/*! 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 An airplane with room for 100 pa... [FREE SOLUTION] | 91Ó°ÊÓ

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An airplane with room for 100 passengers has a total baggage limit of 6000 pounds. Suppose that the total weight of the baggage checked by an individual passenger is a random variable \(x\) with a mean value of 50 pounds and a standard deviation of 20 pounds. If 100 passengers will board a flight, what is the approximate probability that the total weight of their baggage will exceed the limit? (Hint: With \(n=100,\) the total weight exceeds the limit when the average weight \(\bar{x}\) exceeds \(6000 / 100 .)\)

Short Answer

Expert verified
The probability that the total weight of the passengers' luggage will exceed the baggage limit of 6000 pounds is approximately zero.

Step by step solution

01

Calculate the Mean and Standard Deviation

The mean weight of luggage per person, \( \mu \), is already given as 50 pounds, and the standard deviation, \( \sigma \), is 20 pounds. Since there are 100 passengers, the mean weight of luggage for all passengers, \( \mu_t \), is \( \mu \times n = 50 \times 100 = 5000 \) pounds. The standard deviation for all passengers, \( \sigma_t \), is \( \sigma \times \sqrt{n} = 20 \times \sqrt{100} = 200 \) pounds.
02

Calculate the Z-score

The Z-score is defined as the number of standard deviations a data point is from the mean. It is calculated as follows: \( Z = \frac{x - \mu_t}{\sigma_t} \). Here \( x = 6000 \) pounds (baggage limit). So, \( Z = \frac{6000 - 5000}{200} = 5 \).
03

Determine the Probability

In a standard normal distribution table, a Z-score of 5 falls outside the table as it exceeds the maximum data ranges - it represents a probability that is practically one (almost certain). So, the probability that the total weight of baggage exceed the limit of 6000 pounds is practically zero.

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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 quantitative variable that assigns numerical values to outcomes of a random experiment, where the values are subject to chance. In statistics, random variables are classified into two types: discrete and continuous.

Discrete random variables take on a countable number of distinct values, such as the number of heads when flipping a coin multiple times. Continuous random variables, like the weight of luggage described in our exercise, can take on any value within a range. The baggage weight of an individual passenger, denoted as variable \(x\), is a good example of a continuous random variable because it can theoretically be any positive value.

In the context of the exercise, each individual's baggage weight is a random variable with its own distribution based on the airline’s historical data, but together they form a distribution for the total weight which can be analyzed statistically.
Mean and Standard Deviation
The mean is the average value of a set of numbers and provides us with a measure of the central tendency of a distribution. Standard deviation, on the other hand, is a measure of the amount of variation or dispersion of a set of values. A low standard deviation indicates that the values tend to be close to the mean, while a high standard deviation indicates that the values are spread out over a wider range.

In our exercise, the mean weight of baggage per passenger is given as 50 pounds, which represents the center of the distribution of individual weights. The standard deviation of 20 pounds indicates that the individual weights vary around the mean. When considering the total weight of baggage for multiple passengers, we use the properties of the mean and standard deviation to predict collective outcomes, like the likelihood of exceeding the plane's baggage limit.
Z-score
The Z-score is a statistical measurement that describes a value's relationship to the mean of a group of values, measured in terms of standard deviations from the mean. It’s a way to standardize scores on different scales to be comparable. The formula for calculating a Z-score is given by:
\[ Z = \frac{x - \mu}{\sigma} \]
where \(x\) is the value from the sample, \(\mu\) is the mean, and \(\sigma\) is the standard deviation.

In the exercise problem, we applied the Z-score calculation to find out how many standard deviations the baggage limit (6000 pounds) is from the mean total weight of all passengers’ luggage. We found that it's 5 standard deviations away, implying a very rare event in terms of probability.
Normal Distribution
Normal distribution, often called the bell curve because of its shape, is a continuous probability distribution where approximately 68% of values fall within one standard deviation of the mean, 95% within two standard deviations, and nearly all values within three standard deviations. It is symmetrical, with the highest probability at the mean, decreasing as you move away from the center.

Many real-world variables are normally distributed, and it is a fundamental assumption in many statistical tests. Since the Z-score is based on the normal distribution, it allows us to calculate the probability of a random variable falling within a certain range. The Z-score of 5 in our problem far exceeds the typical range of a normal distribution, signaling a very improbable occurrence that the total weight exceeds the baggage limit.

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Most popular questions from this chapter

Suppose that the mean value of interpupillary distance (the distance between the pupils of the left and right eyes) for adult males is \(65 \mathrm{~mm}\) and that the population standard deviation is \(5 \mathrm{~mm}\). a. If the distribution of interpupillary distance is normal and a random sample of \(n=25\) adult males is to be selected, what is the probability that the sample mean distance \(\bar{x}\) for these 25 will be between 64 and \(67 \mathrm{~mm}\) ? at least \(68 \mathrm{~mm}\) ? b. Suppose that a random sample of 100 adult males is to be obtained. Without assuming that interpupillary distance is normally distributed, what is the approximate probability that the sample mean distance will be between 64 and \(67 \mathrm{~mm}\) ? at least \(68 \mathrm{~mm} ?\)

Explain the difference between a population characteristic and a statistic.

Explain the difference between \(\sigma\) and \(\sigma_{\bar{x}}\) and between \(\mu\) and \(\mu_{\bar{x}}^{-}\)

Suppose that a particular candidate for public office is in fact favored by \(48 \%\) of all registered voters in the district. A polling organization will take a random sample of 500 voters and will use \(\hat{p}\), the sample proportion, to estimate \(p\). What is the approximate probability that \(\hat{p}\) will be greater than .5 , causing the polling organization to incorrectly predict the result of the upcoming election?

Consider the following population: \(\\{1,2,3,4\\} .\) Note that the population mean is \(\mu=\frac{1+2+3+4}{4}=2.5\) a. Suppose that a random sample of size 2 is to be selected without replacement from this population. There are 12 possible samples (provided that the order in which observations are selected is taken into account): \(\begin{array}{rrrrrr}1,2 & 1,3 & 1,4 & 2,1 & 2,3 & 2,4 \\ 3,1 & 3,2 & 3,4 & 4,1 & 4,2 & 4,3\end{array}\) Compute the sample mean for each of the 12 possible samples. Use this information to construct the sampling distribution of \(\bar{x}\). (Display the sampling distribution as a density histogram.) b. Suppose that a random sample of size 2 is to be selected, but this time sampling will be done with replacement. Using a method similar to that of Part (a), construct the sampling distribution of \(\bar{x}\). (Hint: There are 16 different possible samples in this case.) c. In what ways are the two sampling distributions of Parts (a) and (b) similar? In what ways are they different?

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