0 Next 5 questions (3-7) are based on the following: In the table below x denote
ID: 3300777 • Letter: 0
Question
0 Next 5 questions (3-7) are based on the following: In the table below x denotes the X-Tract Company’s projected annual profit (in $1,000), along with the probability of earning that profit. The negative value indicates a loss. x f(x) -100 0.02 0 200 0.24 250 0.30 300 0.28 350 0.07 400 0.04 3 The probability that X-Tract will break even is, f(x = 0) = ________. a 0.11 b 0.09 c 0.07 d 0.05 4 The probability that X-Trac will be profitable, that is, f(x > 0) = _________. a 0.95 b 0.93 c 0.90 d 0.88 5 The mean or expected value of X-Trac’s profit is = E(x) = _________. a $229.0 thousand b $233.5 thousand c $238.0 thousand d $245.5 thousand 6 The variance of profit is var(x) = ________ a 8433.05 b 8443.90 c 8454.75 d 8465.60 7 On average, profit (loss) amounts deviate from the expected profit by ______ thousand. a 91.95 b 90.75 c 89.55 d 88.35 0 Next 5 questions (3-7) are based on the following: In the table below x denotes the X-Tract Company’s projected annual profit (in $1,000), along with the probability of earning that profit. The negative value indicates a loss. x f(x) -100 0.02 0 200 0.24 250 0.30 300 0.28 350 0.07 400 0.04 3 The probability that X-Tract will break even is, f(x = 0) = ________. a 0.11 b 0.09 c 0.07 d 0.05 4 The probability that X-Trac will be profitable, that is, f(x > 0) = _________. a 0.95 b 0.93 c 0.90 d 0.88 5 The mean or expected value of X-Trac’s profit is = E(x) = _________. a $229.0 thousand b $233.5 thousand c $238.0 thousand d $245.5 thousand 6 The variance of profit is var(x) = ________ a 8433.05 b 8443.90 c 8454.75 d 8465.60 7 On average, profit (loss) amounts deviate from the expected profit by ______ thousand. a 91.95 b 90.75 c 89.55 d 88.35Explanation / Answer
sOLUTION3:
total probability =1
0.02+x+0.24+0.30+0.28+0.07+0.04=1
x+0.95=1
x=1-0.95
x=0.05
P(X=0)=0.05
OPTION D
soLUTION4:
X=200,250,300,350,400
totalp(x)=0.24+0.3+.28+0.07+0.04
=0.93
ANSWER 0.93
OPTION B
sOLUTIN5:
MEAN=xf(x)
=-100*0.02+0*0.05+200*0.24+250*0.30+300*0.28+350*0.07+400*0.04
MEAN =245.5
OPTION D
sOLUTION6:
Variance=total x^2f(x)-mean2
=(-100)2 .(0.02)+02 (0.05)+2002 (0.24)+2502 (0.30)+3002 (0.28)+3502 (0.07)+4002(0.04)-(245.5)2
=$8454.75
OPTION C
The probability that X-Tract will break even is, f(x = 0) =Related Questions
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