Solved =]

Thanks ANDS!

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- Jul 28th 2009, 08:23 AMSummerCalculating Percentages using Mean and Standard Deviation.
Solved =]

Thanks ANDS! - Jul 28th 2009, 09:32 AMANDS!
My first question would be, how far into statistics are you? If you have already covered Z-scores and the like, then you can find an exact percentage to this situation.

If you are just starting out, and covering normal distributions of a data set, then you could approximate using Chebyshev's Theorem to see where you should be.

I'm guessing that they want you to use Z-scores here.

For example: My monthly grocery bill had a mean of $150 bucks, with a standard deviation of $25, and whose distribution is normally distributed. If I wanted to find the percentage of my bills that were greater than 200 bucks I would simply have to convert 200 bucks to a z-score [(200-150/25)], where 200 is my raw value, 150 is the population mean, and 25 is the population standard deviation (this assumes you know the cost of all your monthly bills, if not then you would have to take sample means/standard deviations). This turns into a z-score of 2.

Now going back to the problem, I want to find the percentage of bills GREATER THAN 200, so I would take my z-score and the corresponding percentage and subtract it from 1.0 (as a percentage can only be between 0 and 1). If I wanted to know the percentage that are LESS THAN 200, then I would simply take the percentage itself. Now this assumes that the table you are using is a CUMMULATIVE table. If not, then you would need to either subtract from 0.5 (for greater than), or add to 0.5 for 200 or less. Hope this helps. - Jul 28th 2009, 06:08 PMSummer
Hi there

ANDS!

Thanks for your reply =]

Unfortunately we haven't been using Z numbers or Chebyshev's Theorem.

We've been given the homework and told to figure out a way to solve it ourselves.

Unfortunately I'm unfamiliar with both methods =S

I've had a look at what you've written and read over it several times but I'm still having trouble understanding how to find the solution...

I understand how to get the Z score of 2. However I'm not entirely sure what you mean by 'corresponding percentage' and why/how I would exactly subtract that from 1.0 =S

I'm not using a cumulative table. What I've typed on my original post is the exact thing it says in the paper I've been given. =S

I think you have explained it quite well, I'm just having a lil bit trouble understanding.

Could you show/explain how to do it again using a different set of values?

Would be entirely grateful

Summer =] - Jul 28th 2009, 07:17 PMANDS!Quote:

I've tried taking the 130 away from 220 and dividing it by the 76...

But I still can't get the final percentage answer =S

What you have described in the quote is a z-score. A z-score is:

Quote:

a quantity derived by subtracting the population mean from an individual raw score dividing the difference by the population standard deviation.

$\displaystyle z=\frac{x-\mu}{\sigma}$

In your example X is 130, $\displaystyle \mu$ is 220, and $\displaystyle \sigma$ is $\displaystyle 76$. Plugging them into the above formula (which you seem to have a grasp of, even if you didn't know what it refers to), yields a value of $\displaystyle -1.18$

Now, on it's own, this number is absolutely meaningless. We need some way of connection it to a Z-table (or as wikipedia calls it, a standardized score). The reason we convert these values into Z-score, or standardized scores, is so that we can have a means of comparing different data sets against OTHER data sets. For example, in the problem you have, suppose you want to compare TWO stores, and see whose monthly costs are the most stable. One store might have a monthly cost of 200, while another may have a cost of 100,000. However, by converting a particular number into a z-score, and using that as a comparison, we can compare the two data sets and we may well find out that the monthly cost of Store A (at lets say 500 dollars) is WAY off, while the cost of Store B (at maybe 110 000) is well within the mean.

This:

http://www.pinkmonkey.com/studyguide...pendix/s58.gif

is a z-score graph and table. When I mentioned subtracting/adding, it was because of charts like this, that do not have a COMPLETE z-score table, which goes from -3.60 to 3.60 on the z-score scale. When you have a z-score, what you do is look at your chart. So our z-score was -1.18. If you look at the z-score table you have a vertical column and a horizontal column. The vertical column is the leading two digits of your z-score; the horizontal will be the hundredths place of your z-score.

So for our score, we need to first find -1.18, as it is our leading two digits. The problem with this chart, is that there are no negative numbers. Thats fine as a z-score graph is symmetric about the y-axis. Lets just take the positive of that number, which is 1.18. We then want to go over nine (0.00, 0.01, 0.02. . .0.08) to 0.02, which corresponds to 1.18 which is "close enough" to the z-score we want.

We see that the percentage that corresponds to a z-score of 1.18 is 0.3810 (or think of us going 0.318 to the left of 0 on our graph). However, we aren't done yet. This isn't a cummulative chart, and doesn't list ALL values from -3.60 to 3.60. We know that a percentage can go from 0 to 1. Our z-score says that our cummulative percentage at 1.18 is 0.3180. We have to add 0.5 to that numberto account for the 0.5 on the RIGHT side of the graph (we are on the left side of the graph, since the value you chose, 130 is to the LEFT of the population mean, 220). So what we need to do, is take $\displaystyle 1 \mbox{(the value of all percentages)}-(0.381+0.5)$, which is $\displaystyle 1-0.881=0.119.$

Therefore, there is a 0.119 chance, that a customer will have a bill LESS than $130.

Hopefully this helps. Perhaps you can make an attempt at the second question, and then we can do the 1st one as it involves a little bit of work. Here is another graph of a z-graph which will help you visualize what all this is better:

http://upload.wikimedia.org/wikipedi...bution.svg.png

http://upload.wikimedia.org/wikipedi...stribution.svg - Jul 29th 2009, 05:29 PMtopic
Correct me if i'm lost here.. They want an approximate percentage from these two amounts?

Quote:

Monthly cost on electronics of consumers at a store is normally distributed with a mean(average month) of £220 and standard deviation(one month) of £76.

- Jul 29th 2009, 05:37 PMANDS!
Thats not how it works. The standard deviation of a sample/population is the amount that a particular value will "deviate" from the mean. It is the square-root of the variance, with is talked about usually in abstract units that have no real application to reality.

In the chart above, you can see standard deviations labeled along the X-axis along with Z and T scores (which are related to two types of distributions).

I see where you would think we are taking percentages, but that is not the case here. The percentage we are looking at her, is the probability that a particular statistic will be a certain distance from the mean. In this example we have found (privately) that a monthly bill of 130 pounds, is about 1 standard deviation away from the population mean (assuming we know all the values so far). We then found (if I remember correctly) that the percentage of monthly bills that would be less than that was about .20 or so. These are the percentages we are looking at. - Jul 29th 2009, 05:39 PMtopic
what kind of math is this exactly??

- Jul 29th 2009, 05:49 PMANDS!
Just math.

Forumla wise, we define the standard deviation "s" as:

$\displaystyle s=\sqrt{\frac{\Sigma(x-\bar{x})^{2}}{n-1}}$

where $\displaystyle s$ is the standard deviation, $\displaystyle x$ is our individual values in the sample, $\displaystyle \bar{x}$ is our mean of the sample, and $\displaystyle n$ is the total number of values in our sample data set.