Sheldon Ross 10: Example 2.52

Question: The lifetime of a special type of battery is a random variable with mean 40 hours and standard deviation 20. hours A battery is used until it fails, at which point it is replaced by a new ne. Assuming a stockpile of 25 such batteries, the lifetimes of which are independent, approximate the probability that over 1100 hours of use can be obtained.

For the convenience, I will slightly deviate from the numbers in the question

  • That the random variable has an exponential distribution with a lifetime of 40 hours. This means that the rate of the exponential distribution is λ = 140
    • This makes sure that we can simulate the variable with a non-negative lifetime
    • Hence making us make inferences about the physical entities like batteries, bulbs and essentially anything that has a lifetime > 0
  • This in-turn means that both the mean the the standard deviation of the of the distribution would be 40 minutes

Numerical Solution

Essentially the idea is to use the mean and the standard deviation in the central limit theorem to make calculations on the probability. For applying the central limit theorem, we need

  • the value of the x (which here is the lifetime)
    • this is usually a dummy placeholder until we have an objective
    • Since the objective is clear for this variable, x = 1100 hours
  • the expected value of the random variable
    • The expected value of the sum of the 25 random variables is the sum of the expected value of the 25 random variables
    • Symbolically, E[x] = E[Σ xi] = Σ E[xi] where 1 ≤ i ≤ 25 for the problem
  • the variance of the sum of the variables, is the sum of the variances of the random variables
    • hence the standard deviation of the sum of the random variable would be the square root of the sum of the variances of the individual random variables

Needed: P { X ≥ 1100 } where X is the sum of 25 independent and identically distributed exponential random variables

Using the central limit theorem we get

P { (x1 + x2+.+.+x25)(40√25) > (1100 – 1000)(40√25) } ≈ 1 – Φ(0.5) ≈ 0.308538


Simulation (Calculation)

We can also use simulation to see how closely it matches up to the calculated value. This should be a fairly simple calculation and shown in the images below

Reading the image

  • Each of the Histogram density is an experiment done for 1000 trials
  • What is changing between the different densities?
    • First density: 100 x 25 batteries tested 1000 times and the proportion of > 1100 cumulative times calculated in each case and plotted as a distribution
    • Second density: 1000 x 25 batteries tested 1000 times and the proportion of > 1100 cumulative times calculated in each case and plotted as a distribution
    • Third density: 2000 x 25 batteries tested 1000 times and the proportion of > 1100 cumulative times calculated in each case and plotted as a distribution


Simulation (Visualization)

Reading the image.

  • Each horizontal row of the dots and represents a set of 25 batteries
  • Each dot in this row is an individual battery
    • When the battery (dot) is red, it means that that it was replaced before the cumulative time was below 1100 hours
    • When the battery (dot) is green, it means that it is a part of the set of the batteries whose cumulative time has extended beyond the intended 1100 hours
  • The vertical green line is the intended lifetime limit of 1100 hours

It also appears (visually) that roughly one third tend to escape the limits of out requirements thus acing our test 😉


Code: Code for both the sections is shown below


End of the post 🙂


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