standard deviation

Sue0113

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You have a six sided die with the numbers 1 to 6 written on the faces.
a)In order to have the smallest possible standard deviation what number(s) would have to show up if you rolled the die five times?
b)In order to have the largest possible standard deviation what number(s) would have to show up if you rolled the die five times?
c)Redo (a) or (b) if no repeats allowed. You ignore a repeated number and roll again

Totally confused
a) the number 1 would have to come up 5 times
b) the number 6 would have to come up 5 times

Can someone help me figure this out.
 
You have a six sided die with the numbers 1 to 6 written on the faces.


a)In order to have the smallest possible standard deviation what number(s) would have to show up
if you rolled the die five times?



b)In order to have the largest possible standard deviation what number(s) would have to show up
if you rolled the die five times?



c)Redo (a) or (b) if no repeats allowed. You ignore a repeated number and roll again

Totally confused


a) the number 1 would have to come up 5 times

Could it not be a repetition of any of the other five numbers? If all of the data equal their mean,
then would it matter?


b) the number 6 would have to come up 5 times The differences (or more specifically, the squares
of the differences from each data to their mean) has to be as large as possible. What about checking
into 1, 1, 1, 6, 6 or 1, 1, 6, 6, 6 to see what standard deviations each of those sets of data give?


Can someone help me figure this out.
...
 
Don't understand on how to calculate

Is there a formula I would use to calculate the smallest possible deviation? and the largest deviation? and if no repeats are allowed?
 
Still Don't understand how to answer the question

I don't understand how to answer this question.
I need someone to explain it in real simple language. What exactly am I suppose to be trying to calculate?
How do I find the smallest deviation and largest deviation?
 
I don't understand how to answer this question.
I need someone to explain it in real simple language. What exactly am I suppose to be trying to calculate?
How do I find the smallest deviation and largest deviation?

Sue, do you know what standard deviation is? Knowing this will help to answer the question. The standard deviation is the square root of the variance. Thus:

\(\displaystyle \sqrt{\sum(\frac{(x_{i}-\mu)^2}{n-1})}\)

Keep in mind, the \(\displaystyle \mu\) is usually noted as "x-bar" as that is the sample mean and \(\displaystyle \mu\) is usually reserved for population mean. I just didn't know how to write "x-bar" in LaTex. Also, the denominator is "n-1" for sample variance/standard deviation and "n" for population variance/standard deviation.

That all being said, for (a) the standard deviation would be smallest if we minimize the numerator. Thus, the smallest the numerator could be is 0 which would occur if each \(\displaystyle x_{i}\) value equalled the mean. Do you see that? So let's find the sample mean. The sample mean is the weighted average of all the possible results that x could be when you roll the dice, thus:

\(\displaystyle \mu=\frac{1+2+3+4+5+6}{6}=\frac{21}{6}=3.5\) Each roll has a 1/6 probability of occurring.

So, since we can't roll a 3.5, the next closest value to the mean would be either a 3 or a 4. Thus, the smallest standard deviation would be:

\(\displaystyle \sqrt{\frac{(4-3.5)^2+(4-3.5)^2+(4-3.5)^2+(4-3.5)^2+(4-3.5)^2}{5-1}}\)

\(\displaystyle \sqrt{\frac{5(0.5)^2}{4}}\)


\(\displaystyle \sqrt{\frac{5(0.25)}{4}}\)

\(\displaystyle \sqrt{\frac{1.25}{4}}\)

\(\displaystyle \sqrt{0.3125}\approx0.559\)

See if you can get (b).

 
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part b

the value would be the same as the above we have a 1/6 chance of each number showing up. which would make the mean the same 4.

That is about as far as I understand. I guess even simple explanation is not getting through.
The standard deviation is the square root of the mean.....this I understand. What I don't understand is where you get the possible outcomes of
rolling the dice. For part a I understand that you roll the dice and 1 to 6 will come up. I'm rolling the dice 5 times so there is a chance that any of these 6 numbers will come up. From there I'm lost.
 
the value would be the same as the above (NO) we have a 1/6 chance of each number showing up. which would make the mean the same 4 (NO. The mean is still 3.5).

That is about as far as I understand. I guess even simple explanation is not getting through.
The standard deviation is the square root of the mean (NO. The standard deviation is the square root of the VARIANCE).....this I understand. What I don't understand is where you get the possible outcomes of
rolling the dice (When you roll a dice, you will either roll a 1, 2, 3, 4, 5 or a 6 all with a 1/6 probability). For part a I understand that you roll the dice and 1 to 6 will come up. I'm rolling the dice 5 times so there is a chance that any of these 6 numbers will come up (YES, but we are trying to determine what values of the rolls would give the smallest or largest standard deviation). From there I'm lost.

