Answer:
7000
Step-by-step explanation:
Answer:
a) P=0.0225
b) P=0.057375
Step-by-step explanation:
From exercise we have that 15% of items produced are defective, we conclude that probabiity:
P=15/100
P=0.15
a) We calculate the probabiity that two items are defective:
P= 0.15 · 0.15
P=0.0225
b) We calculate the probabiity that two of three items are defective:
P={3}_C_{2} · 0.85 · 0.15 · 0.15
P=\frac{3!}{2!(3-2)!} · 0.019125
P=3 · 0.019125
P=0.057375
Answer:
(a) E(X) = -2p² + 2p + 2; d²/dp² E(X) at p = 1/2 is less than 0
(b) 6p⁴ - 12p³ + 3p² + 3p + 3; d²/dp² E(X) at p = 1/2 is less than 0
Step-by-step explanation:
(a) when i = 2, the expected number of played games will be:
E(X) = 2[p² + (1-p)²] + 3[2p² (1-p) + 2p(1-p)²] = 2[p²+1-2p+p²] + 3[2p²-2p³+2p(1-2p+p²)] = 2[2p²-2p+1] + 3[2p² - 2p³+2p-4p²+2p³] = 4p²-4p+2-6p²+6p = -2p²+2p+2.
If p = 1/2, then:
d²/dp² E(X) = d/dp (-4p + 2) = -4 which is less than 0. Therefore, the E(X) is maximized.
(b) when i = 3;
E(X) = 3[p³ + (1-p)³] + 4[3p³(1-p) + 3p(1-p)³] + 5[6p³(1-p)² + 6p²(1-p)³]
Simplification and rearrangement lead to:
E(X) = 6p⁴-12p³+3p²+3p+3
if p = 1/2, then:
d²/dp² E(X) at p = 1/2 = d/dp (24p³-36p²+6p+3) = 72p²-72p+6 = 72(1/2)² - 72(1/2) +6 = 18 - 36 +8 = -10
Therefore, E(X) is maximized.
E^x^2 - 9 = 1
e^x^2 = 1 + 9 = 10
Ln e^x^2 = Ln 10
x^2 = Ln 10
x = sqrt (Ln 10) = 1.517
x = 1.517
