The manufacturer of an extended-life lightbulb claims the bulb has an average life of 12,000 hours, with a standard deviation of 500 hours. if the distribution is bell shaped and symmetrical, what is the approximate percentage of these bulbs that will last between 11,000 and 13,000 hours?

95.44%

Step-by-step explanation:

Mean(μ) = 12000 hrs

Standard deviation (σ) = 500 hrs

Let X be a random variable which is a measure of the extended life of the light bulb.

The probability that these bulbs will last between 11,000 and 13,000 hours = Pr(11000 ≤ X ≤ 13000)

For normal distribution

Z = (x - μ)/ σ

Pr[(11000 - 12000)/500 ≤ (x - μ) /σ ≤μ

≤ (13000 - 12000)/500]

= Pr(-1000/500 ≤ z ≤ 1000/500)

= Pr(-2 ≤ z ≤ 2)

Pr(-2 ≤ z ≤ 2) = Pr(-2 ≤ z ≤ 0) + Pr(0 ≤ z ≤ 2)

By symmetry

Pr(-2 ≤ z ≤ 0) = Pr(0 ≤ z ≤ 2)

Therefore

Pr(-2 ≤ z ≤ 2) = 2Pr(0 ≤ z ≤ 2)

= 2(0.4772)

= 0.9544

Pr(-2 ≤ z ≤ 2) = 0.9544*100

= 95.44%

i need a picture

step-by-step explanation:

you forgot an image

step-by-step explanation: