#$&*
MTH 152
Your 'question form' report has been received. Scroll down through the document to see any comments I might have inserted, and my final comment at the end.
** Question Form_labelMessages **
Math Inquiry RE: MTH152-QA15-1
** **
** **
** **
In the assignment listed in the subject, MTH152-QA15-1, you had responded to my question about why one uses (sqd deviations)_n/(n-1) when working to get the radicand to use in solving for the standard deviation:
@&
There were 6 numbers in th eoriginal distribution, so there are 6 deviations to be squared and added.
The reason we divide by 5 has to do with the way the binomial distribution approaches the normal distribution. Up to n = 30 we get a more accurate result if we divide by 1 less than n. The reasons are too deep to go into here, relying on multivariable calculus and other gems of more advanced mathematics.
*@
If you would care to, I'd love to hear the precise reason why the binomial distribution approaches the normal distribution. That sounds a lot like a positively infinite limit...Don't worry about losing me with your explanation, if there's anything I don't follow I'll either research it or mull it over until I understand. Most likely, both.
CS
@&
Briefly:
The number of ways to get r heads on n flips is C(n, r).
Consider, say, 5 coin flips. The numbers of ways to get 0, 1, 2, 3, 4, 5, Heads are respectively C(5, 0), C(5, 1), C(5, 2), C(5, 3), C(5, 4) and C(5, 5). Working these out, on a large number of repeititions of 32 flips you would expect the average outcome to be
0
1
1
1
1
1
2
2
2
2
2
2
2
2
2
2
3
3
3
3
3
3
3
3
3
3
4
4
4
4
4
5
The mean of these numbers is 2.5. The standard deviation is close to sqrt(5/4).
A historgram representing this distribution consists of five 'bars' with the middle two 10 times as high as the two on the ends, and the other two 5 times as high. It is easy to sketch a bell-shaped curve on top of this histogram and see how the two don't differ too drastically.
If you did this for 100 flips, you would obtain a mean of 50 and a standard deviation of 5. The actual numbers involved would be astronomical (e.g., C(100, 50) is very large). A normal curve with this mean and standard deviation, having the same area as the histogram for this distribution, would pass through the top of each 'bar' of the histogram, pretty much in the center of each. In this sense the histogram and the normal curve are practically indistinguishable.
This is the sense in which the binomial distribution approaches the normal distribution.
It would take several pages to explain the rest, but you can search. I would start with something like
binomial distribution and normal distribution
*@