A question for you math/stats whizzes...

[Fat Elvis]

OK, so it has been many, many (many) years since I've had a statistics course so I've forgotten how to figure some pretty simple problems out. Because of my nearing senility, I was wondering if some of you folks could help clear the cobwebs.

Suppose I have a four sided die that is equally weighted; each side has a .25 probability of showing up on any roll. ( I know that I'm wording this wrong) How rolls would it take to say that a number did not appear for 2.57 standard deviations?

(I have a couple more questions as well, but I will wait for an explaination for this question first)

Thanks

Fat E

[Bugeater]

48÷2(9+3)

[Fat Elvis]

Originally Posted by Bugeater

48÷2(9+3)

Could you explain how you arrived at that? Thanks.

[loochy]

Ask Frankie. He has a super IQ and he knows everything.

[Fat Elvis]

I came up with 16 rolls. At 2.57 standard deviations, on a standard bell curve, there is a 99% probability a given outcome would be expected. If there is a 25% chance that a number would appear (four sided die), wouldn't that mean that there is a 75% chance that that one of the other numbers appears-- and that there is a 75% chance on each successive roll that one of the other numbers appear?

In other words, .75 x .75 x .75...until there is only a .01 probability that the next roll will be one of the three numbers without the original number being rolled?

Does that make sense? Am I completely off base?

[DanT]

Originally Posted by Fat Elvis

I came up with 16 rolls. At 2.57 standard deviations, on a standard bell curve, there is a 99% probability a given outcome would be expected. If there is a 25% chance that a number would appear (four sided die), wouldn't that mean that there is a 75% chance that that one of the other numbers appears-- and that there is a 75% chance on each successive roll that one of the other numbers appear?

In other words, .75 x .75 x .75...until there is only a .01 probability that the next roll will be one of the three numbers without the original number being rolled?

Does that make sense? Am I completely off base?

Hi Fat Elvis,

The problem involves a discrete distribution that puts 25% probability on each of the integers from 1 to 4. The course materials should have provided you a formula for computing the standard deviation for such a distribution, right? If you were to plot the density of that distribution, you will see that it looks nothing like a normal curve. Instead, the distribution puts the same amount of weight on each of only 4 points. You can compute the standard deviation for the problem's discrete uniform distribution using the formula you have for a standard deviation for a discrete probability distribution. (Alternatively, you can simply use one of the two Excel's functions for calculating POPULATION standard deviations and apply that to the four integers, like this:

=stdevp(1,2,3,4)
)

If you do that, you will learn what the population standard deviation is for that distribution. Once you know that, then you just need to compute what 2.57 of those standard deviations would equal. Then you just need to find the next greatest integer, aka the ceiling, aka the minimum integer whose value is at least as big as 2.57 * SD, where SD is the standard deviation whose value you computed at the beginning of the problem.

If I'm interpreting that problem correctly, I suspect that the intent of the problem is to help the student get familiar with translating a word problem into a math problem and then apply concepts you learned early in the class about standard deviations and other parameters that describe key features of distributions.

Hope this helps gets you started on the other problems!

[CrazyPhuD]

42 rolls.

[DanT]

Originally Posted by DanT

Hi Fat Elvis,

The problem involves a discrete distribution that puts 25% probability on each of the integers from 1 to 4. The course materials should have provided you a formula for computing the standard deviation for such a distribution, right? If you were to plot the density of that distribution, you will see that it looks nothing like a normal curve. Instead, the distribution puts the same amount of weight on each of only 4 points. You can compute the standard deviation for the problem's discrete uniform distribution using the formula you have for a standard deviation for a discrete probability distribution. (Alternatively, you can simply use one of the two Excel's functions for calculating POPULATION standard deviations and apply that to the four integers, like this:

=stdevp(1,2,3,4)
)

If you do that, you will learn what the population standard deviation is for that distribution. Once you know that, then you just need to compute what 2.57 of those standard deviations would equal. Then you just need to find the next greatest integer, aka the ceiling, aka the minimum integer whose value is at least as big as 2.57 * SD, where SD is the standard deviation whose value you computed at the beginning of the problem.

If I'm interpreting that problem correctly, I suspect that the intent of the problem is to help the student get familiar with translating a word problem into a math problem and then apply concepts you learned early in the class about standard deviations and other parameters that describe key features of distributions.

Hope this helps gets you started on the other problems!

I should also note that the wording for that problem could be crucial. I've never seen a problem worded quite that way, so because of that, I'd caution you to make sure that you understand what the problem is asking for. The approach you described in your most recent post is off-track for the problem you described in the topic thread header, but it would not be too far afield for attacking another sort of problem in elementary probability courses, one that concerns what are called negative binomial distributions, so make sure you know what the problem really is!

[Bugeater]

Originally Posted by DanT

I should also note that the wording for that problem could be crucial. I've never seen a problem worded quite that way, so because of that, I'd caution you to make sure that you understand what the problem is asking for. The approach you described in your most recent post is off-track for the problem you described in the topic thread header, but it would not be too far afield for attacking another sort of problem in elementary probability courses, one that concerns what are called negative binomial distributions, so make sure you know what the problem really is!

That's exactly what I was thinking.

[Shaid]

Originally Posted by Bugeater

48÷2(9+3)

[Ming the Merciless]

18.41 rolls to be near 2.51 standard deviations--assuming a normal distribution


2.51 std deviations = ~.005 (half of a percent)

I am reading the question as this: how many times would I have to roll the dice to be at 2.51 std devs for not seeing that number come up..


so

.75% ^ x = .005


18.41 rolls in a row without N number showing up




disclaimer: im drunk , tired and going to bed LOL

[Ming the Merciless]

Originally Posted by Fat Elvis

I came up with 16 rolls. At 2.57 standard deviations, on a standard bell curve, there is a 99% probability a given outcome would be expected. If there is a 25% chance that a number would appear (four sided die), wouldn't that mean that there is a 75% chance that that one of the other numbers appears-- and that there is a 75% chance on each successive roll that one of the other numbers appear?

In other words, .75 x .75 x .75...until there is only a .01 probability that the next roll will be one of the three numbers without the original number being rolled?

Does that make sense? Am I completely off base?

No, just slightly since its more like 99.5 and change %


oh crap i dint read dan T's.......i guess it isnt a 'normal' distribution?

[Third Eye]

Originally Posted by Pawnmower

18.41 rolls to be near 2.51 standard deviations--assuming a normal distribution


2.51 std deviations = ~.005 (half of a percent)

I am reading the question as this: how many times would I have to roll the dice to be at 2.51 std devs for not seeing that number come up..


so

.75% ^ x = .005


18.41 rolls in a row without N number showing up




disclaimer: im drunk , tired and going to bed LOL

As mentioned earlier, this is a uniform discrete distribution, not a normal distribution.

[Ming the Merciless]

Originally Posted by Third Eye

As mentioned earlier, this is a uniform discrete distribution, not a normal distribution.

caught that 60 seconds b4 you posted, thx

[Third Eye]

My best guess, given the wording, is that we are working with a geometric distribution with p=.25 and trying to determine the minimum number of trials it will take for a first success that occurs more than 2.57 standard deviation away from the mean. The mean for a geometric distribution is 1/p, so 1/.25=4. The variance is given by (1-p)/p^2 or .75/.0625=12. Thus the sd is 12^.5 or approximately 3.464. 2.57 sds is approximately equal to 8.9. So, to be at least 2.57 sds away from the mean you need at least 4 + 9 rolls, or 13.