MHB How Do You Calculate the Probability in an Exponential Distribution?

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To calculate the probability for a random variable Y with an EXP(2) distribution, the cumulative distribution function (CDF) can be used. The CDF is defined as P[X ≤ x] = 1 - e^(-λx). For P(Y > 1), the calculation involves finding P(Y ≤ 1) first, which is 1 - e^(-2*1), resulting in approximately 0.8467. Consequently, P(Y > 1) is calculated as 1 - 0.8467, yielding a final probability of about 0.1353. This method effectively demonstrates the application of the CDF in determining probabilities in exponential distributions.
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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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