Question

Length of $40$ bits of wire, correct to the nearest centimetre are given below. Calculate the variance.

Length cm	$1-10$	$11-20$	$21-30$	$31-40$	$41-50$	$51-60$	$61-70$
No. of bits	$2$	$3$	$8$	$12$	$9$	$5$	$1$

Answer 1

Let the assumed mean A be $35.5$.

Length	mid value x	no. of bits(f)	$d=x-A$	fd	$f{d}^2$
$1-10$	$5.5$	$2$	$-30$	$-60$	$1800$
$11-20$	$15.5$	$3$	$-20$	$-60$	$1200$
$21-30$	$25.5$	$8$	$-10$	$-80$	$800$
$31-40$	$35.5$	$12$	$0$	$0$	$0$
$41-50$	$45.5$	$9$	$10$	$90$	$900$
$51-60$	$55.5$	$5$	$20$	$100$	$2000$
$61-70$	$65.5$	$1$	$30$	$30$	$900$
		$\sum f=40$		$\sum fd=20$	$\sum f{d}^2=7600$

Variance, ${\sigma }^2=\frac{\sum f{d}^2}{\sum f}-{\left(\frac{\sum fd}{\sum f}\right)}^2=\frac{7600}{40}={\left(\frac{20}{40}\right)}^2$
$=190-\frac{1}{4}=\frac{760-1}{4}=\frac{759}{4}$
$\therefore {\sigma }^2=189.75$.