In the math (shapes,sizes),third option is true and in the second math,(town x and Y)
The mean of Y is greater than that of X’s.
Why is it so? My reasoning is when the curves are narrow SD is small and average is big. Isn’t it?
(Sorry if I’ve asked a basic question)
SD is a measure of dispersion (how spread out a dataset is), and the average/mean is a measure of central tendency (where the “centre” is for a dataset).
IMO, it doesn’t make sense to use the narrowness/wideness of a curve to figure out the average, because you can have a curve that is wide and a curve that is narrow, but both curves could still have the same average.
If you look at these two curves below with identical narrowness/wideness, but one is to the left and the other is to the right, which do you think has the greater average/mean?
If you can answer this, then I think you’ll be able to figure out whether “Y has a greater mean than X” is true or false.
Seems equal to me
If you plot some numbers on the x-axis, let’s say from 0 to 10 (left to right), what is the x-value of the black curve’s “centre”? And what is the x-value of the red curve’s “centre”? Which centre has the bigger x-value?
Uhmmm…I’m sorry,I’m not getting it.
Actually I have raw foundation when it comes to geometry especially coordinate geometry. Manhattan’s chapter was a breeze but whenever I try to do these maths from other sites,I can’t understand anything no matter how many videos I watched on this particular chapter.
Btw, thanks for the response
Do you agree that the normal distribution is a perfectly symmetric distribution? It is easy to visually see this because the left half is just a reflection of the right half over the midline.
Yes, so aren’t both equal?
However, my question was actually on what basis we should assess SD when these kinds of bellcurves are given?
We’ll get to this later, but for now we’re focused on the mean part.
“equal” in what sense? They have equal area but okay i don’t think that’s very relevant to us.
What i was going for here is that a normal distribution is perfectly symmetric (0 skewness and/or kurtosis), and thus mean = median = mode = location of midline, right? Location of midline is where the area is split into two halves and that’s by definition the median. Thus by transitivity, it must also be where the mean and mode is.
Owing to the above, can you reason out why B’s distribution has a greater mean relative to A?
Hint: Look at the “position” of the midline for both cases. As discussed, the position of the midline is also the mean. Additionally, the “position” of the midline is well basically just the location of the midline on the brown horizontal axis.
Because the area is widely spread out. I don’t know.
No, I didn’t talk about this anywhere in my mini write up though.
Okay the main thing what I mentioned was that mean = median = mode = “position of midline” along the horizontal axis (or the X-coordinate).
Just use this much to compute the mean for all three curves:
So your answer should look like:
Blue has mean …
Red has mean …
Green has mean …
(Forget whatever you know about this topic and only use the bolded information I’ve provided you with)
0,0,2.5?
Please don’t give yourself trouble anymore. Now, the problem is getting on my nerves
I think I am not paying full attention to details or not even trying to get it. Sorry for disappointing you.
But thanks for walking me through it.I sincerely apologize for wasting your time.
The first two are fine, but i’m not sure how you got the 2.5. Try again and look carefully at the x coordinate of the leftmost peak.
It’s okay, I don’t mind.
Really? Well, I forgot to add (-) but it doesn’t have to be 2.5, it can be- 2.6,-2.9, right?
The peak looks to me to be around -2, so that’s approximately the mean. But if it’s not clear where the peak happens (x coordinate) then yeah you can give a range.
Now, can you answer the question of which distribution has the greater mean? Exactly the same terminology applies.
It is A
Reasoning? I want to remind you again that mean = mode = median = x coordinate of the peak.
