Question

Find the missing frequencies and the median for the following distribution if the mean is 1.46 and the sum of all frequencies is 200

No. of accidents	0	1	2	3	4	5
No. of days	46	$f_1$	$f_2$	25	10	5

• A. $f_1=76$ ,$f_2=38$
• B. Median $=1$
• C. All of the above
• D. None of these

Answer 1

Answer: (C)

All of the above

$x_i$	$f_i$	$f_i{x}_i$
0	46	0
1	$f_1$	$f_1$
2	$f_2$	$2f_2$
3	25	75
4	10	40
5	5	25
	$\sum f_i=86+f_1+f_2$	$\sum f_i{x}_i=140+f_1+2f_2$

Given $\overline{x}=1.46,\sum f_i=200\Rightarrow \overline{x}=\frac{\sum f_i{x}_i}{\sum f_i}$
$1.46=\frac{140+f_1+2f_2}{200}\Rightarrow 292=140+f_1+2f_2\Rightarrow f_1+2f_2=\mathrm{152...........}\left(i\right)$
Also $86+f_1+f_2=200\Rightarrow f_1+f_2=\mathrm{114.........}\left(ii\right)$
Solving equation (i) and (ii) $f_1=76and{f}_2=38$ For Median

$x_i$	$f_i$	c.f.
0	46	46
1	76	122
2	38	160
3	25	185
4	10	195
5	5	200

$N=200$ So, Median $=\frac{{\left(\frac{N}{2}\right)}^{th}obs+{(\frac{N}{2}+1)}^{th}obs}{2}$
$=\frac{{100}^{th}obs+{101}^{th}obs}{2}=\frac{1+1}{2}=1$
$\therefore$ Median = 1