Answer to Question #235300 in Statistics and Probability for Hastings Mambwe

Question #235300

b) The life of an electronic device is known to have the exponential distribution

with parameter  = 1

1000 .

(i) What is the probability that the device lasts more than 1000 hours?

[2 marks]

(ii) What is the probability it will last less than 1200 hours?

[2 marks]

(iii) Find the mean and variance of the life of the electronic device.

[4 marks]


1
Expert's answer
2021-09-10T09:58:59-0400

Let X=X= the life of an electronic device: XPo(λ).X\sim Po(\lambda).

Given λ=1000.\lambda=1000.

For sufficiently large values of λ,λ, (say λ>1,000λ>1,000 ), the Normal(μ=λ,σ2=λ)Normal(μ = λ,σ^2 = λ) Distribution is an excellent approximation to the Poisson(λ)(λ) Distribution.

(i)


P(X>1000)=1P(X100)P(X>1000)=1-P(X\leq 100)

1P(Z1000+0.510001000)\approx1-P(Z\leq\dfrac{1000+0.5-1000}{\sqrt{1000}})

1P(Z0.01581139)0.49369243\approx1-P(Z\leq 0.01581139)\approx0.49369243

(ii)


P(X<1200)=P(Z<12000.510001000)P(X<1200)=P(Z<\dfrac{1200-0.5-1000}{\sqrt{1000}})

P(Z<6.30874393)\approx P(Z<6.30874393)

0.999999999861\approx0.99999999986\approx1

(iii)


μ=λ=1000\mu=\lambda=1000

Var(X)=σ2=λ=1000Var(X)=\sigma^2=\lambda=1000



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