Median will be 5. The logic stated above is correct. But the 61st element happens to be 5.

Frequency Distribution for List X

Number : Frequency
1 10
2 20
3 19
5 12

Can you check the frequency values again..sum 10+20+19+12 gives 61.

Brent@GMATPrepNow wrote:

MBA.Aspirant wrote:Frequency Distribution for List X

Number : Frequency
1 10
2 20
3 19
5 12

Frequency Distribution for List X

Number : Frequency
6 24
7 17
8 10
9 9

List X and Y each contain 60 numbers. The average of the numbers in list X is 2.7, and the average of list Y is 7.1. List Z contains 120 numbers: the 60 numbers in both list X and Y.

What's the mean of the 120 numbers in list Z?

What's the median of the 120 numbers in list Z?

Mean is easy to get by either getting the sum of the 2 lists and dividing by 120, or as I learned now by getting the mean of the means of both lists. my question is how to calculate the median?

Thanks

There are 120 numbers altogether.
Since there is an even number of values, the median will equal the mean (average) of the two middlemost values when all 120 numbers are arranged in ascending order.

In other words, the median will equal the average of the 60th term and 61st terms of the 120 numbers arranged in ascending order.

Well, from the tables, we can see that the 60th number will be 5 and the 61st number will be 6

So, the median will equal (5+6)/2 = 5.5

Cheers,
Brent

how do you calcualte that?

krishnasty wrote: how do you calcualte that?

Hi,
I will solve a simpler example so that it will be easy to understand
Let's say the frequency distribution is:
N F
1 2
2 4
3 5
5 3
frequency of a number 'p' is 5 means p is repeated 5 times in the series
So, if we write the the table in the form of a series. It will be
1,1,2,2,2,2,3,3,3,3,3,5,5,5 --> there are 14 elements in total. They are arranged in increasing order. So, median is average of 7th and 8th terms
It you count, 7th term is 3, and 8th terms is 3 as well.
So, median is (3+3)/2 = 3
Now, if the frequency is high, we cannot write all the numbers as a series as it would be tiem-consuming. so, what we do is add another column(cumulative frequency), where each term is cumulative frequency of the previous term + frequency of the current term. So, the table for the same case will be
N--F--CF
1--2--2
2--4--6(2+4)
3--5--11(6+5)
5--3--14(11+3)
Median is avg of 7th and 8th terms. Consider the CF column. 7th and 8th are between CF=6 and CF= 11.
So, 7th and 8th terms have values 3 and 3, as every term after 6th term upto 11th term is 3.

Thanks Brent for your help.

Info. corrected now

Frankenstein wrote:

krishnasty wrote: how do you calcualte that?

Hi,
I will solve a simpler example so that it will be easy to understand
Let's say the frequency distribution is:
N F
1 2
2 4
3 5
5 3
frequency of a number 'p' is 5 means p is repeated 5 times in the series
So, if we write the the table in the form of a series. It will be
1,1,2,2,2,2,3,3,3,3,3,5,5,5 --> there are 14 elements in total. They are arranged in increasing order. So, median is average of 7th and 8th terms
It you count, 7th term is 3, and 8th terms is 3 as well.
So, median is (3+3)/2 = 3
Now, if the frequency is high, we cannot write all the numbers as a series as it would be tiem-consuming. so, what we do is add another column(cumulative frequency), where each term is cumulative frequency of the previous term + frequency of the current term. So, the table for the same case will be
N--F--CF
1--2--2
2--4--6(2+4)
3--5--11(6+5)
5--3--14(11+3)
Median is avg of 7th and 8th terms. Consider the CF column. 7th and 8th are between CF=6 and CF= 11.
So, 7th and 8th terms have values 3 and 3, as every term after 6th term upto 11th term is 3.

Awesome explaination...the concept is clear like distilled water now...
thnx a lot..