A hotel records the number of break downs of its lift per week, over a period of 26 weeks. The data is shown in a continuous table below. No. of Breakdowns Freq. of Breakdowns 0-1 18 2-3 7 4-5 1 Use interpolation to find the median number of break downs.

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A hotel records the number of break downs of its lift per week, over a period of 26 weeks. The data is shown in a continuous table below. No. of Breakdowns Freq. of Breakdowns 0-1 18 2-3 7 4-5 1 Use interpolation to find the median number of break downs.

Mathematics
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Now every time I do this I get: Let m be the median. Then, \[\frac{26}{2} = 13\] which lies in the interval 0-1.5 (as there are gaps in the data anything below 1.5 is rounded into 0-1). We can then set up the following proportion, with m as the median: \[\frac{m-0}{1.5-0} = \frac{13-0}{18-0}\] \[\frac{m}{1.5} = \frac{13}{18}\] \[m = 1.08333 = 1.08\] The answer in the book is 0.72 - which is suspect is obtained by just doing the 13/18. If that's the case, why haven't they considered the 1.5 and just called it 1?
  • phi
If you treat breakdowns as discrete events, there is no such thing as 1.5 breakdowns. so the interval really is 1-0 and not 1.5-0
In an earlier example the book was treating discrete data as continuous data when it was calculating medians too...this was my initial thought but then the logic doesn't apply to that previous example.

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  • phi
It would be nice to see the previous example. Here we are trying to estimate the median with binned data. If we know that if the median falls within the first bin ( zero or one), the median is either 0 or 1. The closer we get to the total number of entries in the first bin, the more likely the median is the ending value (1 in this case). So we use interpolation to estimate the median. In this problem, if the 18th number is the median, you would estimate the median as 1*18/18= 1 This has to be a better estimate than 1.5*18/18 = 1.5

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