frac{\sum_{} (x_{i} - \bar{x})}{n} = \frac{0}{5} = 0 .

μ_{2} = \frac{\sum_{} {(x_{i} - \bar{x})}^{2}}{n} = \frac{54}{5} = 10.8

μ_{3} = \frac{\sum_{} {(x_{i} - \bar{x})}^{3}}{n} = \frac{42}{5} = 8.4

μ_{4} = \frac{\sum_{} {(x_{i} - \bar{x})}^{4}}{n} = \frac{978}{5} = 195.6

Therefore, first four moments about mean are

0, 10.8, 8.4, 195.6 .