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Typei light bulbs function for a random amount of time having meanμi and standard deviationσi,i=1,2. A light bulb randomly chosen from a bin of bulbs is a type1bulb with probabilityp and a type2bulb with probability1−p. Let X denote the lifetime of this bulb. Find

(a) E[X];

(b) Var(X).

Short Answer

Expert verified

The values of E[X]and Var(X)is

a)E[X]=pμ1+(1−p)μ2

b)Var(X)=pσ12+(1−p)σ22+p(1−p)μ1−μ22

Step by step solution

01

Step 1:Given Information(part a)

Given that Xdenotes the lifetime of the bulb.

Y=1 â¶Ä…â¶Ä…â¶Ä…type1bulb with probabilityp2 â¶Ä…â¶Ä…â¶Ä…type2bulb with probability(1−p)

role="math" localid="1647349813765" P(Y=1)=p,P(Y=2)=(1−p)

02

Step 2:Explanation(part a)

E[X]=E(X∣Y=1)⋅P(Y=1)+E(X∣Y=2)⋅P(Y=2)

=pμ1+(1−p)μ2

03

Step 3:Final Answer(part a)

TheE[X]is

E[X]=pμ1+(1−p)μ2
04

Step 4:Given Information(part b)

Given that Xas the lifetime of this bulb.

Var(X)=EX2−(E[X])2

05

Step 5:Explanation(Step 5)

EX2=EX2∣Y=1⋅P(Y=1)+EX2∣Y=2⋅P(Y=2)

=pμ12+σ12+(1−p)μ22+σ22

=pσ12+(1−p)σ22+pμ12+(1−p)μ22

Var(X)=pσ12+(1−p)σ22+pμ12+(1−p)μ22−pμ1+(1−p)μ22

=pσ12+(1−p)σ22+p(1−p)μ12+μ22+2μ1μ2

=pσ12+(1−p)σ22+p(1−p)μ1−μ22

06

Step 6:Final Answer(part b)

TheVar(X)is

Var(X)=pσ12+(1−p)σ22+p(1−p)μ1−μ22

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