Yes correct you can't have a median of 0. Nobody is saying the median is 0. by your working, the frequency of the median is 0. You can't make up some fantasy numbers on the x axis when each bar could mean anything. Have you ever answered those histogram questions where it asks "how many...
hahah these little jabs are just gonna make you feel like more stupid when you realise why it's C. HISTOGRAM = frequency of something. you've just assumed that the x axis is numbers when it may not be. it could be absolutely anything. e.g. if your "4" actually represents "amount of people in...
Yes...you skip to the next data point. then there are 7 values to the right and 8 to the left of the median. that means that the data is no longer symmetrical. that means that it cannot be matched with a symmetrical box plot. that means I don't think you understand how this graph works
yes, the average of the 8th and 9th value is 8.5. therefore the median is in the 8.5th place for all graphs. using this logic you can figure out the median for A,B,D which is in the middle column and is symmetrical(like the box plot). but if this is used for C, then the median would be 0 because...
people are overthinking it...it's C because you can count the median on each graph (which is at place 8.5 on all 4 - all total to 16), but on C there is nothing for where the median is supposed to be. so it just doesn't work irrespective of the box plot
1. i worked out the total amounts that were taken out over the 15 and 10 years
2. added both together, then used that (i think it was 360,000) as the FV in 360,000=PV(1+0.02) power of 300 or whatever the periods were
3. then just algebred shit up and found the minimum PV to have at least the sum...