The lifetime of a lightbulb can be modeled with an

Ответ:

ronniethefun

10.01.2022

a) 0.13533

b) 0.36788

Step-by-step explanation:

Let X be the random variable that measures the lifetime of a bulb.

If the random variable X is exponentially distributed and X has an average value of 1000 days, then its probability density function is

$\bf f(x)=\frac{1}{1000}e^{-x/1000}\;(x\geq 0)$

and its cumulative distribution function (CDF) is

$\bf P(X\leq t)=\int_{0}^{t} f(x)dx=1-e^{-t/1000}$

(a) Find the probability that the light bulb will function for more than 2000 days

$\large P(X2000)=1-P(X\leq 2000)=\\\\=1-(1-e^{-2000/1000})=e^{-2}=0.135335$

(b) Find the probability that the light bulb will function for more than 2000 days, given that it is still functional after 500 days

Let A and B be the events,

A = “The bulb will last at least 2000 days”

B = “The bulb will last at least 500 days”

We want to find P(A | B).

By definition P(A | B) = P(A∩B)P(B)

but B⊂A, so A∩B = B and

$\large P(A | B) = P(B)P(B) = (P(B))^2$

We have

$\large P(B)=P(X \geq 500)=1-P(X\leq 500)=1-(1-e^{-500/1000})=e^{-0.5}=0.60653$

hence,

$\large P(A | B)=(P(B))^2=(0.60653)^2=0.36788$

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