The standard deviation is the measure of the spread of the data from the mean. If you have data points very close to the mean then the standard deviation will be small. If you have data points far away from the mean then you will have a large standard deviation. So if the mean is 3.5, what values of the roll would produce the largest standard deviation, those values furthest from the mean? They would be rolls of either a 1 or a 6 as both of these values are 2.5 away from the mean, the furthest you can get in this scenario.
 
so then

All I can roll would be 1's or 6's.
the equation would then be
square root of (6-35)+(6-3.5)+(6-3.5)+(6-3.5)+(6-3.5) / 5-1
 
All I can roll would be 1's or 6's.
the equation would then be
square root of (6-35)+(6-3.5)+(6-3.5)+(6-3.5)+(6-3.5) / 5-1

Not quite.

\(\displaystyle \sqrt{\frac{(6-3.5)^2+(6-3.5)^2+(6-3.5)^2+(6-3.5)^2+(6-3.5)^2}{5-1}}\)

 
well then I'm not seeing this
You forgot to square the differences from the mean.

The sum of the differences from the mean is ALWAYS 0.

Example. The mean of 3, 7, and 14 is 8.

The sum of the differences from the mean = (3 - 8) + (7 - 8) + (14 - 8) = - 5 - 1 + 6 = - 6 + 6 = 0.

But the sum of the squared differences from the mean need not and usually do not = zero.

(3 - 8)^2 + (7 - 8)^2 + (14 - 8)^2 = 25 + 1 + 36 = 62.

By the way, I am not sure whether this problem is about a sample or a population. It probably makes no difference to the answer.
 
part b

so then the equation is as you wrote, which is what I was trying to write but thought I could just say the whole top was squared. devided by 5-1.
 
so then

I would say that the answer is 3.953 if that is not correct then I don't know. but what I don't understand is for part a answer was .559 and part b is 3.953 so these are the smallest and largest deviations but that does not answer the questions as to what numbers would have to show up? My bottle of tylenol is almost empty thinking my brain does not understand statistics.
 
so then the equation is as you wrote, which is what I was trying to write but thought I could just say the whole top was squared. devided by 5-1.
Because the whole top unsquared would be zero, squaring it would always result in zero. It is the difference of the individual terms that must be squared, then summed, then divided, and then the root taken.
 
You have a six sided die with the numbers 1 to 6 written on the faces.
a)In order to have the smallest possible standard deviation what number(s) would have to show up if you rolled the die five times?
b)In order to have the largest possible standard deviation what number(s) would have to show up if you rolled the die five times?
c)Redo (a) or (b) if no repeats allowed. You ignore a repeated number and roll again

Totally confused
a) the number 1 would have to come up 5 times
b) the number 6 would have to come up 5 times

Can someone help me figure this out.
The answer to a relates to what is the smallest possible standard deviation you can possibly get.

If every instance is equal to the mean, the standard deviation is 0. If you look at the definition of the standard deviation, it can never be less than zero, and it can only be zero if every observation equals the mean.

So the answer to part a is 1 five times, or 2 five times, or 3 five times, or 4 five times, or 5 five times, or 6 five times. Those are the only ways that every instance can equal the mean.

Part b is bit trickier. Here you are asked what makes the standard deviation largest. It should be intuitively clear that further the instances are from the mean, the greater the standard deviation will be. If you roll some combination of twos, threes, fours, and fives, the mean is going to lie between 2.6 and 4.4, correct? Then the differences will never equal 3.

If you roll only 1's and 6's, then the mean may lie anywhere between 1 and 6. However, if the mean lies close to 1, then most instances lie close to 1 and the differences will be small. Make sense? If the mean lies close to 6, then most instances lie close to 6. So then the differences will be small. So the standard deviation will be big only if the instances are ones and sixes, but not too many of each.

If you roll 1 three times and 6 twice, the mean will be 3, and the sum of the squares will be 3(1 - 3)^3 + 2(6 - 3)^2 = 3 * 8 + 2 * 9 = 42.

If you roll 1 twice and 6 thrice, the mean will be 4, and the sum of the squares will be 2(1 - 4)^2 + 3(6 - 4)^3 = 2 * 9 + 3 * 8 = 42.

So the answer is either 1 three times and 6 two times or 1 two times and 6 three times.
 
well now I'm totally mixed up

Jeff you didn't answer my question only confused me more.
Because the whole top unsquared would be zero, squaring it would always result in zero. It is the difference of the individual terms that must be squared, then summed, then divided, and then the root taken. Because I don't understand what you mean. what did the others have me figuring out. I've been working on this question for a long time and still have not come any closer to understanding. Just when I think i'm on my way to understanding and completing the equation you stated this. Could you explain in simpler terms what you mean. I'm a women returning to school after 30years and I'm so fustrated.
 
