Statistics Assignment #3
4.24
Bridge network reliability
(a)
P (x > 1) for a1 = 0.85
P (x > 1) for a2 = 0.60
Assuming they are independent events, P (x>1) = 0.85 * 0.60
= 0.51
(b) p(1) for a1 = 0.10
p (1) for a3 = 0.90
p(1) for a6 = 0.25
Therefore p(1) = 0.10 * 0.90 * 0.25
= 0.0225
4.38 Likelihood of flood damage
(a)
Probability Distribution
p(x)
x
(b)
Loss
x
Sample Point
Probability
300,000
Measurable amoun of rain
0.30
0
No Rain
0.70
E(x) = (300,000)(0.3) + (0)(0.7)
= $90,000
Therefore the firm’s expected losses due to flood are $90,000. This however is only gives an idea
of the central tendency and is not actually a possible value of
x,
because either it will rain and the
company will lose $300,000, or it will not rain and the company will not lose anything.
4.60 Parents who condone spanking
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n = 3 , p = 0.6
X ~ Bin (3, 0.6)
(a)
P (X=0) =( n ) p^x * q^(nx)
x
= ( 3 ) 0.6^0 * 0.4^3
0
= 0.064
(b)
P (X ≥ 1) = 1 – P(X=0)
= 1 – 0.064
= 0.936
(c)
mean = μ = np
= 3 x 0.6
= 1.8
standard deviation = σ = sqrt (npq)
= sqrt (3 x 0.6 x 0.4)
= 0.85
These results show the central tendency and variation of the sample results. Of course these
values cannot be actual values of x as there cannot exact fractions of people.
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 Fall '08
 Dr.JoseCorrea
 Statistics, Normal Distribution, 10%, 1%, $300,000, 10 minutes, 5.068 mL

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