You are not logged in.
Hello everyone, I have been an avid viewer of these forums for a while and have finally decided to join to ask a question.
Any help on the following question will be appreciated.
Wages (£000s) class width Frequency Amended Frequency
8<12 4 16 4
12<14 2 30 15
14<15 1 60 60
15<16 1 48 48
16<18 2 42 21
18<26 8 24 3
The first part of the question asked to calculate the median (n/2) and quartile (3n/4 and n/4) wages. I used the formula L + (k/f *w) to obtain the corresponding wages.
The next part of the question asks:
Calculate the number of graduates starting on salaries between £11000 and £21000.
I don't know how to do these as these wages lie inbetween the class widths.
Any help? Thank you
Last edited by cherokeexp (2008-05-11 06:44:05)
Offline
I sort of think you just add up all
the amended frequencies for 11, 12, 13, 14, 15, 16, 17, 18 , 19, and 20.
I get 193.
Let me check:
4 + 15 + 15 + 60 + 48 + 21 + 21 + 3 + 3 + 3 = ??
19 + 75 + 48 + 42 + 9 = ??
94 + 99 = ??
Yes, 193 I get.
But I don't know if you have to estimate even more accurately
by drawing a curve thru the top of the bars in each wage range?
Probably not, so 193 is probably right # of gradz from 11K to 20.9999K.
If you want to include the 21000 to 21999.999, then add 3 more graduates for 196.
Last edited by John E. Franklin (2008-05-11 07:01:29)
igloo myrtilles fourmis
Offline
Welcome to the forums! I'd say John's way is the most accurate you can get. Since you don't know anything else about the classes, you just have to assume that the students are distributed evenly within them. (eg. the same amount of people earn from 12-13 as those who earn from 13-14)
Just so you can check, I worked out the LQ, median and UQ to be about 14.2, 15.1 and 16.3 respectively.
Why did the vector cross the road?
It wanted to be normal.
Offline
thank you both! I really appreciate the help. Yeah I don't think I'm required to use any other advanced technique so the above solution should be fine.
And thanks mathsyperson, I got the same values for the Q and median.
Offline