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anonymous
 4 years ago
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
01 18
23 7
45 1
Use interpolation to find the median number of break downs.
anonymous
 4 years ago
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 01 18 23 7 45 1 Use interpolation to find the median number of break downs.

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anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Now every time I do this I get: Let m be the median. Then, \[\frac{26}{2} = 13\] which lies in the interval 01.5 (as there are gaps in the data anything below 1.5 is rounded into 01). We can then set up the following proportion, with m as the median: \[\frac{m0}{1.50} = \frac{130}{180}\] \[\frac{m}{1.5} = \frac{13}{18}\] \[m = 1.08333 = 1.08\] 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
 4 years ago
Best ResponseYou've already chosen the best response.1If you treat breakdowns as discrete events, there is no such thing as 1.5 breakdowns. so the interval really is 10 and not 1.50

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0In 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.

phi
 4 years ago
Best ResponseYou've already chosen the best response.1It 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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