Question

The time(in seconds) taken by a group of people to walk across a pedestrian crossing is given in the table below.

Time (in sec.)	$5-10$	$10-15$	$15-20$	$20-25$	$25-30$
No. of people	$4$	$8$	$15$	$12$	$11$

If variation is 36.76, so calculate standard deviation of the data.

Answer 1

$Time$	$Mid-point$ $x$	$f$	$xf$	$(x-\overline{x})$	$(x-\overline{x}{)}^2$	$f(x-\overline{x}{)}^2$
$5-10$	$7.5$	$4$	$30$	$-11.8$	$139.24$	$556.96$
$10-15$	$12.5$	$8$	$100$	$-6.8$	$46.24$	$369.92$
$15-20$	$17.5$	$15$	$262.5$	$-1.8$	$3.24$	$48.6$
$20-25$	$22.5$	$12$	$270$	$3.2$	$10.24$	$122.88$
$25-30$	$27.5$	$11$	$302.5$	$8.2$	$67.24$	$739.64$
$Total$		$50$	$965$			$1838$

Here, $\sum xf=965$ and $\sum f=50$

$\Rightarrow$ $\overline{x}=\frac{\sum xf}{\sum f}=\frac{965}{50}=19.3$

Now, $\sum f(x-\overline{x}{)}^2=1838$

$\Rightarrow$ Variance $\left({\sigma }^2\right)=\frac{\sum f(x-\overline{x}{)}^2}{\sum f}$

$=\frac{1838}{50}$

$=36.76$

$\Rightarrow$ Standard deviation $\left(\sigma \right)=\sqrt{\frac{\sum f(x-\overline{x}{)}^2}{\sum f}}$

$=\sqrt{\frac{1838}{50}}$

$=6.06$