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you have 6 quarters, 10 dimes, 4 nickles, 2 pennies. One coin is selected at random. What is the expected value of this experiment?
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multiply the amount you would get times the probability you get it, and add them up
other way is to take the average of all the money, total the amount of money you have, divide by the total number of coins it is the same thing
E[X] = (6/22) * 0.25 + (10/22) * 0.10 + (4/22) * 0.05 + (2/22) * 0.01.
... value of E[X] is in dollars as expressed above.
rulinck wrote first way. second way is \[\frac{6\times .25+10\times .1+4\times .05+2\times .01}{22}\]
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rounded to the nearest cents is .12.....is this correct?
Yes!
$0.123636 ... or about 12 cents.
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