Answer to Question #234050 in Statistics and Probability for Nona

Question #234050
The life of an electronic device is known to have the exponential distribution
with parameter λ =
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.
[
1
Expert's answer
2021-09-16T07:55:45-0400

"P(X>x) = e^{-\u03bbx}"

a) "P(X>1000) = exp(-\\lambda x)"

"= exp(-0.001 \\times 1000) \\\\\n\n= 0.3678"

b) "P(X<1200)= 1 - exp(-0.001 \\times 1200)"

"= 1 -0.3012 \\\\\n\n= 0.6988"

c) Mean

"\\mu = \\frac{1}{\u03bb} \\\\\n\n\\mu = \\frac{1}{0.001} = 1000"

Variance

"\\sigma^2 = \\frac{1}{\u03bb^2} \\\\\n\n\n\\sigma^2 = \\frac{1}{(0.001)^2} \\\\\n\n\n= 10^6"


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