Statistics 1

The median number is the 13th datum.
13/18=Q2-0/1.5-0
Q2=1.08

I have a doubt with how the book gets the answer 1.08 for the following question.

A hotel is worried about the reliability of its lift. It keeps a weekly record of the number of times it breaks down over a period of 26 weeks. The data collected are summarized in the table opposite.
Use interpolation to estimate the median number of breakdowns.

Number of breakdowns Frequency
0 - 1 18
2 - 3 7
4 - 5 1

Interpolation is done on continuous data, so they've treated this as such and did 13th value in 0-1.5, so (13/18)*1.5 = 1.08

Poor example of interpolation, in reality the median here is 13.5th value so 0, 0.5 or 1 as data is discrete.

Can you explain a bit further, please?

If the classes were time (min) with 0-1, 1-2 etc then interpolation fits much better as data is continuous. 1.4 min is sensible, goes into 0-1. Classes also join up smoothly (1.4 to 1.5 to 1.6) etc.

For continuous data the median is 50% mark, for 26 items we'd look at 13th value. So in 0-1, we have a spread of 0-1.5 min with 18 items. We'd need to go (13/18)th of the way into it, so the median is (13/18)*1.5 = 1.08 min

But in this example the data is discrete, only have 0's, 1's etc. So median is (n + 1)/2 = 13.5th value. So in 0-1, median is between two 0's, two 1's or 0 & 1; thus value of 0, 0.5 or 1 (interpolation doesn't work here).

So basically, when am I supposed to use (n+1)/2 and just n/2 to find the median in these cases?
I am confused.

Read again: continuous (50% mark, n/2) vs discrete [(n+1)/2 rule]. In this Q we applied a technique for continuous data, on discrete! Hence confusion.