70% of the batteries last at least how long? (Round your answer to two decimal places.)

Question

Suppose that the useful life of a particular car battery, measured in months, decays with parameter 0.02. We are interested in the life of the battery.

Answer 1

To determine how long 70 percent of the batteries last, we need to find the corresponding lifespan at the 70th percentile of the battery life distribution.

Given that the useful life of the battery decays with a parameter of 0.02, we can use the exponential distribution to model the battery life. The exponential distribution has a probability density function (PDF) of f(x) = λe^(-λx), where λ is the decay parameter and x is the lifespan of the battery.

To find the lifespan at the 70th percentile, we need to solve for the value of x at which the cumulative distribution function (CDF) of the exponential distribution is equal to 0.70.

The CDF of the exponential distribution is F(x) = 1 - e^(-λx).

Setting F(x) = 0.70, we have:

0.70 = 1 - e^(-0.02x)

Rearranging the equation:

e^(-0.02x) = 1 - 0.70

e^(-0.02x) = 0.30

Taking the natural logarithm of both sides:

-0.02x = ln(0.30)

x = ln(0.30) / (-0.02)

Using a calculator, we can calculate the value of x approximately:

x ≈ 57.11

Therefore, 70 percent of the batteries will last for approximately 57.11 months.