Let X be a random variable representing the length of time (in years) that a lap

a) P(Y 3) = (e^(-0.3*3))^2 = 0.1652988 Please rate my solution

ID: 3275286 • Letter: L

Question

Let X be a random variable representing the length of time (in years) that a laptop lasts.

From past experience, I nd X is well-modeled by an Exponential(0:3) random variable. That means the CDF of X is:

F(y) = { 0 when y < 0, 1-e^(-0.3y) when y >= 0}

(a) What is the probability a laptop lasts less than three years?

(b) What is the probability a laptop lasts between two years and four years?

(d) Find the variance and standard deviation of X. (Again, for this and the previous part, it's easier to use formulae than calculus.)

(e) Suppose I buy two laptops. What is the probability that both laptops last more than three years?

a) P(Y < 3) = F(3) = 1 -e^(-0.3*3)

= 0.5934303

b) P(2 <Y< 4) = F(4) - F(2) = e^(-2) - e^(-4) = 0.11701964

c)E(X) = 1/0.3 = 3.333

d) variance = 1/0.3^2 = 100/9

sd = 1/0.3 = 3.333

e) P(X1 > 3 , X2> 3) = (e^(-0.3*3))^2

= 0.1652988

Please rate my solution

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