The lifespan of a 60-watt lightbulb produced by a company is normally distributed with a mean of 1450 hours and a standard deviation of 8.5 hours. if a 60-watt light bulb produced by this company is selected at random, what is the probability that its lifespan will be between 1440 and 1465 hours

To evaluate the probability that the lifespan will be between 1440 and 1465 hours will be given by:
P(1440<x<1465)
using the z-score formula we obtain:
z=(x-μ)/σ
where:
μ=1450
σ=8.5
hence
when x=1440
z=(1440-1450)/8.5
z=-1.18
P(z<-1.18)=0.1190

when x=1465
z=(1465-1450)/8.5
z=1.77
P(z<1.77)=0.9625

hence:
P(1440<x<1465)
=0.9625-0.1180
=0.8445

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Martebi Martebi · Answer 2 · 2022-06-01T17:06:23+02:00

The probability that its lifespan will be between 1440 and 1465 hours is known to be 0.8445.

What is the lightbulb about?

To be able to solve for the probability that the lifespan will be between 1440 and 1465 hours, we say that:

P (1440 < x < 1465)

The when we make use of the z-score formula we see that:

z = (x-μ) ^ σ

Note that :

μ = 1450

σ = 8.5

Therefore,

If x = 1440

Then: z = (1440 - 1450) / 8.5

z = - 1.18

P (z<-1.18) = 0.1190

Then x = 1465

Since z = (1465-1450)/ 8.5

Then z = 1.77

P (z < 1.77) = 0.9625

Therefore,

P (1440 < x < 1465)

= 0.9625-0.1180

So the answer will be =0.8445:

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