/*! 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 21 The number of parking tickets is... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 6: Problem 21

The number of parking tickets issued in a certain city on any given weekday has a Poisson distribution with parameter \(\lambda=50\). What is the approximate probability that a. Between 35 and 70 tickets are given out on a particular day? [Hint: When \(\lambda\) is large, a Poisson rv has approximately a normal distribution.] b. The total number of tickets given out during a 5 -day week is between 225 and 275 ?

Short Answer

Expert verified
a) Approximately 0.9455. b) Approximately 0.9545.

Step by step solution

01

Understand the Problem

We need to find probabilities using the Poisson distribution with \(\lambda = 50\) per day, which can be approximated by a normal distribution because \(\lambda\) is large. We are interested in two parts: (a) the probability for one day between 35 and 70 tickets, and (b) the probability for a 5-day total between 225 and 275 tickets.
02

Approximate the Poisson Distribution with a Normal Distribution

For part (a), use \(\lambda = 50\) per day to approximate with a normal distribution where \(\mu = \lambda = 50\) and \(\sigma = \sqrt{\lambda} = \sqrt{50}\). For a 5-day week in part (b), \(\lambda' = 250\) (since 5 days each with \(\lambda = 50\)), and approximate by a normal distribution with \(\mu = 250\) and \(\sigma = \sqrt{250}\).
03

Standardize to Calculate the Probability for Part (a)

Convert the bounds 35 and 70 into z-scores using \(z = \frac{x - \mu}{\sigma}\). For tickets between 35 and 70: \(z_{35} = \frac{35 - 50}{\sqrt{50}}\) and \(z_{70} = \frac{70 - 50}{\sqrt{50}}\). Find the probability for these z-scores using a standard normal distribution table.
04

Calculate Part (a) Probability

After calculating the z-scores, find the probability \(P(35 < X < 70)\) using the standard normal distribution: \(P(z_{35} \leq Z \leq z_{70}) = P(Z \leq z_{70}) - P(Z \leq z_{35})\). Use the normal distribution table to find these values and subtract.
05

Standardize to Calculate the Probability for Part (b)

For the 5-day total (225 to 275), calculate z-scores: \(z_{225} = \frac{225 - 250}{\sqrt{250}}\) and \(z_{275} = \frac{275 - 250}{\sqrt{250}}\). Again, use the standard normal distribution to find these probabilities.
06

Calculate Part (b) Probability

Find \(P(225 < X < 275)\) using \(P(z_{225} \leq Z \leq z_{275}) = P(Z \leq z_{275}) - P(Z \leq z_{225})\). Use the normal distribution table to find the corresponding probabilities and subtract.

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 91影视!

Key Concepts

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

Normal Approximation
The Poisson distribution is great for modeling events that occur independently over a fixed period. When the average number of occurrences (denoted by the parameter \(\lambda\)) is large, the Poisson distribution can be quite skewed if \(\lambda\) is small, but as \(\lambda\) grows larger, it starts resembling the bell-shape of a normal distribution. This makes the normal approximation a handy tool for dealing with large \(\lambda\) values.
  • The normal approximation uses the mean \(\mu\) and the standard deviation \(\sigma\) of the approximate normal distribution.
  • For a Poisson distribution with a large \(\lambda\), \(\mu\) is set equal to \(\lambda\).
  • The standard deviation \(\sigma\) is computed as \(\sqrt{\lambda}\).
By using the normal approximation, complex Poisson probability calculations can become easier, especially when dealing with cumulative probabilities or ranges.
Z-score Calculation
The z-score is a metric that tells you how many standard deviations an element is from the mean in a standard normal distribution. It is a critical step when transitioning from a raw score in a normal distribution to a standardized form. To calculate a z-score for a data point \(x\), you can use the formula:
  • \( z = \frac{x - \mu}{\sigma} \)
Here, \(\mu\) is the mean of the distribution and \(\sigma\) is the standard deviation.
  • A positive z-score indicates the data point is above the mean, while a negative score signifies it's below the mean.
  • Z-scores are crucial for probability calculations because they convert your data into the standard normal distribution, where you can easily find probabilities using z-tables or computational tools.
Standard Normal Distribution
The standard normal distribution is a special normal distribution where the mean \(\mu\) is set to 0 and the standard deviation \(\sigma\) is 1. It's also known as the Z-distribution.
  • In this format, the x-axis represents the z-score, and you can look up probabilities for these z-scores using the standard normal distribution table, also known as the z-table.
  • Because it's standardized, any normal distribution can be transformed into the standard normal distribution through z-score calculation.
  • This transformation simplifies finding the probability of a data point or the area under the curve since the standard normal distribution is well-tabulated and widely referenced in statistical analyses.
Converting data into the standard normal form not only makes statistical calculations easier but also more consistent, allowing for universal applications and interpretations of results.

