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**Zeeshan 01****Member**- Registered: 2016-07-22
- Posts: 746

Find Quartiles for Ungrouped data // Exercising this type of questions

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

M.M.Zeeshan.K

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**Bob****Administrator**- Registered: 2010-06-20
- Posts: 9,221

hi Zeeshan 01

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

Children are not defined by school ...........The Fonz

You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei

Sometimes I deliberately make mistakes, just to test you! …………….Bob

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**elonjigar****Member**- Registered: 2021-06-06
- Posts: 1

Any set of data can be described by its five-number summary. These five numbers, which give you the information you need to find patterns and outliers, consist of (in ascending order). The minimum or lowest value of the dataset The first quartile Q1, which represents a quarter of the way through the list of all data The median of the data set, which represents the midpoint of the whole list of data The third quartile Q3, which represents three-quarters of the way through the list of all data The maximum or highest value of the data set.

IQR = Q3 - Q1. The interquartile range [link removed by moderator] shows how the data is spread about the median. It is less susceptible than the range to outliers and can, therefore, be more helpful.

*Last edited by zetafunc (2021-06-06 19:23:08)*

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