Applied Mathematics for Electrical Engineering (3130908)

Question-5c ( ii )-OR

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

Q5) (c ( ii ))

Skewness =

β_{1} = \frac{μ_{3}^{2}}{μ_{2}^{3}}

Where,

μ_{2} = \frac{\sum_{} f_{i} {(x_{i} - \bar{x})}^{2}}{\sum_{} f_{i}}

and

μ_{3} = \frac{\sum_{} f_{i} {(x_{i} - \bar{x})}^{3}}{\sum_{} f_{i}}

Here, We need to find

\bar{x}

$Class$	$x_{i}$	$f_{i}$	$f_{i} x_{i}$
$50 - 55$	$52.5$	8	420
$55 - 60$	$57.5$	10	575
$60 - 65$	$62.5$	15	937.5
$65 - 70$	$67.5$	17	1147.5
$70 - 75$	$72.5$	8	580
$\sum_{} x_{i} = 3660$	$\sum_{} x_{i} = 3660$	$\sum_{} f_{i} = 58$	$\sum_{} f_{i} x_{i} = 3660$

\bar{x} = \frac{\sum_{} f_{i} x_{i}}{\sum_{} f_{i}} = \frac{3660}{58} = 63.1034

$Class$	$x_{i}$	$x_{i} - \bar{x}$	${(x_{i} - \bar{x})}^{2}$	${(x_{i} - \bar{x})}^{3}$
$50 - 55$	$52.5$	$- 10.6034$	$112.4321$	$- 1192.1624$
$55 - 60$	$57.5$	$- 5.6034$	$31.3981$	$- 175.9361$
$60 - 65$	$62.5$	$- 0.6034$	$0.3641$	$- 0.2197$
$65 - 70$	$67.5$	$5.3966$	$29.1233$	$157.1668$
$70 - 75$	$72.5$	$9.3966$	$88.2961$	$829.6831$
$\sum_{} x_{i} = 3660$	$\sum_{} x_{i} = 3660$	$\sum_{} {(x_{i} - \bar{x})}^{3} = - 381.4683$	$\sum_{} {(x_{i} - \bar{x})}^{2} = 261.6137$	$\sum_{} {(x_{i} - \bar{x})}^{3} = - 381.4683$

Then

μ_{2} = \frac{\sum_{} f_{i} {(x_{i} - \bar{x})}^{2}}{\sum_{} f_{i}} = \frac{261.6137}{58} = 4.5106

μ_{3} = \frac{\sum_{} f_{i} {(x_{i} - \bar{x})}^{3}}{\sum_{} f_{i}} = - \frac{381.4683}{58} = 6.5770

Skewness =

β_{1} = \frac{μ_{3}^{2}}{μ_{2}^{3}} = \frac{{(4.5106)}^{2}}{{(6.5770)}^{3}} = 0.0715