Question #152008

1.    The lifetime of light bulbs produced by a company are normally distributed with mean 1000 hours and standard deviation 130 hours.

a)    The top 30% of all light bulbs should last at least how many hours?

b)    What is the probability that a randomly selected light bulb will last at least 900 hours?

c)    If we randomly select 8 light bulbs, what is the probability that the average of these light bulbs will last more than 1050 hours?


Expert's answer

Let X=X= the lifetime of light bulb in hours: X∼N(μ,σ2)X\sim N(\mu, \sigma^2)

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

Given μ=1000 h,σ=130 h\mu=1000\ h, \sigma=130\ h

a)

P(X≥x1)=0.3P(X\geq x_1)=0.3

P(X≥x1)=1−P(X<x1)P(X\geq x_1)=1-P(X<x_1)

=1−P(Z<x1−1000130)=0.3=1-P(Z<\dfrac{x_1-1000}{130})=0.3

P(Z<x1−1000130)=0.7P(Z<\dfrac{x_1-1000}{130})=0.7

x1−1000130≈0.524401\dfrac{x_1-1000}{130}\approx0.524401

x1=1000+130(0.524401)=1068x_1=1000+130(0.524401)=1068

The top 30% of all light bulbs should last at least 1068 hours.


b)

P(X≥900)=1−P(X<900)P(X\geq 900)=1-P(X<900)

=1−P(Z<900−1000130)=1-P(Z<\dfrac{900-1000}{130})

≈1−P(Z<−0.76923077)\approx1-P(Z<-0.76923077)

≈1−0.220878=0.779122\approx1-0.220878=0.779122

0.7791220.779122


c) Let Y=Y= the average lifetime of these bulbs in hours: Y∼N(μ,σ2/n)Y\sim N(\mu, \sigma^2/n)

Then Z=Y−μσ/n∼N(0,1)Z=\dfrac{Y-\mu}{\sigma/\sqrt{n}}\sim N(0, 1)

Given μ=1000 h,σ=130 h,n=10\mu=1000\ h, \sigma=130\ h, n=10


P(Y>1050)=1−P(Y≤1050)P(Y>1050)=1-P(Y\leq1050)

=1−P(Z≤1050−1000130/10)=1-P(Z\leq\dfrac{1050-1000}{130/\sqrt{10}})

≈1−P(Z≤1.21626064)\approx1-P(Z\leq1.21626064)

≈1−0.888057=0.111943\approx1-0.888057=0.111943

0.1119430.111943



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