Question

The lifetime of light bulbs produced by a company are normally distributed with mean 1500 hours and standard deviation 125 hours.

(c) If three new bulbs are installed at the same time, what is the probability that exactly two will be burning after 1400 hours?

(d) If three new bulbs are installed at the same time, what is the probability that at least two will be burning after 1400 hours?

Enter your answer as a decimal, not a percentage. Round to four decimal places.

Answer 1

Given, mean μ = 1500 hours

stdev σ = 125 hours

Probability that a bulb will have a lifetime less than 1400 = P(z)

where, z = (X - μ) / σ = (1400 - 1500) / 125 = -0.8

from the z table, P(z) = 0.2119

Hence, Probability that the bulb will last for more than 1400 hours = 1 - 0.2119 = 0.7881 = 78.81%

(c) Probability that exactly 2 will be burning after 1400 hours = P1 burning * P2 burning * P3 not burning = 0.7881 * 0.7881 * 0.2119 = 0.1316

(d) Probability that atleast 2 will be burning after 1400 hours = P1 burning * P2 burning * P3 not burning + P1 burning * P2 burning * P3 burning = 0.7881 * 0.7881 * 0.2119 + 0.7881 * 0.7881 * 0.7881 = 0.6211