If the frequency distribution's mean is 1.46, then select the appropriate options –

No. of Accidents	0	1	2	3	4	5	total
Frequency (f)	46	$f_1$	$f_2$	25	10	5	200

The sum of the values of all the observations $=f_1{x}_1+f_2{x}_2+\dots +f_n{x}_n$, and the number of observations $=f_1+f_2+\dots +f_n$.

So, the mean $\overline{x}$of the data is given by

$\overline{x}=\frac{f_1{x}_1+f_2{x}_2+\cdots +f_n{x}_n}{f_1+f_2+\cdots +f_n}$

According to the table,

$N=\sum f_1=86+f_1+f_2;\sum f_i{x}_i=140+f_1+f_2$

It is given that $\mathrm{N}=200$.

Therefore, $86+f_1+f_2=200\Rightarrow f_1+f_2=114$

Now, $\overline{x}=1.46\Rightarrow 1.46=\frac{\sum f_i{x}_i}{N}$

$\Rightarrow 1.46=\frac{140+f_1+2f_2}{200}$

$\Rightarrow 292=140+f_1+2f_2$

$\Rightarrow f_1+2f_2=152$

Solving (1) and (2),we get$f_1=76$ and $f_2=38$