Exam P Practice Problem 1

Problem 1a
For a certain brand of light bulbs, the time (in thousands of hours) until the instant a light bulb burns out follows an exponential distribution. From past experience, we know that \frac{1}{4} of the light bulbs will go out within 500 hours of use. What proportion of the light bulbs can be expected to go out within 1000 hours of use?

Problem 1b
In a tech support call center for a brand of laser printers, the length of time (in minutes) for a technical support call follows an exponential distribution. From past experience, 60% of the calls last 3 minutes or less. What proportion of the calls can be expected to last more than 15 minutes?

Solution is found below.

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Solution of Problem 1a
Let X be the lifetime of this brand of light bulb. We have P(X \le t)=1-e^{\alpha t}. Plugging in 0.5 (500 hours) for t, we have P(X \le 0.5)=1-e^{-0.5 \alpha}=0.25. This leads to e^{-0.5 \alpha}=0.75. The following is the answer:

\displaystyle \begin{aligned}P(X \le 1)&=1-e^{-\alpha} \\&=1-\biggl[e^{-0.5 \alpha}\biggr]^2 \\&=1-\biggl[\frac{3}{4}\biggr]^2 \\&=\frac{7}{16} \end{aligned}

Answer to Problem 1b
\displaystyle \frac{32}{3125}=0.01024

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One response

  1. When I first attempted problem 1a, I solved for the exponential parameter (I use lambda). This requires one to use ln as the inverse and then plug this value back in to the equation a second time for t=1. Your solution is far more elegant– don’t solve for lambda, use the whole expression as a factor! Much quicker than the “textbook” way.

    I then used your method for the second problem. Elegant indeed.

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