Let X = X= X = the lifetime of light bulb in hours: X ∼ N ( μ , σ 2 ) X\sim N(\mu, \sigma^2) X ∼ N ( μ , σ 2 )
Then Z = X − μ σ ∼ N ( 0 , 1 ) Z=\dfrac{X-\mu}{\sigma}\sim N(0, 1) Z = σ X − μ ∼ N ( 0 , 1 )
Given μ = 1000 h , σ = 130 h \mu=1000\ h, \sigma=130\ h μ = 1000 h , σ = 130 h
a)
P ( X ≥ x 1 ) = 0.3 P(X\geq x_1)=0.3 P ( X ≥ x 1 ) = 0.3
P ( X ≥ x 1 ) = 1 − P ( X < x 1 ) P(X\geq x_1)=1-P(X<x_1) P ( X ≥ x 1 ) = 1 − P ( X < x 1 )
= 1 − P ( Z < x 1 − 1000 130 ) = 0.3 =1-P(Z<\dfrac{x_1-1000}{130})=0.3 = 1 − P ( Z < 130 x 1 − 1000 ) = 0.3
P ( Z < x 1 − 1000 130 ) = 0.7 P(Z<\dfrac{x_1-1000}{130})=0.7 P ( Z < 130 x 1 − 1000 ) = 0.7
x 1 − 1000 130 ≈ 0.524401 \dfrac{x_1-1000}{130}\approx0.524401 130 x 1 − 1000 ≈ 0.524401
x 1 = 1000 + 130 ( 0.524401 ) = 1068 x_1=1000+130(0.524401)=1068 x 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) P ( X ≥ 900 ) = 1 − P ( X < 900 )
= 1 − P ( Z < 900 − 1000 130 ) =1-P(Z<\dfrac{900-1000}{130}) = 1 − P ( Z < 130 900 − 1000 )
≈ 1 − P ( Z < − 0.76923077 ) \approx1-P(Z<-0.76923077) ≈ 1 − P ( Z < − 0.76923077 )
≈ 1 − 0.220878 = 0.779122 \approx1-0.220878=0.779122 ≈ 1 − 0.220878 = 0.779122 0.779122 0.779122 0.779122
c) Let Y = Y= Y = the average lifetime of these bulbs in hours: Y ∼ N ( μ , σ 2 / n ) Y\sim N(\mu, \sigma^2/n) Y ∼ N ( μ , σ 2 / n )
Then Z = Y − μ σ / n ∼ N ( 0 , 1 ) Z=\dfrac{Y-\mu}{\sigma/\sqrt{n}}\sim N(0, 1) Z = σ / n Y − μ ∼ N ( 0 , 1 )
Given μ = 1000 h , σ = 130 h , n = 10 \mu=1000\ h, \sigma=130\ h, n=10 μ = 1000 h , σ = 130 h , n = 10
P ( Y > 1050 ) = 1 − P ( Y ≤ 1050 ) P(Y>1050)=1-P(Y\leq1050) P ( Y > 1050 ) = 1 − P ( Y ≤ 1050 )
= 1 − P ( Z ≤ 1050 − 1000 130 / 10 ) =1-P(Z\leq\dfrac{1050-1000}{130/\sqrt{10}}) = 1 − P ( Z ≤ 130/ 10 1050 − 1000 )
≈ 1 − P ( Z ≤ 1.21626064 ) \approx1-P(Z\leq1.21626064) ≈ 1 − P ( Z ≤ 1.21626064 )
≈ 1 − 0.888057 = 0.111943 \approx1-0.888057=0.111943 ≈ 1 − 0.888057 = 0.111943 0.111943 0.111943 0.111943
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