What's the best Approximation? :3
1 answer:
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Answer:
a) Expected amount of the gambler's win = $0.209
b) SD = 2.26
c)P (X >1) = P(z >0.35) = 0.36317
Step-by-step explanation:
The probability of winning, p = 12/38 =6/19
Probability of losing, q = 1 -p = 1-6/19
q = 13/19
Win amount = $5
Loss amount = $2
a) Expected total amount of win = ((6/19)*5) - ((13/19)*2)
Expected total amount of win = 1.579 - 1.369
Expected amount of win, E(X) = $0.209
b) Standard Deviation for the total amount of the gambler's win
E(X²) = (6/19)*5² - (13/19)*2²
E(X²) = 5.158
SD = 2.26
c) probability that, in total, the gambler wins at least $1.
P(X >1)
μ = E(x) = 0.209
z = (1-0.209)/2.26
z = 0.35
P( X >1) = P(z >0.35)
P(z >0.35) = 1 - P(z <0.35)
P(z >0.35) = 1 - 0.63683
P(z >0.35) = 0.36317