Let X= the lifetime of light bulb in hours: X∼N(μ,σ2)
Then Z=σX−μ∼N(0,1)
Given μ=1000 h,σ=130 h
a)
P(X≥x1)=0.3
P(X≥x1)=1−P(X<x1)
=1−P(Z<130x1−1000)=0.3
P(Z<130x1−1000)=0.7
130x1−1000≈0.524401
x1=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)
=1−P(Z<130900−1000)
≈1−P(Z<−0.76923077)
≈1−0.220878=0.779122 0.779122
c) Let Y= the average lifetime of these bulbs in hours: Y∼N(μ,σ2/n)
Then Z=σ/nY−μ∼N(0,1)
Given μ=1000 h,σ=130 h,n=10
P(Y>1050)=1−P(Y≤1050)
=1−P(Z≤130/101050−1000)
≈1−P(Z≤1.21626064)
≈1−0.888057=0.111943 0.111943
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