M02-28

Question

The given table displays the outcomes of a quality check conducted on a batch of 15 mirrors.
 
 https://gmatclub.com/gmat-practice-tests/resources/import/img/m02-28.png 
 
The median number of defects minus the average number of defects in the examined batch 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

Bunuel · Accepted Answer

Official Solution:

The given table displays the outcomes of a quality check conducted on a batch of 15 mirrors.
 
 https://gmatclub.com/gmat-practice-tests/resources/import/img/m02-28.png 
 
The median number of defects minus the average number of defects in the examined batch 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

First, let's arrange the observed number of defects in ascending order: {0, 0, 0, 0, 0, 0, 1, 2, 2, 2, 2, 3, 3, 3, 4}
 
The median is the middle value, which in this case is 2.
 
We can calculate the average by:  \frac{0*6 + 1*1 + 2*4 + 3*3 + 4*1}{15} = \frac{22}{15} 
 
The difference between the median and the average is  2 -\frac{22}{15} = \frac{8}{15} = 0.53 .

Answer: C

Bunuel · Answer

I have edited the question and the solution by adding more details to enhance its clarity. I hope it is now easier to understand.

BottomJee · Answer

I think this is a  high-quality  question and I agree with explanation.

manntri · Answer

I think this is a  high-quality  question and I agree with explanation. question can be solved more quickly by below approach: no of items : 15 median would be 8th term.

frequency of zero is 6, it will include terms upto 6th. next frequency of 1 is 1 it would cover 7th term that means 8th term would be 2 hence median is 2. now simply find average and subtract.

Pranavp1 · Answer

I think this is a poor-quality question and the explanation isn't clear enough, please elaborate. The question asks for the median no. of defects - Avg. no. of defects. So ideally, we shouldn't include the 0 defect mirrors in the tally as they are not defective. The question is not clear

Bunuel · Answer

The approach you suggest would not be the "ideal" way, it would actually be incorrect. The question is clear and entirely correct. We have the following data of defects across 15 mirrors:

{0, 0, 0, 0, 0, 0, 1,  2 , 2, 2, 2, 3, 3, 3, 4}

This means, for example, that 6 mirrors had 0 defects each, and 1 mirror had 4 defects. The average is calculated as:

\frac{0*6 + 1*1 + 2*4 + 3*3 + 4*1}{15} = \frac{22}{15} .

The median (middle value) is 2. Therefore, the difference between the median and the average is:

2 -\frac{22}{15} = \frac{8}{15} = 0.53 .

Hope now it's clear!

Pranavp1 · Answer

The wording is a bit different. the question is saying  The given table displays the outcomes of a quality check conducted on a batch of 15 mirrors.  Which will include both - mirrors with and without defects. And from this data we have to calculate median and average on. of defects.

While solving the question I had the dilemma of including zeros or not. I chose according to the wording of the question. I wanted to know if such question comes in the exam which direction to go.

Thanks.

Bunuel · Answer

The question wording is not different from what the solution suggests. The question clearly asks for the median and average number of defects across all 15 mirrors, including those with 0 defects. The approach shown in the solution is correct.

PocusFocus · Answer

I like the solution - it’s helpful.

M02-28

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