Question

A lecture hall is having wiring issues, and the projector occasionally powers down randomly following an exponential distribution with rate λ (power downs per minute).

(a) A professor plans to give an 80 minute lecture. What is the minimal power-down rate λ that the projector could have so that there is a probability of 0.05 that the lecture will finish before the projector shuts down?

(b) If the lecture is only 50 minutes, how does this change the minimal rate needed to get a probability of 0.05 of finishing the lecture without an outage?

(c) Given the rates λ that you solved for in parts (a) and (b), what is expected value of how much time occurs between power-down events under the exponential distribution mode?

Answer 1

To find the minimal power-down rate λ that the projector could have so that there is a probability of 0.05 that the lecture will finish before the projector shuts down, we can use the exponential distribution. The answer to part (a) is λ ≈ -0.0033 (rounded to 4 decimal places). For part (b), the minimal power-down rate λ is λ ≈ -0.0092 (rounded to 4 decimal places). The expected value of the time between power-down events for part (a) is approximately -302.43 minutes and for part (b) is approximately -108.77 minutes.

Step-by-step explanation:

To find the minimal power-down rate λ that the projector could have so that there is a probability of 0.05 that the lecture will finish before the projector shuts down, we can use the exponential distribution. The exponential distribution has a probability density function of f(x) = λe^(-λx), where λ is the rate parameter and x is the time. The cumulative density function (CDF) for the exponential distribution is F(x) = 1 - e^(-λx). In this case, we want to find the rate parameter λ that gives a CDF of 0.05 at the lecture time. Let's solve for λ:

1. Since the lecture time is 80 minutes, we have F(80) = 0.05
2. Substituting the CDF formula, 1 - e^(-80λ) = 0.05
3. Now, solve for λ by taking the natural logarithm of both sides: -80λ = ln(0.95)
4. Finally, divide by -80 to get the minimal power-down rate λ: λ = ln(0.95) / -80

The answer to part (a) is λ ≈ -0.0033 (rounded to 4 decimal places).

For part (b), we repeat the same steps but with a lecture time of 50 minutes. The only difference is in step 2, where we have F(50) = 0.05. Solving for λ gives us λ ≈ -0.0092 (rounded to 4 decimal places).

In part (c), we are asked to find the expected value of the time between power-down events. This is simply the reciprocal of the power-down rate λ. So, for part (a), the expected value is 1 / (ln(0.95) / -80), and for part (b), the expected value is 1 / (ln(0.95) / -50).