Question

A certain high quality commercial light bulb has an estimated lifetime of µ = 750 hours with a variance σ 2 = 100. If 20,000 bulbs are installed in city street lights. (a) Use Chebyshev’s inequality to estimate how many of these bulbs will live between 735 and 765 hours. (b) Use the normal distribution to determine how many of these bulbs will live between 735 and 765 hours. (c) Does your answer in (b) verify Chebyshev’s inequality results?

Answer 1

1. Let's recall Chebyshev’s inequality: P(|x-µ| > a) <= σ^2/a^2; x must be between 735 and 765 hours, so a = 15, µ = 750 to satisfy |x-µ| <= a , with which find P(|x-µ| <= a) = 1 - P(|x-µ| > a). Substituting the numbers P(|x-µ| > a) <= 100/(15*15) = 100/225. And P(|x-µ| <= a) = 1 - 100/225 >= 125/225 = 5/9. So we can find that al least 20000* 5/9 = 11111.111... bulbs will live between 735 and 765 hours.
2. P(735<X<765) = Ф((765-750)/10) - Ф((735-750)/10) = Ф(1.5) - (-Ф(1.5)) = 2*Ф(1.5) = 2* 0.43 = 0.86, where Ф(.) - Laplace function
3. Yes, we find from Chebyshev’s inequality that probability is above than 5/9 = 0.55... And the probability from normal distribution 0.86 verify it, because 0.86 > 0.55