How Do You Calculate the Probability in an Exponential Distribution?

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SUMMARY

The discussion focuses on calculating the probability of a random variable Y with an exponential distribution, specifically EXP(2). The probability P(Y > 1) is derived using the cumulative distribution function (CDF), which is defined as P[X ≤ x] = 1 - e^(-λx). By substituting λ = 2 and x = 1 into the CDF, the calculation yields P(Y > 1) = 1 - (1 - e^(-2*1)), resulting in a final probability of approximately 0.1353.

PREREQUISITES
  • Understanding of exponential distributions
  • Familiarity with cumulative distribution functions (CDF)
  • Knowledge of probability calculations
  • Basic calculus for evaluating exponential functions
NEXT STEPS
  • Study the properties of exponential distributions in depth
  • Learn how to derive probabilities using the probability density function (PDF)
  • Explore applications of the exponential distribution in real-world scenarios
  • Investigate other types of distributions and their CDFs
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This discussion is beneficial for statisticians, data analysts, and students studying probability theory, particularly those focusing on exponential distributions and their applications in statistical modeling.

das1
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Help?

Suppose the random variable Y has an EXP(2) distribution. What is P(Y > 1)? (Round to four decimal places as appropriate.)
 
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das said:
Help?

Suppose the random variable Y has an EXP(2) distribution. What is P(Y > 1)? (Round to four decimal places as appropriate.)

Are you allowed to use the CDF for this distribution or should you calculate this purely from the pdf? Either way you'll need to also use this fact: $$P[Y>2]=1-P[Y \le 2]$$
 
I don't think there are any restrictions on what functions I can or can't use
 
das said:
I don't think there are any restrictions on what functions I can or can't use

Ok, then this should be very useful. For the exponential distribution, the CDF is the following: $$P[X \le x]=1-e^{-\lambda x}$$. How can you use this to answer your question?
 
So would I plug in 1 for x and 2 for λ?
Then we would get 1-e^(-2*1)? Which is .8467 but we'd actually want 1-.8467 again because we're looking for P(Y > 1) right? So about .1353?
 
That looks good to me! (Yes)
 

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