/*! 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 13 Luggage Weight Suppose the weigh... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Luggage Weight Suppose the weight of pieces of passenger luggage for domestic airline flights follows a normal distribution with \(\mu=40\) pounds and \(\sigma=10.63\) pounds. a. Calculate the probability that a piece of luggage weighs less than 45 pounds. b. Calculate the probability that the total weight of the luggage for 80 passengers on a particular flight is between 1200 and 2400 pounds. (Assume each passenger has one piece of luggage.) c. Calculate where the probability density function for the weight of passenger luggage is decreasing most rapidly.

Short Answer

Expert verified
a) 0.6808, b) Probability is essentially zero, c) Inflection at 50.63 lbs.

Step by step solution

01

Understand the Given Normal Distribution

The weight of luggage is normally distributed with a mean \( \mu = 40 \) pounds and a standard deviation \( \sigma = 10.63 \) pounds. This implies that the distribution is defined as \( N(40, 10.63^2) \).
02

Find Probability for Luggage Weighing Less than 45 Pounds

To find the probability that a luggage piece weighs less than 45 pounds, calculate the z-score using the formula \( z = \frac{x - \mu}{\sigma} \):\[ z = \frac{45 - 40}{10.63} = \frac{5}{10.63} \approx 0.47 \]Using the standard normal distribution table, find the probability corresponding to \( z = 0.47 \), which is approximately 0.6808.
03

Calculate Standard Error for Total Weight of 80 Pieces

When considering the total weight of luggage for 80 passengers, the standard deviation of the sample sum is \( \sigma_{total} = \sqrt{n} \cdot \sigma \) where \( n = 80 \). Thus:\[ \sigma_{total} = \sqrt{80} \cdot 10.63 \approx 94.86 \]
04

Calculate Z-Scores for Luggage Weight Range of 80 Passengers

Convert the range 1200 to 2400 pounds into z-scores:\[ z_{1200} = \frac{1200 - (80 \cdot 40)}{94.86} = \frac{1200 - 3200}{94.86} \approx -21.08 \]\[ z_{2400} = \frac{2400 - (80 \cdot 40)}{94.86} = \frac{2400 - 3200}{94.86} \approx -8.45 \]
05

Determine the Probability for the Weight Range

The z-scores calculated are extremely low, indicating the probabilities are near zero. Hence, it's virtually impossible (probability is essentially zero) for the total weight to fall between 1200 and 2400 pounds given typical circumstances.
06

Locate the Inflection Point of the Normal Distribution

The point where the probability density function is decreasing most rapidly is the inflection point of the normal distribution, located one standard deviation from the mean.Find this point:\[ x_{inflection} = 40 + 10.63 \approx 50.63 \] (this is where the curve changes concavity).

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.

Mean and Standard Deviation
In statistics, the mean and standard deviation are key measures to understand data distribution, especially when examining normal distribution. The mean, denoted by \( \mu \), represents the average, or central value, around which the data is spread. In the context of airline luggage, this tells us that, on average, each piece of luggage weighs 40 pounds.

The standard deviation, denoted by \( \sigma \), measures the amount of variation or dispersion in a set of values. For the luggage problem, a standard deviation of 10.63 pounds means most luggage weights tend to vary 10.63 pounds from the 40 pound mean.

Understanding these two parameters is crucial because they define the shape and spread of the normal distribution:
  • The further from the mean, the less likely a value is found.
  • Approximately 68% of data falls within one standard deviation of the mean.
  • A normal distribution is symmetrical around the mean.
Probability Calculations
Calculating probabilities in a normal distribution often involves measuring the likelihood of a value falling within a specific range. This requires converting the value to a z-score, a concept we'll cover later. Let's look at how these probability calculations work in practice.

In the luggage weight problem, part (a) involves finding the probability that luggage weighs less than 45 pounds. By calculating the z-score first, we then refer to a standard normal distribution table to find this probability, about 0.6808. This means there's roughly a 68.08% chance a piece of luggage weighs less than 45 pounds.

