I was wondering do we need to order the values to find mean and median in X,Y and Z?

Z = 40,30,20,10,20,30,40

So do we order from lowest value to highest?

If so we get 10,20,20,30,30,40,40

Thus mean = 190/7 = 27.14

and median = 30

and so mean is not equal to median ???

Please explain when do we order the values and when do we not order the values ?

Also, OG explanation doesnt take frequencies into consideration. It just looks at the symmetry and says mean= median for X and Z whcih doesnt make any sense

why dont we consider the frequencies given in this problem too ??? whats the use then of the frequencies given?how do we know we shouldnt consider the frequencies and re-arrange them ?

Responding to a pm:
Go back to the basics. What is mean? It's the average, the single value that can represent all the values. How do you find it? You multiply each value by its frequency, add them all up and divide by the sum of the frequencies.
What is the mean of: 10, 20, 30 - we know it is 20
What is the mean of: 10, 10, 20, 30, 30 - again 20. Why? Because if there is one value '10' which is 10 less than 20, then there is also a value '30' which is 10 more than 20. So effectively, both 10 and 30 give us two 20s.
Similarly, here X is: 10, 20, 20, 30, 30, 30, 40, 40, 40, 40, 50, 50, 50, 60, 60, 70
40 is the mean because we have three 30s and three 50s to balance out. We have two 20s and two 60s to balance out and we have a 10 and a 70 to balance out again.
What is median? Median is the middle value when you arrange the numbers in increasing/decreasing order. We can see that 40 will be the middle value too since there are equal number of total elements on both sides of 40. We have 6 elements smaller than 40 and 6 elements greater than 40. Hence median = 40.

Hence, mean = median for X.

We can reason out the same thing for Z too in exactly the same way.

The only possible answer is (E). I don't have to worry about Y.
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They mention in the question that it is a frequency distribution so anyway there is no doubt about what the shaded squares mean. The diagram tells you how frequently the value appears in the data e.g. in X, 2 squares are shaded above 60 which means that 60 appears twice.
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Incorrect calculation. When you do [10+20+20+30+30+30+40+40+40+40+50+50+50+60+60+70]/16, you get 40 as expected. You can figure it out without doing this calculation. Go back to the explanation I provided above of what average means. It helps you deduce that average in this case has to be 40 without calculating. (or check out a standard Arithmetic book to understand this.) The formula you have given just helps you calculate average if you need to calculate it.
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Karishma,

thanks, you are right. I had a brain fart. My calculations are incorrect.
Had to use excel to actually calculate the problem

I understand now.


thanks!

Great Explanation ! however there is one minor subset that we are all missing here. Symmetry results in "mean equals median" however symmetry ( the kinda visual symmetry that we are referring to here) is not a prerequisite.

Consider the following dist that is a slight modification of the Y graph....


value f vf
10 1 10
20 3 60
30 5 150
40 4 160
50 2 100
60 3 180
70 2 140


sum f 20 vf 800 mean 40
median 40(10th and 11th )

so while X and Z can be easily chosen. A simple residual analysis(sum of variation of value over/below mean) of Y wouldnt hurt.( in this case it is not req coz there is only one option with x and z).

Am i thinking on the right lines or have i totally lost it (john nash style) since i have been doin quants for 8 hrs straight....

True! This problem doesn't warrant an analysis of Y .But on the 700 plus range it is safer to do a quick residual analysis on similar problems that include the asymmetrical option too.Thanks a ton for your response !




You are right that symmetrical data implies mean = median but it is not a prerequisite. But in this question, we see that for X and Z, "mean = median" so it is not possible that for Y "mean = median" because there is no option giving all three (X, Y and Z). So answer has to be (D) and that is why we can ignore Y. Had there been an option which included Y too, we would have to find out whether this relation holds for Y.[/quote]