[Jean Pearson]

The weekly incomes of a sample of 60 employees of a fast-food eatery were organized into the following frequency distribution.

Weekly Incomes Number of Employees
$100 up to $150 5
150 up to 200 9
200 up to 250 20
250 up to 300 18
300 up to 350 5
350 up to 400 3

What is the quartile deviation? A. about $61, B. about $103, C. about $39, D. about $299, E. none of the above

[ebaines]

The quartile deviation is half the difference between the upper and lower quartiles in a distribution. So - estimate what the 15th and 45th person make, find the difference, and divide by 2. Post back what you get for an answer.

[Jean Pearson]

I guess I am having a brain freeze. I understand about using the 15th and 45the person, but I do not know how to estimate the 15th and 45th person using the grouped data. How do I do that?

[ebaines]

You can't get an exact answer, but you can "guestimate" it.

Note that the 15th person is the lowest paid person in the $200-$250 bucket. There are 19 people paid more than him, so he's probably pretty close to $200. A good guess might be 1/20th of the way between $200 and $250, or about $202. Similarly the 45th person is a bit above the middle of the $250 - $300 bucket. So make a guess, calculate the resulting quartile deviation, and see if you get an answer that's reasonably close to one of the choices. Hope this helps.

[Jean Pearson]

Ok. Using Excel, I got the quartiles as 187.50, 250, 312.50 and 400. If I take the first quartile of 187.50 and subtract it from the third quartile, I get 125. Divide that by 2 is 62.50. Is that correct?

[ebaines]

No - the first quartile is clearly greater than $200 and the third is less than $300. What function did you use in Excel, and how did you construct the data set?