Find the missing frequencies in the following frequency distribution if it is known that the mean of the distribution is 1.46. No. of accident (x)    0         1                     2           3         4                     5           Total Frequency (f):           46       ?                     ?           25       10                     5           200

Explanation

Let the missing frequencies be `f_1` and `f_2`.
`46+f_1+f_2+25+10+5=200`
`f_1+f_2=200-86`
`f_1+f_2=114`
`f_1=114-f_2`
Mean`=(sumf_i\x_i)/(sumf_i)`
`1.46=((46xx0)+(1xxf_1)+(2xxf_2)+(3xx25)+(4xx10)+(5xx5))/200`
`292=(f_1+2f_2+75+40+25)`
`f_1+2f_2=292-140`
`f_1+2f_2=152`
`114-f_2+2f_2=152`
`f_2=152-114`
`f_2=38`
Now,
`f_1+f_2=114`
`f_1+38=114`
`f_1=76`
Hence, the values of `f_1` and `f_2` is `76` and `38` respectively.