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InterviewSolution
| 1. |
Find the missing frequencies and the median for the following distribution if the mean is 1.46. No. of accidents:012345 Total Frequency (No. of days):46??25105200 |
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Answer»
Given, N = 200 = 46 + x + y + 25 + 10 + 5 = 200 = x + y = 200 – 46 – 25 – 10 – 5 = x + y = 114 (i) And Mean = 1.46 \(\frac{\sum fx}{N}=1.46\) \(=\frac{x+2y+140}{200}=1.46\) = x + 2y + 140 = 292 = x + 2y = 292 – 140 = x + 2y = 152 (ii) Subtract (i) from (ii), we get X + 2y – x – y = 152 – 114 y = 38 Put the value of y in (i), we get x = 114 – 38 = 76
We have, N = 200 \(\frac{N}{2}=\frac{200}{2}=100\) The cumulative frequency just more than \(\frac{N}{2}\) is 122 so the median is 1 |
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