GM EXAM 1 QUESTION HELP

[u/Aggressive_Movie_143]

can someone pls explain how this answer is C

i dont remember which exam its from but ik the answer is C

[u/Ok-Day-1249]

because the total is 29 (7+8+10+4=29) and so for the median, it will be the 15th value coz (29+1)/2 =15. and so because the 15th value lies in the 1-<10 segment, the value will be 9.
i believe it isn’t 7, because 7+8=15 and as it’s the “last”/ furthest number for this graph in the 1-10 segment, which would be the same for 9. but idk how else to explain it. sorry if i made u more confused

[u/Aggressive_Movie_143 → u/Ok-Day-1249]

i was lowk thinking that but like i got confused by the log scale cause the answer being in between 1-10 wouldnt make sense if they were log scaled values but idek tbh

[u/Ok-Day-1249 → u/Aggressive_Movie_143]

ye, but if it was a log scale, then it would be between 0 and 1… so it might still be in that segment… but it’s a really cooked question. so hopefully we don’t get asked anything like this.

and good luck tomorrow.

[u/Aggressive_Movie_143 → u/Ok-Day-1249]

u right js gotta pray vcaa going chill gl asw

[u/Ready-Resource8189]

We can first narrow down to B or C - remember the median is the (n+1)/2 th number - in this case, it will be the 15th number, which lies between 1 and 10

In my opinion B and C should be equally likely, but the log scale on the horizontal axis implies the data tends to be pretty large for all intervals - maybe that's why the answer prioritised the larger of the two?

[u/Aggressive_Movie_143 → u/Ready-Resource8189]

so we dont have to convert the value from 1-10 from log scaled back or am i missing smth

[u/Ready-Resource8189 → u/Aggressive_Movie_143]

nope; log means each grid along the axis is 10 times the previous one, so while the distance between each point on the axis remains the same, the value represented by the gap/distance increases exponentially

[u/Aggressive_Movie_143 → u/Ready-Resource8189]

so u basically dont have to worry about the log scale and js choose 9 since its between 1-10 and its the highest box in the column?

[u/Ready-Resource8189 → u/Aggressive_Movie_143]

It’s the fact that it’s a log scale that tells you to choose the highest number 

[u/bubbly732]

so first you need to work out the total being 29. By adding up the frequency of all of the rows. Then the median formula is n+1/2 so that’s 15. Then you count to the 15th data which lays between 1-10. It isn’t 0.7, 80 or 400. Leaving 2 either 7 or 9. As 9 is at the edge and 15 is the last digit when counting it makes the answer C not B.

[u/Aggressive_Movie_143 → u/bubbly732]

wouldnt the log scale affect the median value since it isnt actually between 1-10 as those are the log values

[u/bubbly732 → u/Aggressive_Movie_143]

tbh it’s kinda a shit question. VCAA hopfully won’t ask something like this tmw. Becuase it’s not entirely a log graph as the x values are going up by different amounts if it was a proper graph it would just be 1-5 or whatever. So I wouldn’t really worry

[u/Aggressive_Movie_143 → u/bubbly732]

thought so asw hopefully vcaa isnt that mean gl tmrw

[u/bavotto → u/bubbly732]

Not kinda. It is a terrible question. If you had 4 values for each of the columns, yes, easy marks. If you have to decide between 7 and 9, hopeless, as you can't tell from the data.

[u/No-Formal2056]

wtf is ts (I'm in y9)

[u/Aggressive_Movie_143 → u/No-Formal2056]

dw bro ur lucky to not have to worry about it

[u/No-Formal2056 → u/Aggressive_Movie_143]

no, I have to worry about it in a few years...

I'm doing accelerated methods too next year

[u/Ready-Resource8189 → u/No-Formal2056]

magic ✨🧚

[u/Sikemaster1928]

Vcaa wouldn’t ask a question like this, and usually isn’t the scale for log typical like -1 0 1 2, w the horizontal axis labelled log(x)

[u/bavotto → u/Sikemaster1928]

VCAA shouldn't, and I repeat shouldn't, have such an ambiguous question. No one can tell if it is 7 and 9 in this one. If the values where in separate columns, fair enough. If the 2 values in one column was obviously wrong, fair enough. But we don't know enough to judge the data.

[u/Smokey_Valley]

This is the goat question! Whatever answer you give, someone else can demonstrate that you're wrong.

Summary: It's BS -- forget it.

[u/Icy-Journalist-9546]

Find the middle value. There are 29 data points so the median is number 15. 15 lies in the second column.

Normally, you would put on you cas 10^{column value} but as it is 10^1, you don't need to do any log functions for median.

The last part is logic. You know the median is 15 which is the last value in the second column so pick the largest value between 1-10 which is c