Suppose the lifetime of a van has a uniform distribution over [3,10]. Mr Smith buys a new
van as soon as his old van either breaks down or reaches the age of 8 years. A new van costs
$20,000 and an additional cost of $1000 is incurred whenever a van breaks down. Assume
that a T years old car in working order has an expected resale value of 20,000[(1000T)/2].
What is Mr Smith\'s long-run average cost?
Solution
L= Lifetime of the van
L~ U(0,10)
I[A] is the indicator function which takes the value 1 if A is true and 0 otherwise.
Long run cost of Mr. Smith
= Cost incurred* I[ old van breaks down before reaching the age of 8 years] + Cost incurred * [
old van reaches the age of 8 years]
Thus Expected long run cost
= Cost Incurred* P[old van breaks down before reaching the age of 8 years]+ Cost incurred *
P[old van survived reaches the age of 8 years]
= $ 21000* P[ L < 8 ] + ${20000-1000*8/2}* P[ L > 8]
= $ 21000* (8-3)/(10-3) + $16000 * (10-8)/(10-3)
= $ 21000* 5/7 + $ 16000* 2/7
= $ 15000 + $ 4571.4
=$ 19571.4
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suppose the lifetime of x of an electronic components follows distri.pdf
1. suppose the lifetime of x of an electronic components follows distribution with parameter of 1/10
find the probability that the life time is less than 10
Solution
Given X~exp(mean=1/10=0.1) F(X=x)=1-exp(-0.1*x) So the probability is
P(X<10)=F(10) =1-exp(-0.1*10) =0.6321206