For part (b), which looks at a collective weight for 80 passengers, we calculate the standard error first. This is because when you sum many independent variables like luggage weights, their variability accumulates. Here, the standard error for all 80 pieces reflects how much the total weight fluctuates around the expected mean.

These probability calculations help quantify expectations when dealing with real-world uncertainties, guiding decisions with measurable risk assessments.
Z-Score
The z-score is a statistical measurement that describes a value's position relative to the mean of a group of values, measured in terms of standard deviation. Essentially, it tells you how many standard deviations an element is from the mean.

To calculate a z-score, you use the formula:
  • \( z = \frac{x - \mu}{\sigma} \)
Where \( x \) is the value of interest, \( \mu \) is the mean, and \( \sigma \) is the standard deviation. In the luggage example, to find the z-score for a piece of luggage weighing 45 pounds, you plug into the formula resulting in a z-score of approximately 0.47. This indicates 45 pounds is 0.47 standard deviations above the mean luggage weight.

Z-scores are integral in the step involving probability tables, enabling translation of any normal distribution to a standard normal distribution. This normalization facilitates easier calculations by transforming any problem on a common scale.

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

It is estimated that for the first 10 years of production, a certain oil well can be expected to produce oil at a rate of \(r(t)=3.9 t^{3.55} e^{-1.351}\) thousand barrels per year \(t\) years after production begins. a. Write a differential equation for the rate of change of the total amount of oil produced \(t\) years after production begins. b. Use Euler's method with ten intervals to estimate the yield from this oil well during the first 5 years of production. c. Graph the differential equation and the Euler estimates. Discuss how the shape of the graph of the differential equation is related to the shape of the graph of the Euler estimates.

For the first 9 months of life, the average weight \(w\), in pounds, of a certain breed of dog increases at a rate that can be described by the differential equation $$ \frac{d w}{d t}=\frac{33.68}{t} $$ where \(t\) is time given in months. A 1 -month-old puppy weighs 6 pounds. a. Use Euler's method with a 0.25 -month step length to estimate the weight of the puppy at 3 months and at 6 months. b. Use Euler's method with a step length of 1 month to estimate the weight of the puppy at 3 months and at 6 months. c. Is the answer to part \(b\) or the answer to part \(c\) more accurate? Why?

Extraterrestrial Radiation The rate of change in the rate at which the average amount of extraterrestrial radiation in Amarillo, Texas, for each month of the year is changing is proportional to the amount of extraterrestrial radiation received. The constant of proportionality is \(k=-0.212531 .\) In any given month, the expected value of radiation is \(12.5 \mathrm{~mm}\) per day. This expected value is actually obtained in March and September. (Source: Based on data from A. A. Hanson, ed. Practical Handbook of Agricultural Science, Boca Raton: CRC Press, 1990) d. How well does the model estimate the amounts of extraterrestrial radiation in March and September? a. Write a differential equation for the information given. b. In June, the amount of radiation received is approximately \(17.0 \mathrm{~mm}\) per day, and in December, the amount of radiation received is approximately 7.8 \(\mathrm{mm}\) per day. Write a particular solution for this differential equation. c. Change the particular solution into a function giving the average amount of extraterrestrial radiation in Amarillo.

Identify the differential equation as one that can be solved using only antiderivatives or as one for which separation of variables is required. Then find a general solution for the differential equation. \(\frac{d y}{d x}=\frac{k}{y}\)

A demand function and \(a\) supply function for the same commodity is given. a. Locate the shutdown point. Write a sentence of interpretation for this point. b. Locate the point of market equilibrium. Write a sentence of interpretation for this point. \(D(p)=3.6 p^{-0.8}\) thousand units: \(S(p)=\left\\{\begin{array}{ll}0 & \text { for } p<0.2 \\ 7 p & \text { for } p \geq 0.2\end{array}\right.\) thousand units; \(p\) dollars per unit

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.