A company manufactures light bulbs. The lifetime for these bulbs is 4,000 hours with a standard deviation of 200 hrs. What lifetime should the company promote for these bulbs, so that only 2% of them burnout before the claimed lifetime?

The company should promote a lifetime of 3589 hours for these bulbs, so that only 2% of them burnout before the claimed lifetime.

Step-by-step explanation:

Consider the lifetime of light bulbs is normally distributed. Then, μ = 4000 hours and σ = 200 hours.

Let X be the lifetime of bulbs. We need to find the lifetime before which only 2% (0.02) of the bulbs burn out. Suppose the value of this lifetime is y. then, we need to find out P(X<y) = 0.02.

We will use the z-score formula:

z = (x-μ)/σ

P(X<y) = 0.02

P((X-μ)/σ < (y-μ)/σ) = 0.02

P(z < (y-4000)/200) = 0.02

We can find the value of z at which the probability is 0.02 from the normal distribution (areas under the normal curve) table.

From the table we can see that 0.02 lies between z values -2.05 and -2.06. So,

z = [-2.05 + (-2.06)]/2

  = -4.11/2

z = -2.055

So,

(y-4000)/200 = -2.055

y-4000 = -411

y = 4000 - 411

y = 3589 hours

The company should promote a lifetime of 3589 hours for these bulbs, so that only 2% of them burnout before the claimed lifetime.

not sure but it may be b, as i said i'm not sure.