Question

Suppose that the useful life of a car battery, measured in months, decays with parameter 0.03 . We are interested in the life of the battery. In each appropriate box you are to enter either a rational number in "p/q" format or a decimal value accurate to the nearest 0.01
a. The life of the battery is modeled by a random variable X with X ~ (pick one distribution.
b,The mean of X is μ,-
C The standard deviation of X is 7x =
d. On average, how long would you expect one battery to last? EIX) =
e.On average, given 7 batteries and replacing each at the end of its life with the next, how long would you expect the 7 batteries to last?

Answer 1

Answer:

a) X ~ $e(0.03)$

b) μ = 100/3

c) $\sigma = 100/3$

d) A battery is expected to last 100/3 months (33 months and 10 days approximately).

e) For seven batteries, i would expect them to last 700/3 months (approximately 19 years, 5 months and 10 days).

Step-by-step explanation:

a) The life of a battery is usually modeled with an exponential distribution X ~ $e(0.03)$

b) The mean of X is μ = 1/0.03 = 100/3

c) The standard deviation is $\sigma = 1/0.03 = 100/3$

d) The expected value of the bateery life is equal to its mean, hence it is 100/3 months.

e) The expected value of 7 (independent) batteries is the sum of the expected values of each one, hence it is 7*100/3 = 700/3 months.