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

Let \(X\) represent the amount of gasoline (gallons) purchased by a randomly selected customer at a gas station. Suppose that the mean value and standard deviation of \(X\) are \(11.5\) and \(4.0\), respectively. a. In a sample of 50 randomly selected customers, what is the approximate probability that the sample mean amount purchased is at least 12 gallons? b. In a sample of 50 randomly selected customers, what is the approximate probability that the total amount of gasoline purchased is at most 600 gallons. c. What is the approximate value of the 95 th percentile for the total amount purchased by 50 randomly selected customers.

For males the expected pulse rate is \(70 / \mathrm{m}\) and the standard deviation is \(10 / \mathrm{m}\). For women the expected pulse rate is \(77 / \mathrm{m}\) and the standard deviation is \(12 / \mathrm{m}\). Let \(X=\) the sample average pulse rate for a random sample of 40 men and let \(\bar{Y}=\) the sample average pulse rate for a random sample of 36 women a. What is the approximate distribution of \(X\) ? Of \(Y\) ? b. What is the approximate distribution of \(\bar{X}-\bar{Y}\) ? Justify your answer. c. Calculate (approximately) the probability \(P(-2 \leq \bar{X}-\bar{Y} \leq 1)\). d. Calculate (approximately) \(P(X-Y \leq-15)\). If you actually observed \(X-Y \leq-15\), would you doubt that \(\mu_{1}-\mu_{2}=-7\) ? Explain.

In an area having sandy soil, 50 small trees of a certain type were planted, and another 50 trees were planted in an area having clay soil. Let \(X=\) the number of trees planted in sandy soil that survive 1 year and \(Y=\) the number of trees planted in clay soil that survive I year. If the probability that a tree planted in sandy soil will survive l year is. 7 and the probability of 1 -year survival in clay soil is \(.6\), compute \(P(-5 \leq X-Y \leq 5)\) (use an approximation, but do not bother with the continuity correction).

Let \(A\) denote the percentage of one constituent in a randomly selected rock specimen, and let \(B\) denote the percentage of a second constituent in that same specimen. Suppose \(D\) and \(E\) are measurement errors in determining the values of \(A\) and \(B\) so that measured values are \(X=A+D\) and \(Y=B+E\), respectively. Assume that measurement errors are independent of each other and of actual values. a. Show that $$ \begin{gathered} \operatorname{Corr}(X, Y)=\operatorname{Corr}(A, B) \cdot \sqrt{\operatorname{Corr}\left(X_{1}, X_{2}\right)} \\ \cdot \sqrt{\operatorname{Corr}\left(Y_{1}, Y_{2}\right)} \end{gathered} $$ where \(X_{1}\) and \(X_{2}\) are replicate measurements on the value of \(A\), and \(Y_{1}\) and \(Y_{2}\) are defined analogously with respect to \(B\). What effect does the presence of measurement error have on the correlation? b. What is the maximum value of \(\operatorname{Corr}(X, Y)\) when \(\operatorname{Corr}\left(X_{1}, X_{2}\right)=.8100, \operatorname{Corr}\left(Y_{1}, Y_{2}\right)=\) \(.9025 ?\) Is this disturbing?

The tip percentage at a restaurant has a mean value of \(18 \%\) and a standard deviation of \(6 \%\). a. What is the approximate probability that the sample mean tip percentage for a random sample of 40 bills is between \(16 \%\) and \(19 \%\) ? b. If the sample size had been 15 rather than 40 , could the probability requested in part (a) be calculated from the given information?
See all solutions

Recommended explanations on Math Textbooks

Pure Maths

Read Explanation

Decision Maths

Read Explanation

Probability and Statistics

Read Explanation

Geometry

Read Explanation

Calculus

Read Explanation

Statistics

Read Explanation
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.