3. A company manufactures light bulbs. The average lifetime for these bulbs is 4,000 hours with a standard deviation of 200 hrs. What lifetime should the company promote for these bulbs, whereby only 2% of the bulbs burn out before the claimed lifetime? You may assume that the lives of the bulbs are normally distributed

Answer: the company should promote 4412 hours for these bulbs.

Step-by-step explanation:

Assuming that the lives of the bulbs are normally distributed, we would apply the formula for normal distribution which is expressed as

z = (x - µ) / σ

Where

x = lifetime of the bulbs in hours.

µ = mean hours

σ = standard deviation

From the information given,

µ = 4000 hours

σ = 200 hours

If only 2% burn out, then the company would promote 98% would meet the claimed lifetime. Looking at the normal distribution table, the z score corresponding to a probability of 0.98 is 2.06. Therefore,

2.06 = (x - 4000) / 200

Cross multiplying, it becomes

200 * 2.06 = (x - 4000)

412 = x - 4000

x = 4000 + 412

x = 4412 hours