Find the missing frequencies and the median for the following distribution if the mean is 1.46.
No.of accidents: 0 1 2 3 4 5 Total
Frequency(No.of days): 46 ? ? 25 10 50 200
Answers
The missing frequencies are 76 and 38.
The median is 1.
No.of accidents No.of days fx
(x) (f)
0 46 0
1 x x
2 y 2y
3 25 75
4 10 40
5 5 25
N = 200 ∑fx = x+2y+140
Given,
N = 200
46 + x + y + 25 + 10 + 5 = 200
86 + x + y = 200
x + y = 114 .........(1)
Mean = 1.46
Mean = ∑fx / ∑f
1.46 = (x+2y+140) / 200
292 = x + 2y + 140
x + 2y = 152 ............(2)
Solving (1) and (2), we get,
x = 76
y = 38
Therefore, the missing frequencies are 76 and 38.
No.of accidents No.of days Cumulative frequency
(x) (f) cf
0 46 46
1 76 76+46=122
2 38 38+122=160
3 25 25+160=185
4 10 185+10=195
5 5 195+5=200
N = 200
N/2 = 200/2 = 100
Thus, the cumulative frequency just greater than N/2 = 100 is 122 and the corresponding value of number of accidents is 1.
Therefore the median is 1.