Applied Mathematics for Electrical Engineering (3130908)

Question-5c ( ii )

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

Q5) (c ( ii ))

Two automatic filling machines A and B are used to fill mixture of cement, concrete and beam. A random sample of beam on each machine showed following results.

$A$ $32$ $28$ $47$ $63$ $71$ $39$ $10$ $60$ $96$ $14$

$B$ $19$ $31$ $48$ $53$ $67$ $90$ $10$ $62$ $40$ $80$

Find standard deviation of each machine and also comment on the performance of the two machine.

Mean for machine A

\bar{x} = \frac{\sum_{} x_{A}}{n} = \frac{32 + 28 + 47 + 63 + 71 + 39 + 10 + 60 + 96 + 14}{10} = \frac{460}{10} = 46

Standard Deviation for machine A

$x_{A}$	$x_{A} - {\bar{x}}_{A}$	${(x_{A} - {\bar{x}}_{A})}^{2}$
$32$	$- 14$	$196$
$28$	$- 18$	$324$
$47$	$1$	$1$
$63$	$17$	$289$
$71$	$25$	$625$
$39$	$- 7$	$49$
$10$	$- 36$	$1296$
$60$	$14$	$196$
$96$	$50$	$2500$
$14$	$- 32$	$1024$
$\sum {(x_{A} - \bar{x_{A}})}^{2} = 6500$	$\sum {(x_{A} - \bar{x_{A}})}^{2} = 6500$	$\sum {(x_{A} - {\bar{x}}_{A})}^{2} = 6500$

σ_{A} = \sqrt{\frac{\sum {(x_{A} - {\bar{x}}_{A})}^{2}}{n}} = \sqrt{\frac{6500}{10}} = 25.4951

Mean for machine B

\bar{x} = \frac{\sum_{} x_{B}}{n} = \frac{19 + 31 + 48 + 53 + 67 + 90 + 10 + 62 + 40 + 80}{10} = \frac{500}{10} = 50

Standard Deviation for machine B

$x_{B}$	$x_{B} - {\bar{x}}_{B}$	${(x_{B} - {\bar{x}}_{B})}^{2}$
$19$	$- 31$	$961$
$31$	$- 19$	$361$
$48$	$3$	$4$
$53$	$3$	$9$
$67$	$17$	$289$
$90$	$40$	$1600$
$10$	$- 40$	$1600$
$62$	$12$	$144$
$40$	$- 10$	$100$
$80$	$30$	$900$
$\sum {(x_{A} - \bar{x_{A}})}^{2} = 6500$	$\sum {(x_{A} - \bar{x_{A}})}^{2} = 6500$	$\sum {(x_{B} - {\bar{x}}_{B})}^{2} = 5968$

σ_{B} = \sqrt{\frac{\sum {(x_{B} - {\bar{x}}_{B})}^{2}}{n}} = \sqrt{\frac{5968}{10}} = 24.4295

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