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Chapter 6: Problem 15

The average overseas trip cost is \(\$ 2708\) per visitor. If we assume a normal distribution with a standard deviation of \(\$ 405,\) what is the probability that the cost for a randomly selected trip is more than \(\$ 3000 ?\) If we select a random sample of 30 overseas trips and find the mean of the sample, what is the probability that the mean is greater than \(\$ 3000 ?\)

Short Answer

Expert verified
Single trip probability: 0.2358; Sample mean probability: 0.0002.

Step by step solution

01

Identify the Problem Type

We are dealing with two separate probability problems involving the normal distribution. First, we need to find the probability that a single trip cost is more than $3000. Second, we find the probability that the sample mean of 30 trips exceeds $3000.
02

Find the Probability for a Single Trip

To find the probability that a single trip costs more than $3000, we first calculate the z-score using the formula: \[ z = \frac{X - \mu}{\sigma} \]where \( X = 3000 \), \( \mu = 2708 \), and \( \sigma = 405 \). Substituting these values:\[ z = \frac{3000 - 2708}{405} \approx \frac{292}{405} \approx 0.720 \].
03

Calculate the Probability from the Z-score

Using a standard normal distribution table or calculator, find the probability corresponding to the z-score of 0.720. This gives the probability that a trip costs less than \(3000. For \( z = 0.720 \), \( P(X < 3000) \approx 0.7642 \). Therefore, the probability that a single trip costs more than \)3000 is: \[ P(X > 3000) = 1 - 0.7642 = 0.2358 \].
04

Find the Probability for Sample Mean of 30 Trips

Next, calculate the probability that the mean of a sample of 30 trips exceeds $3000. For a sample mean problem, adjust the standard deviation using the formula:\[ \sigma_{\bar{X}} = \frac{\sigma}{\sqrt{n}} \]where \( n = 30 \). Substituting \( \sigma = 405 \): \[ \sigma_{\bar{X}} = \frac{405}{\sqrt{30}} \approx 73.97 \].
05

Calculate the Z-score for Sample Mean

For the sample mean, compute the z-score for \( \bar{X} = 3000 \) using the standard deviation we found in Step 4:\[ z = \frac{3000 - 2708}{73.97} \approx \frac{292}{73.97} \approx 3.946 \].
06

Calculate the Probability for the Sample Mean Z-value

Consult a z-table or calculator for the z-value \( z = 3.946 \). The probability \( P(\bar{X} < 3000) \) is nearly 1 since 3.946 is very high. Thus,\[ P(\bar{X} > 3000) = 1 - P(\bar{X} < 3000) \approx 1 - 0.9998 = 0.0002 \].
07

Final Result

The probability that the cost of a randomly chosen trip is more than $3000 is approximately 0.2358. The probability that the mean of a sample of 30 trips costs more than $3000 is approximately 0.0002.

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Key Concepts

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

Normal Distribution
A normal distribution, often referred to as a bell curve, is a probability distribution that is symmetric around its mean. This means that data near the mean are more frequent in occurrence than data far from the mean. It is characterized by its bell-shaped curve where:
  • Mean: This is the peak of the curve and divides the distribution into two equal parts.
  • Standard Deviation: This determines the width of the curve. Larger standard deviations result in a wider and flatter curve.
For our travel cost scenario, the normal distribution helps us understand the probability of different costs against the average cost. Most values lie close to the mean, with fewer occurring as the distance from the mean increases.
Understanding this distribution allows us to apply statistical methods, like calculating the Z-score, to find probabilities related to specific data points.
Z-score
The Z-score is a statistical measurement that describes a value's relation to the mean of a group of values. It is measured in terms of standard deviations from the mean:
  • A Z-score of 0 indicates the value is exactly at the mean.
  • A positive Z-score shows the value is above the mean.
  • A negative Z-score indicates it is below the mean.
The formula to calculate the Z-score is:\[ z = \frac{X - \mu}{\sigma} \]where:
  • \( X \) is the value to be standardized.
  • \( \mu \) is the mean of the dataset.
  • \( \sigma \) is the standard deviation.
In our exercise, the Z-score helps in determining the probability that the cost of a trip is more than a given value, which is crucial for making data-driven decisions.
Sample Mean
The sample mean is the average of a set of data points from a larger population. It is a common way to estimate the population mean when it's impractical to study the entire population.
  • The sample mean is crucial when analyzing small groups drawn from a large dataset.
  • It serves the same role in calculations as the population mean in probability problems.
When working with sample means, especially with normal distribution, the Central Limit Theorem assures us that the distribution of the sample mean approaches a normal distribution as the sample size grows, regardless of the population's distribution. For a sample size of 30, as in our problem, this theorem validates our computations by ensuring the sample mean is normally distributed.
Standard Deviation
Standard deviation is a measure of the amount of variation or dispersion in a set of values. A low standard deviation means that the values tend to be close to the mean of the set, whereas a high standard deviation means that the values are spread out over a wider range.
  • It is a crucial statistical tool for determining the spread of a data set.
  • In a normal distribution, about 68% of values lie within one standard deviation of the mean.
When dealing with sample problems, especially in our travel cost scenario, the standard deviation tells us how much the cost of trips varies around the average cost of $2,708. For the sample of 30 trips, the calculation of the standard deviation of the sample mean (\( \sigma_{\bar{X}} \)) is essential to determine how tightly the sample means are clustered around the population mean.

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Most popular questions from this chapter

The average salary of a male full professor at a public four-year institution offering classes at the doctoral level is \(\$ 99,685 .\) For a female full professor at the same kind of institution, the salary is \(\$ 90,330\). If the standard deviation for the salaries of both genders is approximately \(\$ 5200\) and the salaries are normally distributed, find the 80th percentile salary for male professors and for female professors.

A mail order company has an 8% success rate. If it mails advertisements to 600 people, find the probability of getting fewer than 40 sales.

Find the z value to the left of the mean so that a. 98.87% of the area under the distribution curve lies to the right of it. b. 82.12% of the area under the distribution curve lies to the right of it. c. 60.64% of the area under the distribution curve lies to the right of it.

Price of Gasoline The average retail price of gasoline (all types) for 2014 was 342 cents. What would the standard deviation have to be in order for there to be a \(15 \%\) probability that a gallon of gas costs less than \(\$ 3.00 ?\)

Twenty-two percent of work injuries are back injuries. If 400 work-injured people are selected at random, find the probability that 92 or fewer have back injuries.
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