2010 doubt dse

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2010 doubt dse

Ansh malhotra
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Re: 2010 doubt dse

Ansh malhotra
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Re: 2010 doubt dse

Mike
In reply to this post by Ansh malhotra
Hi ansh
For ques 2)
 median(m) of a prob dist is defined as
 Pr(X>=m) = Pr(X<=m) >= 1/2
Here this condition is satisfied for two values of X i.e X=0, 1. So answer is (d)
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Re: 2010 doubt dse

Nikkita
In reply to this post by Ansh malhotra
The answer should be (c), I think as in case of even no. Of data, the median is (0+1)/2=1/2.
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Re: 2010 doubt dse

Ansh malhotra
mike is right nikita because these are not  fixed numbers but discrete random variables so median  is one

which divides distribution into half
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Re: 2010 doubt dse

Mike
In reply to this post by Nikkita
Hi Nikita,
My mistake. I should have written my answer more clearly.
You are right about the median = 1/2
But 1/2 is the best choice for the median value. In fact any Value of X s.t. X belongs to [0,1] will serve as the median of the distribution. I understand your point about 1/2 being the median value. That's how we are taught at the undergraduate level. But consider my point. If you still have any doubts, feel free to ask.
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Re: 2010 doubt dse

Nikkita
In reply to this post by Ansh malhotra

Actually I am little confused now too..
Lets think about the definition of a median which says that its a number which divides the data into 2 equal halves.
So, Acc. to that, the answer should be (d)  as 0 is also dividing data into 2 equal halves, 1 also, and any value between them too!!.. Means there are multiple Medians in this special case..

Amit Sir please help..

On 12 Jun 2016 13:39, "Ansh malhotra [via Discussion forum]" <[hidden email]> wrote:
mike is right nikita because these are not  fixed numbers but discrete random variables so median  is one

which divides distribution into half


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Re: 2010 doubt dse

JMKeynes
In reply to this post by Nikkita
Mike is right. Both 0 and 1 are the median. Pr(X>=0)=.6 and PR(X<=0)=.5
Also Pr(X>=1)=.5 and PR(X<=1)=.8
Since these probabilities are all greater than or equal to .5, both 0 and 1 are the median.
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Re: 2010 doubt dse

Nikkita
Okok..got it.. Thankyou all
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Re: 2010 doubt dse

Mike
In reply to this post by JMKeynes
Thanks JM Keynes