please solve: On average, each profit (loss) amounts deviate from the mean profi
ID: 1179072 • Letter: P
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please solve:
On average, each profit (loss) amounts deviate from the mean profit by thousand. 71.59 74.62 78.81 81.84 Suppose $100 thousand is added to each previously projected profit (loss) level (y = 100 + x), but probabilities are maintained as before. The updated expected value of profit and the variance of profit for Z-Lab, respectively, are: Suppose, instead of adding $100 thousand, each profit (and loss) amount is doubled (y = 2x) and probabilities maintained as before. The new expected value of profit and the variance of profit for Z- Lab, respectively, are:Explanation / Answer
3) Here is the complete answer wit explanations:
The probability of breaking even (ie profit = 0) is 1 - (sum of all other probabilities) = 1 - (.05+.25+.4+.1+.05) = 0.15 = Answer(c)
4) P (f(x)>0) = (.25+.4+.1+.05) = (0.8) = Answer (d)
5) Expected value = ($1000)*[0.05*(-100) + 0.15*0 + .25*80 + .40*140 + .1*200 + .05*240] = $103 thousand = Answer(a)
6) The variance of profit = (1000)^2 * [0.05*(-100-103)^2 + 0.15*(0-103)^2 + .25*(80-103)^2 + .40*(140-103)^2 + .1*(200-103)^2 + .05*(240-103)^2] = $ 6211.00 mn = Answer (a)
7) Average deviation = sqrt (variance) = 78.81 thousand = Answer(c)
8) If $100 thousand profit is added uniformly, the mean changes by $100 thousan but the variance remains the same.
Hence answer = E(y) = 203 thousand , var(y) = 6,211.00 mn = Answer (a)
9) If profit is doubled, the expected value of profit is also doubled. Hence = E(y) = 206 thousand and variance will become 4 times. Hence var(y) = $24844.00 mn = Answer(c)
I am 100% sure of each step. All the best :)
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