Question #278140

A battery lasts, average 4 years with a standard deviation of 1.4 years.the battery life is normally distributed

1
Expert's answer
2021-12-13T08:28:25-0500

Let X=X=the lifetime of a battery: XN(μ,σ2)X\sim N(\mu, \sigma^2)

Given μ=4,σ=1.4\mu=4, \sigma=1.4

a)


P(X<3)=P(Z<3μσ)=P(Z<341.4)P(X<3)=P(Z<\dfrac{3-\mu}{\sigma})=P(Z<\dfrac{3-4}{1.4})

P(Z<0.7143)0.2375\approx P(Z<-0.7143)\approx0.2375

b)


P(1.8<X<5)=P(X<5)P(X1.8)P(1.8<X<5)=P(X<5)-P(X\leq1.8)

=P(Z<541.4)P(Z1.841.4)=P(Z<\dfrac{5-4}{1.4})-P(Z\leq\dfrac{1.8-4}{1.4})

P(Z<0.7143)P(Z1.5714)\approx P(Z<0.7143)-P(Z\leq -1.5714)

0.762480.058040.7044\approx0.76248-0.05804\approx0.7044


c)


P(X<x)=P(Z<x41.4)=0.25P(X<x)=P(Z<\dfrac{x-4}{1.4})=0.25

x41.40.6745\dfrac{x-4}{1.4}\approx-0.6745

x41.4(0.6745)x\approx4-1.4(0.6745)

x3.0557x\approx3.0557


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