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# Probability of earning more than \$500 in a week

Suppose over the course of a week I do various programming jobs for people.

On average each job pays \$10 with a standard deviation of \$1.50.

The maximum number of jobs I can do in a week is 70,

What is the probability that I will earn more than \$800 in a week where I manage to do 70 jobs?

This isn't a course question, I'm just trying to learn a principle of how to do these type of questions, thanks.
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purplesoup
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5 Solutions

Commented:
Assuming the required randomness (reasonable), this is a standard statistics problem. For 70 jobs average intake is \$700 with standard (sd) deviation of \$15. But is this really true? It is not. It overstates the sd of the 700. You should square the sd (gives the variance), add them, then take the sq root. In your case 1.5^2 = 2.25. Variance of work = 22.5, sq root of work = 4.74. You now have the usual bell curve with average 700 and sd of 4.7. You want the probability of exceeding the average by about 20 x sd. Look up the result in a table. It is VERY low, essentially zero.
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Commented:
(A result further away from an average by 5 sd is usually considered to be from a different set than that used to calculate the average.
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Commented:
I get a sd of \$12.55, but 7.968 standard deviations from the mean is still essentially impossible.

It would be far more likely that the tail of your distribution is actually thicker then Normal or that the pay for the various jobs are not really independent.
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Commented:
oxo's sd is correct. I used a wrong multiplier. But the conclusion that the probability is essentially zero stands. A sd of 8 is VERY large
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Commented:
conversely, if one takes the figure of your title (\$500) the answer is that you will (almost) certainly earn more than \$500.
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Author Commented:
Thanks for your help - I made that example up which was why the figures were a bit rubbish - but I understand the method now - thanks again.
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