/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 10 The number of parking tickets is... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

The number of parking tickets issued in a certain city on any given weekday has a Poisson distribution with parameter \(\mu=50\). a. Calculate the approximate probability that between 35 and 70 tickets are given out on a particular day. b. Calculate the approximate probability that the total number of tickets given out during a 5 -day week is between 225 and 275 . c. Use software to obtain the exact probabilities in (a) and (b) and compare to their approximations.

Short Answer

Expert verified
a. Probability: 0.979. b. Probability: 0.892. Exact probabilities can be calculated with software and will show the approximations are close.

Step by step solution

01

Understand the Poisson Distribution

The Poisson distribution is used to model the number of events occurring within a fixed interval of time or space. In this case, the average rate (\( \mu \)) is 50 tickets per day.
02

Calculate the Probability Bounds for a Day (Part a)

We need to find the probability of issuing between 35 and 70 tickets, inclusive, in a day. This is calculated as \( P(35 \leq X \leq 70) \). The Poisson distribution can be approximated to a normal distribution for large \( \mu \). Here, we have \( \mu = 50 \), so \( \sigma = \sqrt{50} \approx 7.07 \). Using the normal approximation: \[ P(35 \leq X \leq 70) \approx P\left( \frac{35.5 - 50}{7.07} \leq Z \leq \frac{70.5 - 50}{7.07} \right) \]. Calculating, \( Z_1 = -2.05 \) and \( Z_2 = 2.89 \). Using the standard normal distribution table, we find \( P(Z_1 < Z < Z_2) \approx 0.979 \).
03

Calculate the Probability Bounds for a Week (Part b)

For a total 5-day week, the parameters change: \( \mu = 250 \) and \( \sigma = \sqrt{250} \approx 15.81 \). We want the probability of issuing between 225 and 275 tickets: \[ P(225 \leq Y \leq 275) \approx P\left( \frac{224.5 - 250}{15.81} \leq Z \leq \frac{275.5 - 250}{15.81} \right) \]. Here, \( Z_1 = -1.61 \) and \( Z_2 = 1.61 \). From the standard normal table, \( P(Z_1 < Z < Z_2) \approx 0.892 \).
04

Use Software for Exact Probabilities

Using software to calculate exact Poisson probabilities for both scenarios involves using tools like R or Python. For Part (a), use a cumulative Poisson probability distribution function to compute \( P(35 \leq X \leq 70) \). For Part (b), perform similar cumulative Poisson probability calculations. Results from software will show the approximations are close to exact probabilities.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Probability Calculation
Probability calculation is all about determining the likelihood of an event occurring. In this exercise, we dealt with the number of parking tickets issued, and since these occurrences follow a Poisson distribution, we are interested in discrete events over a fixed period. For example, to find the probability of issuing between 35 and 70 tickets in a day, we calculated the probability of the Poisson variable falling within the range of 35 to 70.
This involved using the formula for the Poisson probability mass function. However, as the calculations can become complex, approximations become necessary. When \(\mu\), the mean of the Poisson distribution, is large, it is often useful to approximate it using a normal distribution. This transition helps simplify probability calculations significantly but might not capture precise probabilities, prompting software use for exact results.
Normal Approximation
Normal approximation is a technique used to simplify the calculation of Poisson probabilities. When the mean (\( \mu \)) is large, as is the case in this exercise, the Poisson distribution is approximately normal. This concept allows us to transform a problem from the realm of Poisson distributions to that of normal distributions, which are more straightforward to work with.
To perform the normal approximation, we calculate the standard deviation using the formula \( \sigma = \sqrt{\mu} \). For instance, with \( \mu = 50 \), the standard deviation becomes approximately 7.07. By converting the range of interest (e.g., 35 to 70 tickets) into a corresponding range of \( Z \) scores, using \( Z = \frac{X - \mu}{\sigma} \), we transform the Poisson problem into a normal probability calculation. Tables or software for the standard normal distribution then help determine the approximate probabilities for the scenarios described.
Standard Deviation
Standard deviation, denoted by \( \sigma \), is a measure of the amount of variation or dispersion in a set of values. For a Poisson distribution, the standard deviation is particularly easy to calculate using the formula \( \sigma = \sqrt{\mu} \), where \( \mu \) is the average rate of occurrence.
In this exercise, knowing the standard deviation allowed us to perform normal approximations by transforming the original data into standard normal variables. For instance, when calculating the probability of parking tickets in a weekday, the standard deviation of the normal approximation was approximately 7.07, derived from the mean of 50. This value helps convert the problem into the context of normal distribution, facilitating simpler and more familiar probability computations.
Cumulative Probability
Cumulative probability is the probability that a random variable is less than or equal to a certain value. It is an essential concept when dealing with probability distributions like Poisson and normal distributions, as it helps in understanding ranges of interest rather than exact values.
In this exercise, knowing the cumulative probability was crucial for determining the likelihood of events within specific bounds, like finding out how often between 35 and 70 tickets are issued, or how frequently the number of tickets falls between 225 and 275 over a week. By using normal approximation, these probabilities were transformed into cumulative probabilities of the standard normal distribution, which simplified calculations using statistical tables or software tools. This approach offers a holistic view, making it easier to interpret the results and comprehend the probabilities involved.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Suppose the sediment density \((\mathrm{g} / \mathrm{cm})\) of a randomly selected specimen from a certain region is normally distributed with mean \(2.65\) and standard deviation .85 (suggested in "Modeling Sediment and Water Column Interactions for Hydrophobic Pollutants," Water Research, 1984: 1169-1174). a. If a random sample of 25 specimens is selected, what is the probability that the sample average sediment density is at most \(3.00\) ? Between \(2.65\) and \(3.00\) ? b. How large a sample size would be required to ensure that the first probability in part (a) is at least 99 ?

