Question

Based on an analysis of sample data, an article proposed the pdf f(x) = 0.65e−0.65(x − 1) when x ≥ 1 as a model for the distribution of X = time (sec) spent at the median line. (Round your answers to three decimal places.) (1) What is the probability that waiting time is at most 9 sec? More than 9 sec? at most 9 sec P(X ≤ 9) = more than 9 sec P(X > 9) =

Answer 1

To solve for probabilities associated with an exponential distribution, you integrate the pdf for a given range to find P(X ≤ x) or subtract this value from 1 to find P(X > x). The memoryless property of exponential distribution simplifies conditional probabilities such as P(X > 7|X > 4) to P(X > 3).

Step-by-step explanation:

Understanding Exponential Distribution

The question is related to the exponential distribution which is a continuous probability distribution used to model the time between events in a process in which events occur continuously and independently at a constant average rate.

To find the probability that the waiting time is at most 9 seconds for the provided probability density function (pdf) f(x) = 0.65e−0.65(x − 1) when x ≥ 1, we need to integrate the pdf from 1 to 9. This gives the cumulative distribution function (CDF), which will yield P(X ≤ 9). To find P(X > 9), we use the fact that the total probability is 1, so P(X > 9) = 1 - P(X ≤ 9).

For the case where the cumulative distribution function is P(X < x) = 1 -e−0.25x, and we wish to find P(X > 7|X > 4), we can use the memoryless property of the exponential distribution. The memoryless property simplifies this to P(X > 3), and by substituting 3 for x in the CDF, we get P(X > 3) = e−0.75.