Question

Q6. A Light bulb manufacturer warrantees that the life of bulbs has normal distribution with

       average life (μ) of 400 hours and standard deviation (σ) 20 hours. A customer selects one bulb

       randomly from the received shipment and installs it under the ceiling of the house. The true

       statement (s) that the installed bulb will continue to burn for at least for 482 hours is/are:

       a. It is rare but not impossible that bulb will continue to burn beyond 482 hours

       b. It is impossible that bulb continues to burn beyond 480 hours

       c. It is certain that bulb will continue to burn beyond 482 hours

       d. None of the a, b, c statements are true

Answer 1

Given average life (μ) of 400 hours

Standard deviation (σ) 20 hours

Given that that the life of bulbs has normal distribution

Z-score = (X - ) /

Here we need to find P(X482), the probability that the installed bulb will continue to burn for at least for 482 hours

Z-score = (482 - 400) / 20

= 82 / 20

= 4.1

The area to the right of z-score 4.1 will give us P(X482) which is 0.0000207 from the online z-score to p-value calculator for a right tail

So the probability that the installed bulb will continue to burn for at least for 482 hours As P(X482) = 0.0000207

As P(X482) = 0.0000207 , So it is rare but not impossible  that bulb will continue to burn beyond 482 hours

It will be impossibe if the probabiluty is 0

So Answer is Option A