29. A random variable has an exponential distribution with probability function given f(x) = \begin{cases} 3e^{-3x} & \text{for } x > 0\\ 0 & \text{for } x \le 0 \end{cases} What is the probability that x is not less than 4? Find mean and standard deviation. 30 Studies of a single-machine-tool system showed that the time the machine operates before breaking down is exponentially distributed with a mean 10 hours. (i) Find the probability that the machine operates for at least 12 hours before breaking down. (ii) If the machine has already been operating 8 hours, what is the probability that it will last another 4 hours?
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This is given by: $P(X \ge 4) = \int_4^\infty 3e^{-3x} dx$ $= [-e^{-3x}]_4^\infty$ $= -e^{-\infty} + e^{-12}$ $= e^{-12}$ Show more…
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Question 1. (Revision on probability distribution functions from last term). Let X be a random variable with the exponential distribution with parameter 1, which means that the probability density function of X is f_X(x) = { e^-x, for x ≥ 0, 0, for x < 0. (*) i) How do you find P(a ≤ X ≤ b), the probability that a value of X lies between a and b, using the density function? ii) Compute each of P(X < 1), P(1 ≤ X ≤ 2) and P(X ≥ 2). iii) Suppose you sample 1000 independent observations of the random variable X, having the distribution (*). What are the expected numbers lying in the intervals [0, 1), [1, 2) and [2, ∞)?
Adi S.
3. The lifetime of a certain type of electronic device (measured in hours) is a continuous random variable X with probability density function given by f(x) = {a/x^3, x > 9; 0, x <= 9. (a) Find a, then compute the mean and the variance of X. (b) Find P(X > 20). (c) What is the probability that of 5 such types of devices exactly 2 will function for at least 20 hours, with the assumption that all devices function independently?
Hoan N.
Exponential Random Variable (30 pts) 2. (30 pts) An electronic component is known to have a useful life represented by an exponential random variable with failure rate of 1/100 failures per hour (i.e., λ = 1/100). Let X be the time to failure of this component. Then the mean time to a failure, E(X), is 100 hours. The company producing this component earns 10 dollars per unit. However, there is a warranty clause: if the component lasts for fewer than 300 hours, the company must pay a penalty of 4 dollars to the customer. Find the expected earning. Hint 1: The component cost is given according the Expected earning = 10 – 4 = 6, X < 300, = 10, X ≥ 300, Hint 2: e ≈ 2.71828
Madhur L.
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