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What is the diff between median defects and Avg defects?

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by vittalgmat » Tue Dec 30, 2008 11:15 pm
The following table shows results of a quality inspection of a lot of 15 mirrors.
Defects ------Frequency
0 ------------- 6
1 ------------- 1
2 ------------- 4
3 ------------- 3
4 ------------- 1

The difference between the median defects and the average defects in the sample checked is between:

A. 0
B. 0 and 0.5
C. 0.5 and 1
D. 1 and 1.5
E. 1.5 and 2

[spoiler]OA: C[/spoiler]
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by cramya » Tue Dec 30, 2008 11:23 pm
Median is the 8th value since the number of values is 15

So median is 2

Average =

6*0+1*1+4*2+3*3+4*1 / 15 = 1.46

Median - Mean = 2 - 1.46 = .54

Hence C)
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by cramya » Tue Dec 30, 2008 11:27 pm
Vittal,

Mean is equal to the median where the difference between any 2 successive elements is the same.

In this case the difference is not the same since the frequencies are different

0 0 0 0 0 0 1 2 2 2 2 3 3 3 4

Hope this helps!
Last edited by cramya on Tue Dec 30, 2008 11:28 pm, edited 2 times in total.
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by ronniecoleman » Tue Dec 30, 2008 11:27 pm
Mean

0*6 + 1*1 + 2*4 + 3*3 +4*1 / 15 = 16/15

median

0 0 0 0 0 0 1 2 2 2 2 3 3 3 4 = 2

2- 1.6 = 0.4

OA B
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by ronniecoleman » Tue Dec 30, 2008 11:27 pm
ronniecoleman wrote:Mean

0*6 + 1*1 + 2*4 + 3*3 +4*1 / 15 = 16/15

median

0 0 0 0 0 0 1 2 2 2 2 3 3 3 4 = 2

2- 1.6 = 0.4

OA B

Silly mistake :x
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by cramya » Tue Dec 30, 2008 11:28 pm
Ronnie,
No sweat; as long as u dont make one in the real exam u r good.

Good luck!
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by vittalgmat » Tue Dec 30, 2008 11:43 pm
thanks everyone.

Thinking aloud..

I knew there are 2 ways to solve (sort of)
1. Cramya/Ronnie's solution.
2. creating a third column of cumulative freq and solving.

Well!!!! actually 2 ways of representing the data to figure out the median.
I guess both take same time.. so fine.

thanks
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