Question

4. In proof testing of light bulbs, the probability that any light bulb will fail is 0.05. Suppose a box contains 10 light bulbs, all which are independent of one another. A box of light bulbs can be sold only if all its light bulbs work. (a) Let X be the number of failed light bulbs. What distribution does X follow? State and name the parameters (b) How many light bulbs would you expect to fail, and what is the standard deviation of the number that will fail? c) If 10 boxes are shipped to a particular customer, how likely is it that at least 8 of them were good to be sold? (A box is good to be sold only if all of its light bulbs work and assume the boxes are independent of one another.)

Answer 1

The probability that bulb will fail is: p 0.05 The total number of bulbs is: n 10 a) It is observed that all the bulbs are independent of each other. Let X denote the number of failed light bulbs and follows a Binomial distribution with parameters is: (n-10 and p-0.05) That is: X-Binomial(n-10, p-0.05) b) The mean number of bulbs that will fail is, Mean np -10(0.05) -0.5 Standard deviation -Jnp(1-p) 10x005(1-005) 0.69 (Round to 2 decimal place) = 0.6 892

c) The number of boxes is shipped is: 10 Let Y denote the number of boxes which are good The probability that at least 8 boxes were good is: P(Y28) A box is good to sold only if all the light bulbs work. The probability all bulbs works is, = (1-005)10 0.9510 0.5987 Thus, the Y follows a Binomial distribution with (n = 10 and p = 0.5987) The probability that at least 8 bulbs were good is, 10 8(0.5987) (1-0,5987) 10-8 10 10-9 (0.5987) (1-0.5987) + 10 10-10 (0.5987) (10.5987) - 0.119630.03966+0.00592 0.1652 Hence, the required probability is0.1652