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Consider a random variable X with moment generating function
M(t)=e (³t) ,for −[infinity]

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Final answer:

The question discusses the exponential distribution, a type of continuous random variable used to model time between random events. It characterizes the distribution using a moment generating function, mean, standard deviation, probability density function, and cumulative distribution function. The exponential distribution also has a memoryless property.

Step-by-step explanation:

The question involves understanding the exponential distribution, which is a type of continuous random variable often used in scenarios where we want to model the time between random events, such as intervals between patients arriving at a hospital.

The moment generating function provided, M(t) = e³t, is typically part of the characterization of this distribution. However, without further context, the question seems incomplete or contains a typo. Presuming the moment generating function should be M(t) = emt, this would be associated with an exponential distribution where m is the decay parameter. The mean (μ) of the random variable can be found as the inverse of m, which is μ = 1/m, and its standard deviation (σ) is also equal to 1/m. The probability density function for an exponential distribution is f(x) = me-mx, for x ≥ 0, and the cumulative distribution function is P(X ≤ x) = 1 - e-mx.

Furthermore, the exponential distribution possesses the memoryless property, meaning that the probability of an event occurring in a future interval is independent of the time that has already elapsed.

User Erick Fleming
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