Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

Tickets. 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 is 23.4 tickets; standard deviation is approximately 2.97 tickets. Assumes independent ticketing for each truck.

Step by step solution

01

Define Mean for a Single Truck

For a single truck, the mean number of tickets is given as 1.3 tickets per month. This is the average number of tickets each truck is expected to collect.
02

Calculate Total Mean for All Trucks

Since there are 18 trucks, the mean number of tickets for all trucks combined is simply the single truck mean multiplied by the number of trucks: \[ \text{Mean total} = 18 \times 1.3 = 23.4 \] This means collectively, the trucks are expected to receive an average of 23.4 tickets this month.
03

Define Standard Deviation for a Single Truck

For a single truck, the standard deviation of the number of tickets is given as 0.7 tickets per month. This represents the variability in the number of tickets from the mean for a single truck.
04

Calculate Total Standard Deviation for All Trucks

When calculating the total standard deviation for independent events, we use the formula for the standard deviation of a sum: \[ \text{SD total} = \sqrt{N} \times \text{SD single} \] where \( N \) is the number of independent events (trucks). For 18 trucks, \[ \text{SD total} = \sqrt{18} \times 0.7 \approx 2.97 \] This is the standard deviation of the total number of tickets for all trucks.
05

Assumption Made

The key assumption here is that the number of tickets each truck gets is independent from the other trucks. We assume no truck's parking behavior affects another, and each has a similar chance of receiving a ticket, with outcomes following a normal distribution.

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 Vaia!

Key Concepts

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

Mean Calculation
Mean calculation is at the heart of statistical analysis. It helps us find the average, or expected value, of a dataset. In this exercise, the mean number of parking tickets for one truck is 1.3 per month. This represents the typical number of tickets a single truck might collect.

To find the mean for all 18 trucks, we simply multiply the mean for one truck by 18. This gives us an expected total of 23.4 tickets for the month.
  • Mean for one truck: 1.3 tickets.
  • Total mean for 18 trucks: \( 18 \times 1.3 = 23.4 \) tickets.
This calculation is crucial as it sets the foundation for further statistical analysis, like finding the variance and standard deviation.
Standard Deviation
Standard deviation measures how much values in a dataset differ from the mean. In our context, it shows the spread of parking tickets around the average number of tickets per truck. For a single truck, the standard deviation is 0.7.

To calculate the standard deviation for all 18 trucks, because each truck is independent, we use the formula for the standard deviation of a sum.
  • Square root of the number of trucks: \( \sqrt{18} \).
  • Total standard deviation: \( \text{SD total} = \sqrt{18} \times 0.7 \approx 2.97 \).
This means there is a moderate spread in the total number of tickets they might receive, considering all factors as constant.
Independent Events
In probability and statistics, independent events are those where the occurrence of one event does not affect another. Here, each truck getting a ticket does not impact whether another truck gets one.

Understanding independence is crucial as it allows us to calculate probabilities and statistical measures more easily. For example, because the trucks are independent:
  • We can apply the sum of mean calculations across all trucks.
  • We can calculate total standard deviation using the square root rule.
This assumption simplifies our analysis and is fundamental for correctly applying the normal distribution model.
Statistical Assumptions
Statistical assumptions form the basis of our calculations and interpretations. In this scenario, we assumed that the ticketing process is essentially random and normally distributed.

These assumptions were necessary:
  • Each truck operates independently; no external factors influence the number of tickets they receive collectively.
  • The number of tickets follows a normal distribution, which allows us to use tools like mean and standard deviation effectively.
If these assumptions hold true, our statistical conclusions will be accurate. Deviations from these assumptions may require a reassessment of the model.

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

Racehorse. A man buys a racehorse for \(\$ 20,000\) and enters it in two races. He plans to sell the horse afterward, hoping to make a profit. If the horse wins both races, its value will jump to \(\$ 100,000\). If it wins one of the races, it will be worth \(\$ 50,000\). If it loses both races, it will be worth only \(\$ 10,000\). The man believes there's a \(20 \%\) chance that the horse will win the first race and a \(30 \%\) chance it will win the second one. Assuming that the two races are independent events, find the man's expected profit.

Carnival. A carnival game offers a \(\$ 100\) cash prize for anyone who can break a balloon by throwing a dart at it. It costs \(\$ 5\) to play, and you're willing to spend up to \(\$ 20\) trying to win. You estimate that you have about a \(10 \%\) chance of hitting the balloon on any throw. a) Create a probability model for this carnival game. b) Find the expected number of darts you'll throw. c) Find your expected winnings.

Garden. A company selling vegetable seeds in packets of 20 estimates that the mean number of seeds that will actually grow is 18 , with a standard deviation of \(1.2\) seeds. You buy 5 different seed packets. a) How many bad seeds do you expect to get? b) What's the standard deviation? c) What assumptions did you make about the seeds? Do you think that assumption is warranted? Explain.

Weightlifting. The Atlas BodyBuilding Company (ABC) sells "starter sets" of barbells that consist of one bar, two 20-pound weights, and four 5 -pound weights. The bars weigh an average of 10 pounds with a standard deviation of \(0.25\) pounds. The weights average the specified amounts, but the standard deviations are \(0.2\) pounds for the 20-pounders and \(0.1\) pounds for the 5 -pounders. We can assume that all the weights are normally distributed. a) \(\mathrm{ABC}\) ships these starter sets to customers in two boxes: The bar goes in one box and the six weights go in another. What's the probability that the total weight in that second box exceeds \(60.5\) pounds? Define your variables clearly and state any assumptions you make. b) It costs \(\mathrm{ABC} \$ 0.40\) per pound to ship the box containing the weights. Because it's an odd-shaped package, though, shipping the bar costs \(\$ 0.50\) a pound plus a \(\$ 6.00\) surcharge. Find the mean and standard deviation of the company's total cost for shipping a starter set. c) Suppose a customer puts a 20-pound weight at one end of the bar and the four 5 -pound weights at the other end. Although he expects the two ends to weigh the same, they might differ slightly. What's the probability the difference is more than a quarter of a pound?

Cereal. The amount of cereal that can be poured into a small bowl varies with a mean of \(1.5\) ounces and a standard deviation of \(0.3\) ounces. A large bowl holds a mean of \(2.5\) ounces with a standard deviation of \(0.4\) ounces. You open a new box of cereal and pour one large and one small bowl. a) How much more cereal do you expect to be in the large bowl? b) What's the standard deviation of this difference? c) If the difference follows a Normal model, what's the probability the small bowl contains more cereal than the large one? d) What are the mean and standard deviation of the total amount of cereal in the two bowls? e) If the total follows a Normal model, what's the probability you poured out more than \(4.5\) ounces of cereal in the two bowls together? f) The amount of cereal the manufacturer puts in the boxes is a random variable with a mean of \(16.3\) ounces and a standard deviation of \(0.2\) ounces. Find the expected amount of cereal left in the box and the standard deviation.

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.

Sign-up for free