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My knowledge on probability topics is a bit rusty, so I was hoping you guys could help me.

Let X be the amount of products a person buys.

The probability that the person buys 1 to 12 is 60%.

From 13 to 20, 35%.

From 21 to 100, 5%.

I need to find the expected number of products he/she buys.

I did (12-1)*0.6 + (20-13)*0.35 + (100-21)*0.05 = 13.

Is that correct?

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It is not correct. There is not enough information available to compute the mean (average). There is a crude case for using $\frac{12+1}{2}(0.6)+\frac{20+13}{2}(0.35)+\frac{100+21}{2}(0.05)$, but only crude. The thing we multiply by $0.6$ should probably be closer to $9$. The thing we multiply by $0.05$ should probably be in the $35$ range. But it is all guesswork, one needs finer-grained information. –  André Nicolas Nov 4 '13 at 3:56
    
@AndréNicolas, I didn't understand anything you said :s –  l19 Nov 4 '13 at 4:50
    
I suggested another computation, to replace yours. What I said is that the division into just $3$ largish ranges does not tell us enough to make a reliable estimate of the mean. –  André Nicolas Nov 4 '13 at 4:57
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1 Answer

up vote 1 down vote accepted

it should be (1+2+...+12)(0.6/12) + (13+...+20)(0.35/8)+(21+...+100)(0.05/80)

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note: this is done on the assumption that 1 to 12 is each uniformly distributed with probability 60/12 = 5%. e.t.c. –  freak_warrior Nov 4 '13 at 5:28
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