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Question
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 | 5 | 200 |
Solution
| No. of accidents (x) |
No. of days (f) | fx |
| 0 | 45 | 0 |
| 1 | x | x |
| 2 | y | 2y |
| 3 | 25 | 75 |
| 4 | 10 | 40 |
| 5 | 5 | 25 |
| N = 200 | `sumf_1x_1=x+2y+140` |
Given, N = 200
⇒ 46 + x + y + 25 + 10 + 5 = 200
⇒ 86 + x + y = 200
⇒ x + y = 200 - 86
⇒ x+ y = 114 ..............(1)
And Mean = 1.46
`rArr(sumfx)/N=1.46`
`rArr(x+2y+140)/200=1.46`
⇒ x + 2y + 140 = 1.46 x 200
⇒ x + 2y + 140 = 292
⇒ x + 2y = 292 - 140
⇒ x + 2y = 152 ................(2)
Subtract equation (1) from equation (2)
⇒ x + 2y - (x+ y) = 152 - 114
⇒ x + 2y - x - y = 38
⇒ y = 38
Put the value of y in (1),
⇒ x+ y = 114
⇒ x + 38 = 114
⇒ x = 114 - 38
⇒ x = 76
Hence, the missing frequencies are 38 and 76.
(2) Calculation of median.
| No. of accidents (x) |
No. of days (f) | Cumulative Frequency |
| 0 | 45 | 46 |
| 1 | 76 | 122 |
| 2 | 38 | 160 |
| 3 | 25 | 185 |
| 4 | 10 | 195 |
| 5 | 5 | 200 |
| N = 200 |
We have
N = 200
So, `N/2=200/2=100`
Thus, the cumulative frequency just greater than 100 is 122 and the value corresponding to 122 is 1.
Hence, the median is 1.
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