Math Problem Statement

The amount of time passing between successive customer check-ins to a downtown Calgary hotel can be modeled by an exponential distribution with a mean of 12 minutes.

Solution

Let's go through the questions step-by-step based on the exponential distribution model, which has a mean of 12 minutes.

Question (a)

For an exponential distribution with mean μ=12\mu = 12 minutes:

  • The exponential distribution is always right-skewed because it models time between events, where events are memoryless.
  • The mean μ\mu of an exponential distribution is equal to the standard deviation σ\sigma.

Thus, Option A is correct:

  • "The distribution is right-skewed with a mean of 12 minutes and a standard deviation of 12 minutes."

Question (b)

We are asked to compute probabilities for different time intervals. Given the exponential distribution with mean μ=12\mu = 12, the rate λ=1μ=112\lambda = \frac{1}{\mu} = \frac{1}{12}.

The probability density function (PDF) of an exponential distribution is: f(x)=λeλxf(x) = \lambda e^{-\lambda x} And the cumulative distribution function (CDF) is: F(x)=1eλxF(x) = 1 - e^{-\lambda x}

Using these formulas, let's calculate each probability:

(i) Probability that time between check-ins is less than 7 minutes:

P(X<7)=F(7)=1e1127P(X < 7) = F(7) = 1 - e^{-\frac{1}{12} \cdot 7}

(ii) Probability that time between check-ins is between 9 and 20 minutes:

P(9X20)=F(20)F(9)=(1e11220)(1e1129)P(9 \leq X \leq 20) = F(20) - F(9) = \left(1 - e^{-\frac{1}{12} \cdot 20}\right) - \left(1 - e^{-\frac{1}{12} \cdot 9}\right)

(iii) Probability that time between check-ins is more than 26 minutes:

P(X>26)=1F(26)=e11226P(X > 26) = 1 - F(26) = e^{-\frac{1}{12} \cdot 26}

Question (c)

This question involves the memoryless property of the exponential distribution. Given that 14 minutes have passed, the probability that at least another 7 minutes will pass is the same as the probability that the time between check-ins is more than 7 minutes:

P(X>7)=e1127P(X > 7) = e^{-\frac{1}{12} \cdot 7}

Question (d)

To find the time tt such that 92% of check-ins occur within tt minutes, we need to solve for tt in the equation: F(t)=0.92F(t) = 0.92 1e112t=0.921 - e^{-\frac{1}{12} \cdot t} = 0.92 e112t=0.08e^{-\frac{1}{12} \cdot t} = 0.08 112t=ln(0.08)-\frac{1}{12} \cdot t = \ln(0.08) t=12ln(0.08)t = -12 \cdot \ln(0.08)

Would you like detailed calculations for each part, or is there a specific part you'd like to explore further?

Here are some related questions you may find helpful:

  1. What does the memoryless property imply for exponential distributions?
  2. How does the exponential distribution differ from the normal distribution in terms of skewness?
  3. How do you interpret the rate λ\lambda in an exponential distribution?
  4. What are common applications of exponential distributions outside of wait times?
  5. How can you derive the probability that an event will happen within a given interval in an exponential distribution?

Tip: For exponential distributions, remember that the mean is always equal to the standard deviation, which makes it unique among probability distributions.

Ask a new question for Free

By Image

Drop file here or Click Here to upload

Math Problem Analysis

Mathematical Concepts

Probability
Exponential Distribution
Skewness
Memoryless Property

Formulas

Exponential distribution probability density function: f(x) = λ e^(-λx)
Exponential distribution cumulative distribution function: F(x) = 1 - e^(-λx)
Mean of exponential distribution: μ = 1/λ

Theorems

Memoryless Property of Exponential Distribution

Suitable Grade Level

Undergraduate level