9.1.3 Expected Lifetimes: Continuous Case

Now we return to our original problem — that of deciding what should be the expected lifetime of a random bulb — for which we are ready to do a divide-and-conquer argument. If we experiment with `F(t)=1-e^(-0.032 t)`, we find that all the bulbs (except possibly one) should be burned out by day `200`. Specifically, `F(200) = 0.9983`. We choose a large integer `n`, and we imagine the time interval from `0` to `200` divided into `n` subintervals, each of length `Delta t = 200`/`n`. Our real bulbs don't have exact lifetimes, of course, but we will approximate the expected lifetime by pretending that they do: If a bulb burns out between the time `t_(k-1)=(k-1) Delta t` and the time `t_k = k Delta t`, then we will count it as having burned for exactly `t_k` days. We are now in the situation described by the expression

where `p_k = F(t_k) - F(t_(k-1))` for `k = 1, 2, ..., n`. We can thus approximate the expected lifetime of a random bulb by the sum

This sum does not look quite like an approximating sum for an integral, because it is lacking a factor of `Delta t` in each term. Our solution to this problem is the same as it was for other problems in which we failed to find a hoped-for algebraic form: We multiply and divide each term of the sum by `Delta t`, which does not disturb the value of the sum. Then we factor `Delta t` out of each term, which permits us to write the sum in the form

For large values of `n`, `Delta t` is small, and

Note 2 — on approximate value of the difference quotient

If we abbreviate the derivative of `F` by `f`, then our approximate expression for the expected lifetime becomes

As `n` gets large, this sum approaches the integral

Activity 3

1. Given `F(t) = 1 - e^(-0.032 t)`, find `f(t)`.

2. Find a symbolic evaluation of the integral

   ∫ 0 200 t ⁢ f ( t )   d t .

3. Check your calculation with your computer or calculator.

Comment on Activity 3

From Activity 3, we find that the expected lifetime of a typical bulb measured in hours is `30.87 times 24 = 741` hours. At first glance, the `750` on the package seems to be a little exaggerated. However, `[0,200]` may not be a large enough interval.

Nevertheless, that interval is very large. More generally, to find expected lifetimes for things that fail, with a "cumulative failure fraction function" of the form

we need to know how to calculate

possibly for very large values of `T`.

Checkpoint 3