https://file.aiquickdraw.com/mathbot/file/1729163743698nkgarq31.jpg

To solve the problem presented in the image, we first need to understand that it is based on the exponential distribution. The waiting time \( X \) follows an exponential distribution with a mean waiting time of 41 minutes. The probability density function (PDF) of an exponential distribution is given by:

\[
f(x; \lambda) = \lambda e^{-\lambda x} \text{ for } x \geq 0
\]

Where \( \lambda = \frac{1}{\text{mean}} = \frac{1}{41} \).

Now, let's break down the problem into parts:

---

### (a) Provide the standard deviation of this distribution.

For an exponential distribution, the mean and the standard deviation are both equal to \( \frac{1}{\lambda} \). Since the mean waiting time is 41 minutes:

\[
SD(X) = 41 \text{ minutes}
\]

---

### (b) Compute the probability that you will have to wait between 28 and 38 minutes to get in.

We need to calculate \( P(28 \leq X \leq 38) \). The cumulative distribution function (CDF) for an exponential distribution is:

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

Thus, the probability can be calculated as:

\[
P(28 \leq X \leq 38) = F(38) - F(28)
\]

Substitute \( \lambda = \frac{1}{41} \) into the CDF formula:

\[
F(38) = 1 - e^{-\frac{38}{41}}, \quad F(28) = 1 - e^{-\frac{28}{41}}
\]

Now, calculate:

\[
P(28 \leq X \leq 38) = \left(1 - e^{-\frac{38}{41}}\right) - \left(1 - e^{-\frac{28}{41}}\right)
\]

\[
P(28 \leq X \leq 38) = e^{-\frac{28}{41}} - e^{-\frac{38}{41}}
\]

After evaluating the exponents and subtracting, the final probability should be calculated (you can use a calculator or software for accurate results).

---

### (c) What is the chance that you will have to wait at most 53 minutes in total, given that you have already waited 30 minutes?

This is a conditional probability, so we use the formula for the conditional exponential distribution:

\[
P(X \leq 53 \mid X > 30) = \frac{P(30 < X \leq 53)}{P(X > 30)}
\]

Where \( P(X > 30) = 1 - F(30) = e^{-\frac{30}{41}} \), and \( P(30 < X \leq 53) = F(53) - F(30) \).

\[
F(53) = 1 - e^{-\frac{53}{41}}, \quad F(30) = 1 - e^{-\frac{30}{41}}
\]

Now calculate the conditional probability:

\[
P(X \leq 53 \mid X > 30) = \frac{\left(1 - e^{-\frac{53}{41}}\right) - \left(1 - e^{-\frac{30}{41}}\right)}{e^{-\frac{30}{41}}}
\]

Simplify and evaluate the exponents.

---

### (d) 50% of the time, you will have to wait at most how many minutes?

For the 50th percentile, we need to find \( x \) such that \( F(x) = 0.5 \). From the CDF of the exponential distribution:

\[
F(x) = 1 - e^{-\frac{x}{41}} = 0.5
\]

Solving for \( x \):

\[
e^{-\frac{x}{41}} = 0.5 \quad \Rightarrow \quad -\frac{x}{41} = \ln(0.5)
\]

\[
x = -41 \ln(0.5)
\]

Now, calculate \( x \).

---

If you'd like any further explanation or need help with the calculations, let me know!

After 8:00 pm on each Thursday, the amount of time that a person spends waiting in line to get into a well-known pub is a random variable represented by X. Suppose we can model the behavior of X with the exponential probability distribution with an average waiting time of 41 minutes. (a) Provide the standard deviation of this distribution. (b) Suppose you are in line to get into the pub. Compute the probability that you will have to wait between 28 and 38 minutes to get in. (c) It has been at least 30 minutes since you entered the lineup to get into the pub, and you are still waiting. What is the chance that you will have to wait at most 53 minutes in total? (d) 50% of the time, you will have to wait at most how many minutes to get into this pub?

Solve a statistical problem involving an exponential distribution for pub waiting times with an average of 41 minutes. Learn how to calculate the standard deviation, probabilities for waiting intervals, and conditional probabilities based on real-world scenarios.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation