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prachich1987: Legendary Member; Posts: 752; Joined: Sun Sep 12, 2010 2:47 amFollowed by: 10 members; Thanked: 20 times; GMAT Score:700

by prachich1987 » Sun Dec 19, 2010 12:35 am

If the variables A, B, and C take on the values 1, 2, 3, 4, 5, 6, 7, 8, or 9 with the frequencies indicated by the bars above, for which of the frequency distributions is the mean equal to the median?

(A) A only

(B) B only

(C) C only

(D) A and B only your answer

(E) A and C only correct

PLZ NOTE THE FIRST ATTACHMENT IS C, SECOND IS B & THIRD IS A Mark ur answer accordingly

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Source: Beat The GMAT — Problem Solving | Sun Dec 19, 2010 12:35 am

goyalsau: Legendary Member; Posts: 866; Joined: Mon Aug 02, 2010 6:46 pm; Location: Gwalior, India; Thanked: 31 times

by goyalsau » Sun Dec 19, 2010 12:59 am

According to be B only should be the answer.

What is the OA? I m correct, then i will try to explain. otherwise will be looking forward for the explanation.........

Saurabh Goyal
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EveryBody Wants to Win But Nobody wants to prepare for Win.

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prachich1987: Legendary Member; Posts: 752; Joined: Sun Sep 12, 2010 2:47 amFollowed by: 10 members; Thanked: 20 times; GMAT Score:700

by prachich1987 » Sun Dec 19, 2010 1:02 am

goyalsau wrote:According to be B only should be the answer.

What is the OA? I m correct, then i will try to explain. otherwise will be looking forward for the explanation.........

The OA given is E i.e A & C follow the property mean=median

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anshumishra: Legendary Member; Posts: 543; Joined: Tue Jun 15, 2010 7:01 pmFollowed by: 3 members; Thanked: 147 times

by anshumishra » Sun Dec 19, 2010 8:55 am

First of all , how to guess it without solving ?

Check the distributions :

The mean and median would be same in a symmetrical distribution :

A and C are symmetrical (If you place a mirror in the center, the left part will resemble the right part).

However, if you want to confirm Here is how you test it. I am doing it for A and leaving the case for B and C.

A can be represented as :

(1,2,2,3,3,3,4,4,4,4,5,5,5,5,5,6,6,6,6,7,7,7,8,8,9)

Mean = 1*1+2*2+3*3+4*4+5*5+6*4+7*3+8*2+9*1 / (1+2+3+4+5+4+3+2+1) = 125/25 = 5

Median = 5 (which number comes at midpoint ?).

Similarly B can be represented as :

(1,1,1,1,1,1,2,2,2,2,2,2,3,3,3,3,3,3,4,4,4,4,4,4,5,5,5,5,6,6,7,7,8,8,9,9)

and C as :

(1,1,1,1,2,2,2,2,3,3,3,3,3,3,4,4,4,4,4,4,5,5,5,5,6,6,6,6,6,6,7,7,7,7,7,7,8,8,8,8,9,9,9,9)

Thanks

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shovan85: Community Manager; Posts: 991; Joined: Thu Sep 23, 2010 6:19 amFollowed by: 24 members; Location: Bangalore, India; Thanked: 146 times

by shovan85 » Sun Dec 19, 2010 10:00 am

What Anshu has stated is absolutely correct.

When you see a data in a Graph just draw a line in the exact middle of it and see the symmetry.

If both of the sides are mirror image to each other then Mean = Median

If the problem is Easy Respect it, if the problem is tough Attack it

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prachich1987: Legendary Member; Posts: 752; Joined: Sun Sep 12, 2010 2:47 amFollowed by: 10 members; Thanked: 20 times; GMAT Score:700

by prachich1987 » Sun Dec 19, 2010 10:07 am

shovan85 wrote:What Anshu has stated is absolutely correct.

When you see a data in a Graph just draw a line in the exact middle of it and see the symmetry.

If both of the sides are mirror image to each other then Mean = Median

I agree that for A, mean=median
But for B also mean=median though it is not symmetrical.
For B mean=(6*4+4+2*4)/9=4.Also the median is 4

But for C, though it's symmetrical mean is not equal to median
mean=(4*2+6*2+4*1+6*2+4*2)/4=44/9 whereas median is 4.

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anshumishra: Legendary Member; Posts: 543; Joined: Tue Jun 15, 2010 7:01 pmFollowed by: 3 members; Thanked: 147 times

by anshumishra » Sun Dec 19, 2010 10:20 am

prachich1987 wrote:
shovan85 wrote:What Anshu has stated is absolutely correct.

When you see a data in a Graph just draw a line in the exact middle of it and see the symmetry.

If both of the sides are mirror image to each other then Mean = Median

I agree that for A, mean=median
But for B also mean=median though it is not symmetrical.
For B mean=(6*4+4+2*4)/9=4.Also the median is 4

But for C, though it's symmetrical mean is not equal to median
mean=(4*2+6*2+4*1+6*2+4*2)/4=44/9 whereas median is 4.

prachich1987,

Your set B and C is not right.
As mentioned in my previous post they are :

B = (1,1,1,1,1,1,2,2,2,2,2,2,3,3,3,3,3,3,4,4,4,4,4,4,5,5,5,5,6,6,7,7,8,8,9,9)

C = (1,1,1,1,2,2,2,2,3,3,3,3,3,3,4,4,4,4,4,4,5,5,5,5,6,6,6,6,6,6,7,7,7,7,7,7,8,8,8,8,9,9,9,9)

Let me calculate mean/median for C :

Mean = 1*4+2*4+3*6+4*6+5*4+6*6+7*6+8*4+9*4/(4+4+6+6+4+6+6+4+4) = 220/44 = 5
Median = The middle term , here there are 44 terms, So the 22th term would be the middle one = 5.

I guessed you missed the part where I have identified the three sets A, B and C.

The frequency distribution shows, how many times, 1, 2,...9 have appeared.
So, when I write 1 only a single time, in my set, that means 1 has appeared only one time (check the figure).

Hope, now you would get it.

Thanks

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shovan85: Community Manager; Posts: 991; Joined: Thu Sep 23, 2010 6:19 amFollowed by: 24 members; Location: Bangalore, India; Thanked: 146 times

by shovan85 » Sun Dec 19, 2010 10:28 am

prachich1987 wrote: I agree that for A, mean=median
But for B also mean=median though it is not symmetrical.
For B mean=(6*4+4+2*4)/9=4.Also the median is 4

But for C, though it's symmetrical mean is not equal to median
mean=(4*2+6*2+4*1+6*2+4*2)/4=44/9 whereas median is 4.

I am showing it for C

You agree set :
1,1,1,1,2,2,2,2,3,3,3,3,3,3,4,4,4,4,4,4,5,5,5,5,6,6,6,6,6,6,7,7,7,7,7,7,8,8,8,8,9,9,9,9

If yes then how many data are there in the set? 44 Right!! (Calculate the boxes)

What is the sum of data? 10*4+10*4+10*6+10*6+10*2 = 10*22 = 220 (I have taken 9+1, 8+2, 7+3, 4+6 and 5+5)

What is the mean? 220/44 = 5

What is the median? 5

If the problem is Easy Respect it, if the problem is tough Attack it

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prachich1987: Legendary Member; Posts: 752; Joined: Sun Sep 12, 2010 2:47 amFollowed by: 10 members; Thanked: 20 times; GMAT Score:700

by prachich1987 » Sun Dec 19, 2010 11:20 am

Thanks anushmishra ,shovan
I was making a BIG silly mistake..

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anshumishra: Legendary Member; Posts: 543; Joined: Tue Jun 15, 2010 7:01 pmFollowed by: 3 members; Thanked: 147 times

by anshumishra » Sun Dec 19, 2010 11:22 am

prachich1987 wrote:Thanks anushmishra ,shovan
I was making a BIG silly mistake..

No problem prachich1987,

Glad that you got it ! All of us make mistakes.

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