Question #167930

An electrical firm manufactures light bulbs that have a length of life that is normally distributed with mean equal to 778 hours and a standard deviation of 43 hours. Find the probability that a bulb burns between 755 and 892 hours.


1
Expert's answer
2021-03-02T05:11:09-0500

Let X=X= a length of life of the light bulb: XN(μ,σ2).X\sim N(\mu, \sigma^2).

Then Z=XμσN(0,1)Z=\dfrac{X-\mu}{\sigma}\sim N(0, 1)

Given μ=778 hr,σ=43 hr\mu=778\ hr, \sigma=43\ hr


P(755<X<892)=P(X<892)P(X755)P(755<X<892)=P(X<892)-P(X\leq755)

=P(Z<89277843)P(Z75577843)=P(Z<\dfrac{892-778}{43})-P(Z\leq\dfrac{755-778}{43})

P(Z<2.651163)P(Z0.534884)\approx P(Z<2.651163)-P(Z\leq-0.534884)

0.99598920.29636510.699624\approx0.9959892-0.2963651\approx0.699624

The probability that a bulb burns between 755 and 892 hours is 0.699624.



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