Intervals Frequency 0 - 10-----------> 2 10-20 ----------> 5 20-30 ----------->6 30-40 ----------->3 The mean and median are: a) mean = 21.25 Median = 25.00 b) mean = 21.25 Median = 21.67 c) Mean = 25; Median = 21.67 d) Mean = 25 Median = 25 Can anyone tell me why the answer is B, instead of A?
Median would be middle value in the frequency distribution. So that is the 8.5th value which will fall in the 3rd interval (based on Cumulative frequency) Cum Freq Mid point 0-10 2 5 10-20 7 15 20-30 13 25 30-40 16 35 So u arrive at the actual number as 8.5/13 * (25-15) + 15 = 21.54… so approx --> 21.67 I believe 21.67 was arrived by using the 8th value
Why wouldn’t you just do (25+25)/2=25 ?
that is not the definition of median… it is the middle value in the entire set, right…
You have 16 values in total. Using midpoints, if you arrange them out, the 8th and 9th values should be 25 and 25. So based on the 8.5 that we calculated, you would take the average of the two, no ?
no – that value could be anywhere from 20-30 right… given that the count is anything that lies between 20 and 30 would have a tick… (counter++).
Could you possibly explain your calculation cpk (“8.5/13 * (25-15) + 15”). Sorry for the “silly” questions. Either I’m blanking out, or just confused. I do realize what you’re trying to say though. I’m just trying to make sense of the way you calculated it.
Please keep it going… I don’t understand this either and it’s killing me. You would think this would be a snap.
I know!!! It’s frustrating. Seems like such a straightforward concept. I’m just waiting for that classic light bulb moment.
Exactly. I have been hanging out in the weeds, just waiting for someone to lift the fog. I’m not totally convinced that cpk’s calculation is correct. The answer provided as “correct” (A), has such a clean .67 decimal. I think cpk’s “approximate” answer may prove incorrect.
actually my calculation was close, but it was the wrong way. correct way you need the 8th number that falls in the class 20-30 (using the cum freq). L=Lower limit of class containing that frequency = 20 F = Frequency of class interval = 6 Cfb = Cumulative frequency of class before = 7 i = Class interval = 10 Median = L + [(n/2 - Cfb)] / F] * i = 20 + (8-7)/6 * 10 = 20 + 1.67 = 21.67
I hope I have lifted the fog…
Thanks for sticking with it cpk… no more fog.
Sweet. The generalization of that calculation helped. Somehow, I can’t recall seeing this formula in Book 1 of the Schweser study notes. I guess I just missed it. Thanks a lot for that clarification cpk!
If I had to eyeball this question to get to the answer, I would have added one to the first two interval frequencies to bring equal weight between the first two and the second two intervals making the mean and median 20. Then I would realize that by subtracting back one from each interval it would only slightly change the weight in favor of the second half of intervals. 21.whatever would make the most sense for both the mean and median. 25 would require way too much of a swing in weight.
I am assuming the median in the case of intervals assumes a linear distribution of points within that interval. So we know the median is between the 8th and 9th values, right? We know there are 7 values before the 20-30 interval, and there are 6 values within that interval. So breaking that interval into 6 equal sub-intervals, the 8th interval would be from 20 to 21.67, and the 9th interval would be from 21.67 to 23.33. At the edge where these two intervals meet is the point 21.67.
cpk, Why are you taking only 8th value & not average of 8th & 9th? There are 16 observations, L(50)=(16+1)*50/100=8.5 So shouldn’t we take average of 8 & 9 which will be – 8th= 20 + 10/6 = 21.67
Guys can anyone please explain how did we calculated mean? I’m getting different answer. Thanks.
Hari - Sum of (Midpoint of interval * Interval Frequency) for each interval / 16