I agree with your answers.

Why this formula? Let's start with the median.

It'll make more sense if the number of data items is 7. Let's say:

2, 4, 6, 8, 10, 12, 14

The median is the middle item in ordered list so the median is 8.

If the list had 8 items, say 2, 4, 6, 8, 10, 12, 14, 16 then there's no middle number. So statisticians have chosen a rule for this: Choose the number half way between the 4th and the 5th numbers ie 0.5(8+10) = 9. They call this the 4.5th number in the list, just to give it a name. It's just a name for it so don't get too bothered by it.

The quartiles are defined to be the numbers one quarter and three quarters of the way along the list.

So for n=7, list = {2, 4, 6, 8, 10, 12, 14} the lower quartile is the 2nd number which is 4, and the upper quartile is 6th number which is 12.

This works nicely when n is a multiple of 4 minus 1. You can easily find actual numbers in the list 1/4, 1/2 and 3/4 of the way along the list.

But what should you do when n isn't a multiple of 4 minus 1 ? To understand the formulas look at how it works when n=7

(7+1)/4 , (7+1)/2 , 3(7+1)/4 give 2nd, 4th and 6th numbers in the list so that's the formula that has been chosen for any value of n.

(n+1)/4, (n+1)/2 and 3(n+1)/4

Of course, in a list where n is not a multiple of 4 minus 1, this leads to apparently silly positions as you have found 1.5th etc. For small lists it isn't really a useful statistic anyway but if n is large it gives a good estimate of the values 1/4, 1/2 and 3/4 of the way along the list.

The situation where I have found it is really useful is when trying to compare two sets of data. Let's say you have the exam results for two classes and you want to consider which class is better. Probably you'd work out the class averages for this. But it's also useful to know how spread out the results are.

Just comparing the range of marks can be misleading because there might be outlying values that distort the overall comparisons. If you calculate the inter quartile range (upper quartile minus lower quartile) you remove the bottom and top quarters of the data and just look at those in the middle 50%. That statistic is a better way to compare the data. But this calculation has to work whatever 'n' is, which is why a formula is needed.

Bob

]]>The ordered observations 84.60,88.03,94.50,94.90,95.05

n=5

Q1=(n+1)/4 observation ie 1.5th observation

so Q1=84.60+0.5(88.03-84.60) // Why use this formula and what is 1.5th observation and what is formula when ans is 1.75 , 1.06 etc

Q1= 86.315

Q3=3(n+1)/4

= 4.5 observation

so, Q3 = 94.90+0.5(95.05-94.50) // this calculation changed from above

so Q3= 95.175

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