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27 February, 07:21

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

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  1. 27 February, 11:06
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    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
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