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A recent study of the hourly wages of maintenance crew members for major airlines showed that the mean hourly wage was \(\$ 20.50,\) with a standard deviation of \(\$ 3.50 .\) Assume the distribution of hourly wages follows the normal probability distribution. If we select a crew member at random, what is the probability the crew member earns: a. Between \(\$ 20.50\) and \(\$ 24.00\) per hour? b. More than \(\$ 24.00\) per hour? c. Less than \(\$ 19.00\) per hour?

Short Answer

Expert verified
a. 0.3413; b. 0.1587; c. 0.3336.

Step by step solution

01

Understanding the Problem

We know that the hourly wages follow a normal distribution with a mean (\( \mu \)) of \(20.50 and a standard deviation (\( \sigma \)) of \)3.50. We need to find probabilities of earning in certain ranges using this normal distribution.
02

Standardizing the Values

For each part, we convert the specific hourly wage values into corresponding Z-scores using the formula: \( Z = \frac{X - \mu}{\sigma} \). This standardization allows us to use standard normal distribution tables (Z-tables).
03

Calculate for Part A: Between $20.50 and $24.00

1. Find Z-score for \(20.50: \( Z = \frac{20.50 - 20.50}{3.50} = 0 \).2. Find Z-score for \)24.00: \( Z = \frac{24.00 - 20.50}{3.50} \approx 1.00 \).3. Use Z-table to find the probability for Z = 1.00, which is 0.8413.4. Probability (Z=0) is 0.5000 (since it's the mean).5. Probability between \(20.50 and \)24.00: 0.8413 - 0.5000 = 0.3413.
04

Calculate for Part B: More than $24.00

1. Use Z-score for $24.00: \( Z = 1.00 \).2. Probability for Z greater than 1.00: 1 - 0.8413 = 0.1587 (from Z-table).
05

Calculate for Part C: Less than $19.00

1. Find Z-score for $19.00: \( Z = \frac{19.00 - 20.50}{3.50} \approx -0.429 \).2. Use Z-table to find the probability for Z = -0.43. Approximate value from Z-table is 0.3336.

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Key Concepts

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

Mean and Standard Deviation
When dealing with normal distributions, two important values to understand are the mean and standard deviation. The **mean** represents the average value of all data points in the distribution. It is the central point where the data balances out. In the context of the exercise, the mean hourly wage is given as \( \\(20.50 \). This means, on average, a maintenance crew member earns \( \\)20.50 \) per hour.

The **standard deviation** indicates how spread out the data points are around the mean. A smaller standard deviation means the data points are closer to the mean, while a larger standard deviation indicates more variability. Here, the standard deviation is \( \\(3.50 \). This tells us that the wages typically deviate from the mean by \( \\)3.50 \) on either side.

Understanding these two values allows us to grasp how clustered or spread out the wages are in this particular normal distribution. The combination of a mean and standard deviation allows us to describe the overall shape of the wage distribution.
Z-scores
Z-scores allow us to compare individual data points to the mean of the distribution. A **Z-score** is essentially a measure of how many standard deviations an element is from the mean. The formula to calculate a Z-score for a given value \( X \) is:
\[ Z = \frac{X - \mu}{\sigma} \]where \( \mu \) is the mean and \( \sigma \) is the standard deviation.

In our exercise, finding Z-scores helps to standardize various hourly wage values relative to the distribution centered at \( \\(20.50 \). For instance, when determining how far \( \\)24.00 \) is from the mean, we calculate:
\[ Z = \frac{24.00 - 20.50}{3.50} \approx 1.00 \]This Z-score of 1.00 tells us that \( \$24.00 \) is one standard deviation above the mean. By computing Z-scores, we convert different wage estimates to a common scale, enabling us to use Z-tables for finding probabilities associated with those scores.

Understanding Z-scores is key to analyzing individual data in terms of its position in any normal distribution.
Probability Calculation
Calculating probabilities using the normal distribution involves translating Z-scores into probability values. Once Z-scores are determined, we can use a **Z-table** to find the probability associated with those scores. The Z-table lists probabilities that correspond to standard normal distribution Z-scores.

For example, to find the probability of earning between \( \\(20.50 \) and \( \\)24.00 \), we calculate that the Z-scores are 0 and 1, respectively. The Z-table shows that the probability of a Z-score less than or equal to 1 is 0.8413, and it indicates a probability of 0.5000 for the mean (Z-score of 0). The difference between these probabilities gives us the likelihood that a crew member earns within that wage range, calculated as 0.8413 - 0.5000 = 0.3413.

In another scenario, to determine the likelihood of earning more than \( \\(24.00 \), the complement probability (1 - 0.8413) helps us find a value of 0.1587. This approach reverses the probability of earning less than \( \\)24.00 \). Similarly, by using the Z-table for negative Z-scores, we grasp how often a wage is less than \( \$19.00 \), finding a corresponding probability of 0.3336.

Overall, probability calculations using Z-scores and Z-tables allow us to understand the likelihood of various outcomes within normal distributions.

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Most popular questions from this chapter

A normal population has a mean of 80.0 and a standard deviation of \(14.0 .\) a. Compute the probability of a value between 75.0 and 90.0 . b. Compute the probability of a value of 75.0 or less. c. Compute the probability of a value between 55.0 and 70.0 .

The Kamp family has twins, Rob and Rachel. Both Rob and Rachel graduated from college 2 years ago, and each is now earning \(\$ 50,000\) per year. Rachel works in the retail industry, where the mean salary for executives with less than 5 years' experience is \(\$ 35,000\) with a standard deviation of \(\$ 8,000\). Rob is an engineer. The mean salary for engineers with less than 5 years' experience is \(\$ 60,000\) with a standard deviation of \(\$ 5,000\). Compute the \(z\) values for both Rob and Rachel, and comment on your findings.

A normal distribution has a mean of 50 and a standard deviation of 4 . Determine the value below which \(95 \%\) of the observations will occur.

Among the thirty largest U.S. cities, the mean one-way commute time to work is 25.8 minutes. The longest one-way travel time is in New York City, where the mean time is 39.7 minutes. Assume the distribution of travel times in New York City follows the normal probability distribution and the standard deviation is 7.5 minutes. a. What percent of the New York City commutes are for less than 30 minutes? b. What percent are between 30 and 35 minutes? c. What percent are between 30 and 50 minutes?

Explain what is meant by this statement: "There is not just one normal probability distribution but a 'family' of them."

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