Jeff you didn't answer my question only confused me more.
Because the whole top unsquared would be zero, squaring it would always result in zero. It is the difference of the individual terms that must be squared, then summed, then divided, and then the root taken. Because I don't understand what you mean. what did the others have me figuring out. I've been working on this question for a long time and still have not come any closer to understanding. Just when I think i'm on my way to understanding and completing the equation you stated this. Could you explain in simpler terms what you mean. I'm a women returning to school after 30years and I'm so fustrated.
I do not mean to frustrate you. Statistics is hard; there is something about it that most people find very counter-intuitive.

First off, I am not sure whether Sir Michael is correct to have you dividing by 5 - 1. That is the right thing to do if the five rolls are considered a sample, but the divisor should be 5 if the five rolls are considered a population. I am not 100% sure what is intended by the problem about that issue. So I avoided solving your equation. However, the divisor is irrelevant to the answer.

\(\displaystyle (6 - 3.5)^2 + (6 - 3.5)^2 + (6 - 3.5)^2 + (6 - 3.5)^2 + (6 - 3.5)^2 + (6 - 3.5)^2 = 6(2.5)^2 = 37.5.\)

\(\displaystyle \{(6 - 3.5) + (6 - 3.5) + (6 - 3.5) + (6 - 3.5) + (6 - 3.5) + (6 - 3.5)\}^2 = \{6(2.5)\}^2 = 15^2 = 225 \ne 37.5.\)

The sum of the squares does not equal the square of the sums.

In any case, the way I understand the problem (which may be wrong; it is a sort of odd problem) if you roll a 6 five times, the mean will be 6, not 3.5. So I must admit I do not understand that equation. I gave you the answer of what I THINK the question is asking.
 
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Thanks Jeff

Thanks for helping me I actually get what you are saying and I couldn't figure out why they had me finding the smallest deviation and largest when I needed the numbers of the dice. All the question stated was this is a standard deviation contest. and then the previous information I gave.
 
Jeff you didn't answer my question only confused me more.
Because the whole top unsquared would be zero, squaring it would always result in zero. It is the difference of the individual terms that must be squared, then summed, then divided, and then the root taken. Because I don't understand what you mean. what did the others have me figuring out. I've been working on this question for a long time and still have not come any closer to understanding. Just when I think i'm on my way to understanding and completing the equation you stated this. Could you explain in simpler terms what you mean. I'm a women returning to school after 30years and I'm so fustrated.
I hate to jump in so far into the lifetime of this question, but perhaps a completely different perspective will help.

STARTING OVER FROM THE VERY BEGINNING
An example of 5 rolls of a die:
1 6 3 4 4
First step. Take the mean of the five numbers = (1+6+3+4+4)/5 = 3.6
That is a measure of the "central tendency" of the experimental data.

Now we need a measure of the dispersion or the width of the distribution. The statistic we choose is called the "standard deviation," which I take to be defined as "The square root of the mean of the squares minus the square of the mean." Operationally that can be found on the same pass through the data as finding the mean.
Second step. Take the mean of the squares = (1+36+9+16+16)/5 = 15.6

Then sigma = sqrt[15.6 - 3.6^2] = 1.62

Note that if all five numbers are the same - whatever the value - this gives a standard deviation of zero, because every one of the five deviations is zero. That is clearly the smallest.

A different procedure to find sigma is first to find the mean, then the deviations from the mean, and then the squares of those deviations.
mean square deviation = [(1-3.6)^2 + (6-3.6)^2 + (3-3.6)^2 + (4-3.6)^2 + (4-3.6)^2]/5 = 2.64
taking the square root --> sigma = 1.62

When n is small (and n=5 is small), either of these procedures produces a "biased" estimate of the standard deviation. Many people multiply by sqrt[n/(n-1)] for an "unbiased" estimator. The rationale for that is that one of the n degrees of freedom has been used up in finding the mean of the sample. That is a dfine point that will not affect comparing sigmas for different sets of 5 digits.

To make sigma as large as possible, the deviations from the mean have to be large. If repeats are allowed, these two sets have max sigma:
1 1 1 6 6
1 1 6 6 6

If repeats are not allowed, then think about omitting one of the six numbers. Omitting either 1 or 6 results in a set that is as compact as possible, hence will have minimum standard deviation.
Omitting one of the numbers in the middle, either 3 or 4, will be the most spread out you can get, hence maximum sigma.

Hope this helps.
 
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