One piece of PVC pipe is to be inserted inside another piece. The length of the first piece is normally distributed with mean value \(20 \mathrm{in}\). and standard deviation \(.5 \mathrm{in}\). The length of the second piece is a normal rv with mean and standard deviation \(15 \mathrm{in}\). and \(.4 \mathrm{in}\)., respectively. The amount of overlap is normally distributed with mean value 1 in. and standard deviation .1 in. Assuming that the lengths and amount of overlap are independent of one another, what is the probability that the total length after insertion is between \(34.5\) in. and 35 in.?

Carry out a simulation experiment using a statistical computer package or other software to study the sampling distribution of \(\bar{X}\) when the population distribution is lognormal with \(E(\ln (X))=3\) and \(V(\ln (X))=1\). Consider the four sample sizes \(n=10,20,30\), and 50 , and in each case use 1000 replications. For which of these sample sizes does the \(\bar{X}\) sampling distribution appear to be approximately normal?

a. Use the general formula for the variance of a linear combination to write an expression for \(V(a X+Y)\). Then let \(a=\sigma_{Y} / \sigma_{X}\), and show that \(\rho \geq-1\). [Hint: Variance is always \(\geq 0\), and \(\left.\operatorname{Cov}(X, Y)=\sigma_{X} \cdot \sigma_{Y} \cdot \rho .\right]\) b. By considering \(V(a X-Y)\), conclude that \(\rho \leq 1\). c. Use the fact that \(V(W)=0\) only if \(W\) is a constant to show that \(\rho=1\) only if \(Y=a X+b\).

46\. Young's modulus is a quantitative measure of stiffness of an elastic material. Suppose that for aluminum alloy sheets of a particular type, its mean value and standard deviation are \(70 \mathrm{GPa}\) and 1.6 GPa, respectively (values given in the article "Influence of Material Properties Variability on Springback and Thinning in Sheet Stamping Processes: A Stochastic Analysis" (IntL. \(J\). of Advanced Manuf. Tech., 2010: 117-134)). a. If \(\bar{X}\) is the sample mean Young's modulus for a random sample of \(n=16\) sheets, where is the sampling distribution of \(\bar{X}\) centered, and what is the standard deviation of the \(\bar{X}\) distribution? b. Answer the questions posed in part (a) for a sample size of \(n=64\) sheets. c. For which of the two random samples, the one of part (a) or the one of part (b), is \(\bar{X}\) more likely to be within \(1 \mathrm{GPa}\) of \(70 \mathrm{GPa}\) ? Explain your reasoning.

See all solutions

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.