Life Span of a Computer Part The life (in months) of a...

Life Span of a Computer Part The life (in months) of a certain electronic computer part has a probability density function defined by $$f(t)=\frac{1}{2} e^{-t / 2} \text { for } t \text { in }[0, \infty) \text { . }$$ Find the probability that a randomly selected component will last the following lengths of time. a. At most 12 months b. Between 12 and 20 months c. Find the cumulative distribution function for this random variable. d. Use the answer to part c to find the probability that a randomly selected component will last at most 6 months.

Life Span of a Computer Part The life (in months) of a certain electronic computer part has a probability density function defined by
$$f(t)=\frac{1}{2} e^{-t / 2} \text { for } t \text { in }[0, \infty) \text { . }$$
Find the probability that a randomly selected component will last the following lengths of time.
a. At most 12 months
b. Between 12 and 20 months
c. Find the cumulative distribution function for this random variable.
d. Use the answer to part c to find the probability that a randomly selected component will last at most 6 months.

Key Concepts

Cumulative Distribution Function (CDF)

The cumulative distribution function represents the probability that a continuous random variable is less than or equal to a given value. It is obtained by integrating the probability density function up to that value. The CDF is essential for determining the probability that a random variable falls within a specific interval and for understanding the overall probability distribution.

Integration in Probability

Integration is a key mathematical tool in probability for calculating the total probability over a continuous range. It is used to determine probabilities from the pdf by summing up infinitesimal contributions over an interval. This method is crucial for working with continuous probability distributions like the exponential distribution, particularly when finding probabilities over specified time periods.

Exponential Distribution

The exponential distribution is a continuous probability distribution commonly used to model the time between independent events occurring at a constant rate. It is characterized by its probability density function, which for a rate parameter ? is given by f(t) = ?e^(-?t) for t ? 0. This distribution is widely used in reliability and life data analysis, making it ideal for studying the lifespan of components.

Probability Density Function (pdf)

The probability density function is a fundamental concept in probability theory that describes the likelihood of a random variable taking on specific values. For continuous distributions, the pdf must integrate to one over the support of the random variable. In the context of this problem, the pdf of the exponential distribution forms the basis for calculating probabilities over defined time intervals.

Transcript

00:01 For the first part of this question, we want to find the probability that a randomly selected component will last at most 12 months.

00:12 So at most 12 months means it can last.

00:17 We want to find the probability that t basically is not greater than 12.

00:24 So that's the probability that t would be between 0 and 12.

00:29 So to do that, we're going to set up our integral from 0 to 12, 1 over 2, e to negative.

00:35 T over 2 d t and we're going to see that this is negative e to the negative t over 2 evaluated from 0 to 12 plugging into upper bound we would get negative e to the negative 6 playing in the lower bound we would get plus 1 so rewriting this a little cleaner we would have 1 minus e to the negative 6 for the next part, we want to find the probability that it has a lifetime between 12 and 20 months.

01:23 So that's going to be basically the same problem, except just changing our bounds of integration.

01:33 So we're going to get this exact thing here.

01:36 And our bounds then, plighting me in the upper bound of 20, we would get negative e to the negative 10.

01:43 And the lower bound of 12, we would get plus e to the negative 6.

01:49 And again, i would rewrite this to be e to the negative 6 minus e to the negative 10.

01:57 And the last part, well, second to the last part, is to find the cumulative distribution function for this random variable.

02:09 So we're already part of the way there just by doing this integration, right? if we want to find the cdf, we want to do the integration of the pdf with no bounds on our integral.

02:23 And so that's going to give us negative e to the negative t over two plus c and the value for this constant that we want is going to give us either a one when we plug in the right endpoint or a zero when we plug in the left endpoint so let's start by plugging in the left endpoint here if we plug in zero to our function we're going to get to the 0 negative negative to the negative 0 which is just going to be negative 1 plus c so you get negative 1 plus c and we said that we want at the left end point 0 to get a value of actually 0 so that means our c has to equal 1 and so if we go ahead and plug in the c into this function here we would get the that f of t is equal to 1 minus e to negative t over 2...

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