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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. Probability ≈ 0.9838 b. Probability ≈ 0.8926

Step by step solution

01

Understanding the Problem

We need to determine the probability under a Poisson distribution with \( \lambda = 50 \). For large \( \lambda \), the Poisson distribution approximates to a normal distribution with mean \( \mu = \lambda \) and variance \( \sigma^2 = \lambda \). In our problem, the mean \( \mu \) and standard deviation \( \sigma \) for a single day are both 50. For a 5-day week, mean \( \mu = 5 \times 50 = 250 \) and variance \( \sigma^2 = 5 \times 50 = 250 \), so \( \sigma = \sqrt{250} \approx 15.81 \).
02

Convert Poisson to Normal - Part a

For part a, where we need the probability between 35 and 70 tickets in one day, approximate using the normal distribution with \( \mu = 50 \) and \( \sigma = \sqrt{50} \approx 7.07 \). Using the continuity correction, we need \( P(34.5 < X < 70.5) \).
03

Standardize and Find Z-Scores - Part a

Standardize the values for the normal approximation by calculating the z-scores: the lower z-score is \( z = \frac{34.5-50}{7.07} \approx -2.19 \)and the upper z-score is \( z = \frac{70.5-50}{7.07} \approx 2.90 \).
04

Calculate Probability - Part a

Using the standard normal distribution table, find the probability for these z-scores. \( P(-2.19 < Z < 2.90) = P(Z < 2.90) - P(Z < -2.19) \).Approximate values are \( P(Z < 2.90) \approx 0.9981 \) and \( P(Z < -2.19) \approx 0.0143 \). Thus, the probability is \( 0.9981 - 0.0143 = 0.9838 \).
05

Convert Poisson to Normal - Part b

For part b, we need between 225 and 275 tickets over a 5-day week. The mean \( \mu = 250 \) and standard deviation \( \sigma \approx 15.81 \). Using the continuity correction, find \( P(224.5 < X < 275.5) \).
06

Standardize and Find Z-Scores - Part b

Standardize these values: The lower z-score is \( z = \frac{224.5-250}{15.81} \approx -1.61 \) and the upper z-score is \( z = \frac{275.5-250}{15.81} \approx 1.61 \).
07

Calculate Probability - Part b

Using the standard normal distribution table, the probability is \( P(-1.61 < Z < 1.61) = P(Z < 1.61) - P(Z < -1.61) \).Approximate values are \( P(Z < 1.61) \approx 0.9463 \) and \( P(Z < -1.61) \approx 0.0537 \). Thus, the probability is \( 0.9463 - 0.0537 = 0.8926 \).
08

Final Solution

a. The probability that between 35 and 70 tickets are issued is approximately 0.9838. b. The probability that between 225 and 275 tickets are issued over a 5-day week is approximately 0.8926.

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Key Concepts

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

Probability Calculation
Probability calculation in the context of the Poisson distribution is key to understanding how likely a certain number of events will occur within a fixed interval of time. In our example, the city issues parking tickets following a Poisson distribution with a daily average, or mean ( \( \lambda \) ), of 50 tickets. This parameter \( \lambda \) explains the expected number of events, in this case, parking tickets, and is necessary for computing the probability of different outcomes.
The Poisson probability mass function is used to calculate the exact probability of a certain number of events. However, as \( \lambda \) gets larger, calculations can become complex. For such scenarios, we employ approximations, as we'll see in the following sections.
When dealing with the Poisson distribution, remember its two core features:
  • Mean, \( \mu = \lambda \) .
  • Variance, \( \sigma^2 = \lambda \) .
These characteristics guide the transformation from Poisson probabilities to more manageable approximations.
Normal Approximation
The normal approximation allows us to simplify probability calculations when working with a Poisson distribution with a large \( \lambda \) . In our problem, \( \lambda = 50 \), which is substantial enough to use the normal distribution as a good estimate.
To approximate the Poisson distribution using a normal one, we set the mean and variance of the normal distribution equal to those of the Poisson distribution. For example:
  • Mean ( \( \mu \)) = \( \lambda \).
  • Standard deviation ( \( \sigma \)) is \( \sqrt{\lambda} \).
The problem gives \( \lambda = 50 \) for one day or \( \lambda = 5 \times 50 = 250 \) for a 5-day week, providing guidelines for calculating probabilities over both durations.
This approximation helps us find probabilities for ranges, like between 35 and 70 tickets. Importantly, using a normal distribution lets us employ the z-score, where each value can be standardized and referenced against the standard normal distribution table for easy probability lookup.
Continuity Correction
Continuity correction is essential when using a continuous normal distribution to approximate probabilities for a discrete Poisson distribution. It involves adjusting the interval of interest slightly compared to the discrete counts.
Why do we need this? Since the Poisson distribution gives integer results (you can't issue 35.5 tickets), we need this correction to convert to the continuous approximate world.
Here's how it looks in practice:
  • To find the probability of issuing between 35 and 70 tickets, instead of 35 and 70 precisely, we adjust our search to between 34.5 and 70.5 tickets.
  • This adjustment ensures we include all possible discrete values seamlessly into our continuous approximation model.
Continuity correction results in slightly modified z-scores, affecting probability outcomes. This little adjustment is crucial for accuracy, ensuring our normal approximation closely mirrors the Poisson distribution's operation.

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Most popular questions from this chapter

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