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Chapter 15: Problem 43

A delivery company's trucks occasionally get parking tickets, and based on past experience, the company plans that the trucks will average 1.3 tickets a month, with a standard deviation of 0.7 tickets. a) If they have 18 trucks, what are the mean and standard deviation of the total number of parking tickets the company will have to pay this month? b) What assumption did you make in answering?

Short Answer

Expert verified
Mean: 23.4 tickets; Std. Dev.: 2.97 tickets. Assumed independence of tickets among trucks.

Step by step solution

01

Define the Variables

Each truck's parking tickets can be modeled as a random variable. Let's denote the random variable for a single truck as \(X\). We know that for each truck \( \mu_X = 1.3 \) and \( \sigma_X = 0.7 \), where \( \mu_X \) is the mean number of tickets per truck, and \( \sigma_X \) is the standard deviation per truck.
02

Calculate the Mean of Total Tickets

If we have 18 trucks, the total number of tickets, \(Y\), is the sum of the tickets for each truck. Thus, the mean of \(Y\) is \( \mu_Y = 18 \times \mu_X = 18 \times 1.3 \). This simplifies to \( \mu_Y = 23.4 \).
03

Calculate the Standard Deviation of Total Tickets

Assuming that the number of tickets for each truck is independent, the standard deviation for the total number of tickets is the square root of the sum of variances. Therefore, \( \sigma_Y = \sqrt{18} \times \sigma_X = \sqrt{18} \times 0.7 \). Calculating gives \( \sigma_Y = \sqrt{18} \times 0.7 \approx 2.97 \).
04

List Assumptions Made

To calculate these values, we assumed that the number of tickets for each truck is independent of each other. This independent assumption allows us to add the variances to get the variance for the total number of tickets.

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Key Concepts

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

Random Variable
A random variable is a concept in probability and statistics that represents a numeric outcome from a random phenomenon. It maps possible outcomes of a casual occurrence to numerical values. In the context of the exercise, each truck's parking tickets represent a random variable, which we'll call \(X\). This means that for each truck, there is an average (mean) and a variability (standard deviation) of the tickets they will receive each month. For our scenario, the mean, \(\mu_X\), is 1.3, which implies that on average, each truck gets 1.3 tickets per month.
  • Mean \(\mu_X = 1.3\)
  • Standard Deviation \(\sigma_X = 0.7\)
These random variables will help us figure out the overall ticket scenario for the fleet of trucks.
Mean and Standard Deviation
The mean and standard deviation are essential concepts used to summarize the characteristics of a data set or probability distribution. The mean provides a measure of the central tendency, while the standard deviation indicates how much the values differ from the mean.To determine the total number of parking tickets for all trucks, we calculate the mean \(\mu_Y\) and standard deviation \(\sigma_Y\) for all trucks together.

Mean Calculation

We know each truck has a mean of 1.3 tickets. For 18 trucks, the total mean is:\[ \mu_Y = 18 \times 1.3 = 23.4 \]Thus, in a month, we expect about 23.4 tickets for the entire fleet based on average.

Standard Deviation Calculation

The standard deviation requires us to consider the independence of events. Assuming independence, we calculate the total standard deviation as:\[ \sigma_Y = \sqrt{18} \times 0.7 \approx 2.97 \]This value describes how much the total number of tickets could vary from the mean of 23.4.
Independent Events
Independent events are occurrences that are not affected by any other events. This concept is crucial when dealing with probabilistic calculations involving random variables. In our exercise, we assumed that each truck's parking tickets are an independent event. This means:
  • The number of tickets one truck receives doesn't affect the number of tickets any other truck receives.
  • This independence allows us to compute the total variance by summing up the individual variances of each truck and then taking the square root to find the standard deviation.
Without this assumption of independence, it would not be valid to directly sum the variances of each truck to determine the overall variance, which is a key step for determining the total standard deviation of tickets for all trucks.

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