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ps (m02q28)

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ps (m02q28) [#permalink] New post 15 Nov 2007, 16:18
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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) -1 and 0
(B) 0 and 0.5
(C) 0.5 and 1
(D) 1 and 1.5
(E) 1.5 and 2

[Reveal] Spoiler: OA
C

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Re: ps (m02q28) [#permalink] New post 02 Mar 2010, 21:55
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Ravshonbek wrote:
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:

0
0 and 0.5
0.5 and 1
1 and 1.5
1.5 and 2



the arrangement would be as under

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

here the median is 2

The mean is 1.47 (22/15)

the answer is 2-1.47 = 0.53 (C)
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Re: ps [#permalink] New post 15 Nov 2007, 19:06
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Ravshonbek wrote:
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:

0
0 and 0.5
0.5 and 1
1 and 1.5
1.5 and 2


I am missing something here:

Median - 2
Mean - {0(6) + 1(1) + 2 (4) + 3 (3) + 4 (1)}/15 = 1.47
Difference b/c Median & Avg: 2 - 1.47 = 0.53

What is the OA?
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Re: ps [#permalink] New post 16 Nov 2007, 21:07
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A similar problem exists in GMAT club challenge 24 (question 26). Check out its explanation.


rakesh22 wrote:
GMATBLACKBELT wrote:
yogachgolf wrote:
Ravshonbek wrote:
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:

0
0 and 0.5
0.5 and 1
1 and 1.5
1.5 and 2


I am missing something here:

Median - 2
Mean - {0(6) + 1(1) + 2 (4) + 3 (3) + 4 (1)}/15 = 1.47
Difference b/c Median & Avg: 2 - 1.47 = 0.53

What is the OA?


I get the same


I too get the same, 0.53. We are not concerned with the sample size, but, only with the number of defects in the given sample.
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Re: ps (m02q28) [#permalink] New post 24 Sep 2010, 23:13
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GaryDunn wrote:
cnrnld wrote:
15 mirrors, but with a total number of 22 as defective?

I don't understand this:(




It implies that the company is measuring more than one type of defect.


Its 22 defects in 15 lots of 15 mirrors each. I Got C.
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Re: [#permalink] New post 28 Sep 2012, 05:51
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Amardeep Sharma wrote:
B = 0 and 0.5

In the sample of 15 numbers, median will be (7th + 8th)/2, in this case 7th number is 1 and 8th number is 2 so median = 1.5

Average of the number is addition of all the numbers of defects divided by 15 = 3/5,

so Median - avg = 9/10, which lies between 0 and 0.5

Amar


Whoah there tiger.
Median of 15 terms is the 8th term, not the average of 7th and 8th.
Median is 2.

Mean of any set which consists of n odd integers is Quotient (n/2) + 1 = 15/2 + 1 = 8th term
Mean of any set which consists of n even integers is the average of the "n"th and "n+1" term

For any arithmetic progression, Mean = Median = Average = Average of first and last term

Back to the problem..............
Average = 22/15, which is a little bit less than 1.5 (Why ? Because 1.5 x 1.5 = 225. Knowledge of squares comes in handy here. The GMAT never asks you to do busy work.
Always remember that !!! )
Difference = 2 - (a term that is a little bit less than 1.5) = A term that is a little bit more than 0.5
Answer - c
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 [#permalink] New post 15 Nov 2007, 16:24
B = 0 and 0.5

In the sample of 15 numbers, median will be (7th + 8th)/2, in this case 7th number is 1 and 8th number is 2 so median = 1.5

Average of the number is addition of all the numbers of defects divided by 15 = 3/5,

so Median - avg = 9/10, which lies between 0 and 0.5

Amar
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 [#permalink] New post 15 Nov 2007, 16:36
i think i cannot read the table. where are the lots of 15 mirrors. how are they represented in the table?

thanks
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 [#permalink] New post 15 Nov 2007, 16:40
Amardeep Sharma wrote:
B = 0 and 0.5

In the sample of 15 numbers, median will be (7th + 8th)/2, in this case 7th number is 1 and 8th number is 2 so median = 1.5

Average of the number is addition of all the numbers of defects divided by 15 = 3/5,

so Median - avg = 9/10, which lies between 0 and 0.5

Amar


I did it the same way, but, and Im sure Im being a moron, why if there are an odd number of mirrors do we need to average two items - surely the 8th mirror is the median?
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 [#permalink] New post 15 Nov 2007, 16:45
Yep, you are right and I am wrong... thnx for correcting me.... :)

I dont know how I made that mistake, well as you got median as 2, rest will follow the same procedure

thanx once again

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Re: ps [#permalink] New post 15 Nov 2007, 22:33
GMATBLACKBELT wrote:
yogachgolf wrote:
Ravshonbek wrote:
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:

0
0 and 0.5
0.5 and 1
1 and 1.5
1.5 and 2


I am missing something here:

Median - 2
Mean - {0(6) + 1(1) + 2 (4) + 3 (3) + 4 (1)}/15 = 1.47
Difference b/c Median & Avg: 2 - 1.47 = 0.53

What is the OA?


I get the same


I too get the same, 0.53. We are not concerned with the sample size, but, only with the number of defects in the given sample.
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 [#permalink] New post 16 Nov 2007, 03:35
Difference between median & mean is 0.53, Ans. C)
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Re: ps [#permalink] New post 17 Nov 2007, 12:05
i too get the answer as C.....please post the OA....
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 [#permalink] New post 17 Nov 2007, 13:30
OA is C, thanks a million
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Re: ps (m02q28) [#permalink] New post 24 Sep 2010, 05:06
15 mirrors, but with a total number of 22 as defective?

I don't understand this:(
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Re: ps (m02q28) [#permalink] New post 24 Sep 2010, 05:33
Foiled by poor reasoning :(
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Re: ps (m02q28) [#permalink] New post 24 Sep 2010, 07:11
cnrnld wrote:
15 mirrors, but with a total number of 22 as defective?

I don't understand this:(




It implies that the company is measuring more than one type of defect.
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Re: ps [#permalink] New post 14 Oct 2010, 07:57
GMATBLACKBELT wrote:
yogachgolf wrote:
Ravshonbek wrote:
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:

0
0 and 0.5
0.5 and 1
1 and 1.5
1.5 and 2


I am missing something here:

Median - 2
Mean - {0(6) + 1(1) + 2 (4) + 3 (3) + 4 (1)}/15 = 1.47
Difference b/c Median & Avg: 2 - 1.47 = 0.53

What is the OA?


I get the same



I also got the same thing.
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Re: ps (m02q28) [#permalink] New post 27 Sep 2013, 22:55
the median is 2
the avg is 22/15
so, median - avg = 2 - 22/15 = 8/15

hence answer is between 0.5 and 1.
C.
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Re: ps (m02q28) [#permalink] New post 01 Oct 2013, 11:59
Median is 2 (take the middle number in the frequency)
The average is 22/15 which you do not have to calculate just state that it is less than 1.5 (logic because 22.5/15=1.5)! Lets say 1.4 to make the logic simplier.

Therefore, you have the difference 2-1.4: 0.6.

The difference is between 0.5 and 1. Answer C.
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Re: ps (m02q28)   [#permalink] 01 Oct 2013, 11:59
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