The following information is available about defective staplers a… - Vidyadip

Question

The following information is available about defective staplers after testing $50$ packets of $500$ staplers each. Find coefficient of skewness using Karl Pearson’s method.

No. of defective staplers	$19$	$20$	$21$	$22$	$23$	$24$	$25$	$26$
No. of packets	$5$	$18$	$10$	$8$	$4$	$2$	$2$	$1$

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Answer

No. of defective staplers $x$	No. of packets $f$	$d = (x – A) A = 22$	$fd$	$fd^2 = fd.d$
$19$	$5$	$-3$	$-15$	$45$
$20$	$18$	$-2$	$-36$	$72$
$21$	$10$	$-1$	$-10$	$10$
$22$	$8$	$0$	$0$	$0$
$23$	$4$	$1$	$4$	$4$
$24$	$2$	$2$	$4$	$8$
$25$	$2$	$3$	$6$	$18$
$26$	$1$	$4$	$4$	$16$
Total	$n = 50$	-	$18 - 61$ $\sum f d = -43$	$\sum f d^2=173$

Mean :
$s=\sqrt{\frac{\sum fd ^2}{ n }-\left(\frac{\sum fd }{ n }\right)^2}$
$=22-\frac{43}{50}$
$=22-0.86$
$=21.14$ packets
Mode :
$M _0=$ The observation having maximum frequency $18 =20$ packets
Standard deviation :
$s =\sqrt{\frac{\sum fd ^2}{ n }-\left(\frac{\sum fd }{ n }\right)^2}$
$ =\sqrt{\frac{173}{50}-\left(\frac{-43}{50}\right)^2}$
$ =\sqrt{3.46-(-0.86)^2}$
$ =\sqrt{3.46-0.7396}$
$ =\sqrt{2.7204}$
$ =1.65$ packets
Coefficient of skewness by karl pearson's
method:
$j =\frac{\overline{\bar{x}}- M _0}{8}$
$=\frac{21.14-20}{1.65}$
$=\frac{1.14}{1.65}$
$\therefore j =0.69$

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