Welcome, Guest. Please login or register.

November 12, 2019, 11:49:22 pm

Author Topic: Median from cumulative frequency table  (Read 52 times)  Share 

0 Members and 1 Guest are viewing this topic.

dani01

  • Forum Regular
  • **
  • Posts: 82
  • Respect: +1
Median from cumulative frequency table
« on: October 22, 2019, 12:30:42 pm »
0
Hi would someone be able to explain to me why the answer is b?
I worked it out by doing half of 32 which is 16. the 16th score is 19. so why is the answer 19.5??
thanks


RuiAce

  • ATAR Notes Lecturer
  • Honorary Moderator
  • Great Wonder of ATAR Notes
  • *******
  • Posts: 8586
  • "All models are wrong, but some are useful."
  • Respect: +2390
Re: Median from cumulative frequency table
« Reply #1 on: October 22, 2019, 12:42:15 pm »
+2
Hi would someone be able to explain to me why the answer is b?
I worked it out by doing half of 32 which is 16. the 16th score is 19. so why is the answer 19.5??
thanks


If you list out the scores, you get:
17 17 17 18 18 18 18 18 18 19 19 19 19 19 19 19 20 20 21 21 22 22 22 22 22 22 22 22 22 23 23 23

Now try to find the median using the cross-out-numbers-from-both-sides method. You'll find that you end up with both the 19 and the 20 left. So using that old method, you would get 19.5.

So where's the conflict? In reality, half of 32 is not correct. The formula for the median is the \( \frac{n+1}{2} \)-th score, not the \( \frac{n}{2}\)-th score. So you actually have to compute \( \frac{33}{2} \), which is equal to \(16.5\). Hence, you need to compare both the 16th score, and the 17th score.