Solution to 1. The manufacturer of a certain kind of light bulb claims that the lifetime of … - Sikademy
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Peace Awoke

1. The manufacturer of a certain kind of light bulb claims that the lifetime of the bulb in hours, X can be modeled as a random quantity with PDF. fx(x)= {0, x<100 c/x2, x>=100, where c is a constant. i. What value must c take in order for this to define a valid PDF? ii. What is the probability that the bulb lasts no longer than 150 hours? iii. Given that a bulb lasts longer than 150 hours, what is the probability that it lasts longer than 200 hours? iv. Evaluate the cumulative distribution function.

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(i) Any valid PDF must integrates to 1, so

\int_{-\infty}^{+\infty} f(x)dx=c\int_{100}^{+\infty}{\frac 1 {x^2}}dx=-c*{\frac 1 x}|_{x=100}^{x=+\infty}={\frac c {100}}=1\implies c=100

(ii) P(X<150)=100\int_{100}^{150}{\frac 1 {x^2}}dx=-100*{\frac 1 x}|_{x=100}^{x=150}=-100({\frac 1 {150}}-{\frac 1 {100}})={\frac 1 3}

(iii) P(X<200|X>150)={\frac {P(150<X<200)} {P(X>150)}}

P(X>150)=1-P(X<150)={\frac 2 3}

P(150<X<200)=100\int_{150}^{200}{\frac 1 {x^2}}dx=-100*{\frac 1 x}|_{x=150}^{x=200}=-100*({\frac 1 {200}}-{\frac 1 {150}})={\frac 1 6}

So, P(X<200|X>150)={\frac {{\frac 1 6}} {\frac 2 3}}=0.25

(iv) Let F(x) be a cumulative distribution function, then F(x)=\int_{-\infty}^x f(t)dt. So,

F(x) = 0 when x < 100

F(x)=100\int_{100}^x {\frac 1 {t^2}}dt=-100*{\frac 1 t}|_{t=100}^{t=x}=1-{\frac {100} x} when x ≥ 100

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