# If 20% of pea pods contain 80% of the peas, it is possible for the number of peas in a pod to be normally distributed?

Asked by PhiNotPi (12681) January 5th, 2013

To be honest, I’m not sure if the example in the title is very good. Instead of pea pods, have shoppers spending large amounts of money, and instead of peas, let that be the amount of money that they spend.

If the amount that any particular shopper spends follows a normal distribution, is it possible for 20% of the shoppers to spend 80% of the money?

To be even more specific, let’s say that the mean amount of money spent by a customer is \$100. What would the standard deviation have to be in order for 20% of the shoppers to spend 80% of the money?

Assume that the store has a very, very large number of shoppers, and that spending follows a perfect normal distribution, regardless of practicality.

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I’m not doing any math, but this is my instinctive response. I don’t think it’s possible to have a normal distribution under those circumstances, unless those 20% are in the middle of the pack and we have a very narrow and tall bell shape. Just a guess.

wundayatta (58738)

@wundayatta I’m not quite sure what you are saying, but I don’t think that you are interpreting the normal distribution correctly.

PhiNotPi (12681)

No. In order to achieve the normal distribution, you have to have some number of shoppers to have more than the mean, and the problem you have outlined won’t allow that.

Here’s a simple example.

Let’s say that you have 100 shoppers (I like that example better) and of that number, 20 of them spend \$400 each. (We’ll use real simple numbers here.) That means that 80 other shoppers spent an average of \$25 each in order to reach a total of \$10,000. (The \$8000 spent by the “middle 20” represents your 80% parameter.)

In order to maintain a normal distribution, roughly half of the 80 shoppers would have to spend more than the \$400 average spent by the highest spenders. Obviously that can’t happen with this group and still maintain your major parameter.

CWOTUS (26102)

Are you trying to talk about the 80 20 rule by pareto??

What Exactly Is The 80/20 Rule?

By the numbers it means that 80 percent of your outcomes come from 20 percent of your inputs. As Pareto demonstrated with his research this “rule” holds true, in a very rough sense, to an 80/20 ratio, however in many cases the ratio can be a lot higher – 99/1 may be closer to reality.

creative1 (12071)

@CWOTUS I think that you are thinking that the 20% has to be the middle 20%. I never said anything like that. In my opinion, it would have to be the top 20% that is spending 80% of the money.

Say that there are 100 shoppers. Of them, 25 spend \$5, 50 spend \$100, and 25 spend \$195. This distribution is symmetric and a very rough approximation to a normal distribution.

In this example, there is a total of \$10000 spent, or \$100 per person; however, the top 25% spends 48.75% of the total.

PhiNotPi (12681)

If you want a normal curve, then that’s where the 80% (or so) of all spending would be.

I’m assuming that the curve is of “spending by customer”, in which case the stair-step Pareto chart you pose above is not at all “normal”. Normal would be a sort of bell curve: a few customers who spend next to nothing (the left hand leg of the chart), a few customers who spend hugely at the right side of the chart, and the bulk of customers – in the middle of the curve – who spend a little more or a little less than the mean. Somewhere in that group you’re also more likely to find a mode and (assuming that you don’t have some outlier customer spending millions on his visit to the store), the median as well.

CWOTUS (26102)

If it’s a Pareto distribution, it is definitely not a standard distribution. Standard distributions apply to populations where a very small number are at the extreme high and extreme low ends of the distribution. Pareto distributions apply to populations where there is a cluster of high performers at one end of the curve and a steady fall-off of performance as you move down the scale.

ETpro (34605)
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