(10+10+10 points) 1. The lifetime of a car battery in years follows the lognorma
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(10+10+10 points) 1. The lifetime of a car battery in years follows the lognormal distribution, with a log (to the base 10) adjusted mean of 0.8 and log adjusted variance of 0.24. a. What is the probability that a battery lasts more than 4 years? b. What is the time to which 75% of the batteries will last? c. You are trying car batteries one after another and measuring their lifetimes. What is the probability that you will find 2 batteries that last more than 4 years each within your 1st 4 trials?Explanation / Answer
Here cumulative distribution of log-normal distribution is
F(X) = 1/2 + 1/2 erf [ (log10 x - )/ 2]
where x is the lifetime of a car battery
Here erf is error function is inverse error function.
(a) Here probability that battery will lasts more than 4 years.
Pr(X >4) = 1 - Pr(X < 4 years)
Pr(X < 4 years) = 1/2 + 1/2 * erf [ (log104 - 0.8)/ (2 * 0.24)]
= 0.5 + 0.5 * erf [ (0.602 - 0.8)/0.6928)]
= 0.5 + 0.5 * erf [-0.2857]
= 0.5 + 0.5 * (-0.3138)
= 0.3431
(b) Here lets say for time = X , 75% of batteries will last.
so F(X) = 0.75
0.75 = 1/2 + 1/2 erf [ (log10 x - )/ 2]
0.25 = 0.5 * erf [ (log10 x - )/ 2]
0.5 = erf [ (log10 x - )/ 2]
(log10 x - )/ 2 = erf-1(0.5)
(log10 x - )/ 2 = 0.478
(log10 x - ) = 0.478 * 0.48 = 0.3312
log10 x = 0.3312 + 0.8 = 1.1312
x = 13.53 Hours
(c)
Pr( more than 4 hours) = 0.3431
Now it is negative binomial distribution
Pr( 2 battery with more than 4 hours life time in 4 trials) = Pr(2 such battery in 2 trials) + Pr(2 such batttery in 3 trials) + Pr(2 such battery in 4 trials) = 2C2 (0.3431)2 + 2C1(0.3431)2 (1-0.3431) + 3C1(0.3431)2 (1 - 0.3431)2
= 0.4248
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