Given the following frequency distribution: Interval Frequency 10 up to 30 5 30 up to 50 10 50 up to 70 15 70 up to 90 5 Which of the following statements is FALSE? A) The interval size is 20. B) The interval frequency of the third interval is 15. C) The relative interval frequency of the second interval is 14.28. D) The mean of the frequency distribution is 51.43. Can someone compute and explain C and D pleaseeeeeeeeeeeee! Oh and the Median!!! I know this is fifth grade math, but for some reason… i can’t recall it.
OK I got C and the right answer!!!
But how to calc the mean and median?
mean (20 * 5 + 40 * 10 + 60 * 15 + 80 * 5) / (5 + 10 + 15 + 5) = 1800 / 35 = 51.43 Median: Go look at Mid Point, Cum Freq 20 - 5 40 - 15 60 - 30 80 - 35 You need the 17.5th value so now interpolate between 40 + (17.5 - 15)/(30-15) * (60-40) = 40 + 2.5/15(20) = 40+50/15 = 43.33
mean=average median=middle
Houston… no way…
cpk - The problem is that you are estimating in a 40/60/80 kinda way. That 51.43 might be a decent estimator of the mean but (unless someone defines the mean of a frequency distribution that way) is not really the mean of the data or the distribution.
Joey agreed - it is an estimator of the mean of the freq distribution, as you have stated. CP
cpk123 Wrote: ------------------------------------------------------- > mean > > (20 * 5 + 40 * 10 + 60 * 15 + 80 * 5) / (5 + 10 + > 15 + 5) = 1800 / 35 = 51.43 > > Median: > Go look at Mid Point, Cum Freq > 20 - 5 > 40 - 15 > 60 - 30 > 80 - 35 > > You need the 17.5th value > so now interpolate between 40 + (17.5 - > 15)/(30-15) * (60-40) = 40 + 2.5/15(20) = 40+50/15 > = 43.33 Cpk123, Could you pls explain me more detail about the discipline of above median calculation? I understood the median should be the 17.5th observation but i don’t understand why its value should be (17.5 -15)/(30-15) * (60-40) extra? May it has any statistical discipline or any logical maths formula? Normally the median is the midpoint observation in the rank that order ascending or descending. But if the case of too large interval, i mean the range in interval too large your computation may get some distortions. Is there any acceptable limit for above solution? Pls kindly explain. Thanks a lot.
I am doing a simple linear interpolation above 15th value is at 40 (middle point) 30th cumulative value is at 60 (middle point). so at what point will my 17.5th point lie? 17.5 - 15 40 + ----------- * (60 - 40) 30 - 15
Yes agree this is just an estimator of the mean, and I’am assuming that’s because we have rounded the median values of each Intervale to the closest and lowest interger. (20 for 20,5; 40 for 40,5; 60 for 60,5; and 80 for 80,5). So normally the mean should be 51,92. C is not correct because the relative frequency on an interval is the number of observation in that interval divided by the total number of observation so it should be 10/35= 28,57 which is double the answer given for C. So C is not correct
No cpk123 You don’t understand my confusion. I mean the way of estimating median value in intervals. 17.5-15 --------- * (60-40) 30-15 The estimation above seems imply a brand new sample with sample size stretch from 0 to 30-15=15) and sample range stretch from 0 to (60-40=20). And we have to calculate the value of the (17.5-15)th observation in the sample. I don’t mention the formula if it is in the book. I mean the logical of estimating method. Maybe i get some difficult in English terminology?
This is a simple linear interpolation between values. That is an estimation for the median value. This formula is probably not in any text book (at least from a CFA standpoint). 15 = 40 30 = 60 so 17.5 = ? This is what I am calculating above.
Yezzzz, I understood already. To me, it’s not a simple linear interpolation 
Uhm, so stupid i am 
Thanks cpk123 a lot and have a good day!
Thu Thuy, linear interpolation is something which they would teach you in an introductory stats lecture. It is just a standard way of working out this type of problem. If you don’t really understand the logic behind the equation you can just learn it off by heart and you will be fine. Just make sure you know which number gets substituted in where. Do a few practise ones like this and you will easily get it.
Is there a quick way or a short cut in estimating the mean, median, relative freq of a frequency distribution? 20 - 5 40 - 15 60 - 30 80 - 35