Based on an analysis of sample data, an article proposed the pdf

*f*(*x*) = 0.35*e*−0.35(*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.)

(a) What is the probability that waiting time is at most 6 sec? More than 6 sec?

at most 6 sec

*P*(*X* ≤ 6)

=

more than 6 sec

*P*(*X* > 6)

=

(b) What is the probability that waiting time is between 4 and 7 sec?

The given probability density function (pdf) is:

\[
f(x) = 0.35 e^{-0.35(x - 1)}, \quad x \geq 1
\]

This is an exponential distribution shifted by 1 second with a rate parameter \( \lambda = 0.35 \).

### (a) Probability that waiting time is at most 6 sec, and more than 6 sec

To find the probability that \( X \leq 6 \), we need to calculate the cumulative distribution function (CDF) of the given pdf. The CDF for an exponential distribution with rate \( \lambda \) is given by:

\[
F(x) = 1 - e^{-\lambda (x - 1)}
\]

where \( \lambda = 0.35 \) and \( x \geq 1 \).

#### At most 6 seconds:
\[
P(X \leq 6) = F(6) = 1 - e^{-0.35(6 - 1)} = 1 - e^{-0.35(5)}
\]
\[
P(X \leq 6) = 1 - e^{-1.75}
\]
Using a calculator:
\[
P(X \leq 6) = 1 - 0.17377 = 0.82623
\]
So,
\[
P(X \leq 6) \approx 0.826
\]

#### More than 6 seconds:
\[
P(X > 6) = 1 - P(X \leq 6) = 1 - 0.82623
\]
\[
P(X > 6) = 0.17377
\]
Thus,
\[
P(X > 6) \approx 0.174
\]

### (b) Probability that waiting time is between 4 and 7 sec

To find the probability that \( X \) is between 4 and 7 seconds, we calculate the difference between the CDF values at 7 and 4.

\[
P(4 \leq X \leq 7) = F(7) - F(4)
\]

First, calculate \( F(7) \):
\[
F(7) = 1 - e^{-0.35(7 - 1)} = 1 - e^{-0.35(6)} = 1 - e^{-2.1}
\]
Using a calculator:
\[
F(7) = 1 - 0.12246 = 0.87754
\]

Now, calculate \( F(4) \):
\[
F(4) = 1 - e^{-0.35(4 - 1)} = 1 - e^{-0.35(3)} = 1 - e^{-1.05}
\]
Using a calculator:
\[
F(4) = 1 - 0.34994 = 0.65006
\]

Finally, subtract:
\[
P(4 \leq X \leq 7) = 0.87754 - 0.65006 = 0.22748
\]
Thus,
\[
P(4 \leq X \leq 7) \approx 0.227
\]

### Final Answers:
- \( P(X \leq 6) \approx 0.826 \)
- \( P(X > 6) \approx 0.174 \)
- \( P(4 \leq X \leq 7) \approx 0.227 \)

Would you like more details or have any questions?

Here are 5 related questions to expand this topic:
1. What is the mean waiting time for this distribution?
2. How do you calculate the median of an exponential distribution?
3. What is the standard deviation of this distribution?
4. What is the probability that waiting time exceeds 10 seconds?
5. How would the distribution change if the rate parameter were doubled?

**Tip:** For exponential distributions, the rate parameter \( \lambda \) is the inverse of the mean waiting time.

Based on an analysis of sample data, an article proposed the pdf f(x) = 0.35e^−0.35(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.) (a) What is the probability that waiting time is at most 6 sec? More than 6 sec? (b) What is the probability that waiting time is between 4 and 7 sec?

Calculate the probabilities for waiting times using the exponential distribution with PDF f(x) = 0.35e^−0.35(x − 1), including probabilities at most 6 sec, more than 6 sec, and between 4 and 7 sec. This example covers exponential distribution theory and key probability calculations.

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