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A recent study of the hourly wages of maintenance crew members for major airlines showed that the mean hourly salary 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're given a normally distributed set of data with a mean (\(\mu\)) of \(20.50\), a standard deviation (\(\sigma\)) of \(3.50\), and we are asked to find probabilities related to specific hourly wages.
02

Calculating Z-Scores for the Range

To find probabilities, first, convert the hourly wages to Z-scores. The formula for Z-score is \(Z = \frac{X - \mu}{\sigma}\). For \(X = 24.00\):\[Z_1 = \frac{24.00 - 20.50}{3.50} = 1.00\]For \(X = 20.50\), since it is the mean:\[Z_0 = 0\]
03

Finding Probabilities for the Range

Using the Z-table, find the probability corresponding to the Z-scores.\(P(Z_0)\) is 0.5000 because it is the mean.\(P(Z_1)\) corresponding to \(Z = 1.00\) is approximately 0.8413.So, \(P(20.50 < X < 24.00) = 0.8413 - 0.5000 = 0.3413\).
04

Calculating Z-Score for More Than $24.00/hr

We need the probability of a wage greater than \$24.00/hr.From step 2, \(Z_1 = 1.00\).The probability we want is \(P(X > 24.00) = 1 - P(Z \leq 1) = 1 - 0.8413 = 0.1587\).
05

Calculating Z-Score for Less Than $19.00/hr

For \(X = 19.00\):\[Z_2 = \frac{19.00 - 20.50}{3.50} = -0.4286\approx -0.43\]Using the Z-table, \(P(Z < -0.43)\) is approximately 0.3336.So the probability that a crew member earns less than \$19.00/hr is \(0.3336\).

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Key Concepts

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

Z-score
The Z-score is a statistical measure that explains how many standard deviations a data point is from the mean. It helps us understand the position of an individual data point within a distribution. To calculate the Z-score, we use the formula: \( Z = \frac{X - \mu}{\sigma} \), where \( X \) is the value, \( \mu \) is the mean, and \( \sigma \) is the standard deviation.

The Z-score allows us to transform the normal distribution into a standard normal distribution, which has a mean of 0 and a standard deviation of 1. This transformation is beneficial because it enables us to use the Z-table, a tool that provides the probabilities associated with each Z-score. For our specific problem, we calculated Z-scores to determine the probabilities of earning between certain hourly wages. For example, a wage of \\(24.00 leads to a Z-score of 1.00, meaning it is one standard deviation above the mean wage of \\)20.50.
Probability
Probability is the measure of the likelihood of an event happening. In the context of normal distribution, it tells us how likely it is for a certain value of a random variable to occur.

In our exercise, we are looking for the probabilities of different ranges of hourly wages based on a normal distribution. By converting these wages into Z-scores, we can use a Z-table to find the probabilities associated with each Z-score. For example, to find the probability of a random crew member earning between \\(20.50 and \\)24.00, we look up the Z-scores for these amounts. The difference between the probabilities gives us the likelihood of the member earning within this range. This method enables clear understanding and calculation of probabilities in various scenarios.
Standard Deviation
Standard deviation is a key concept in statistics that measures the amount of variation or dispersion in a set of data values. It is the square root of the variance and gives insight into how spread out the data points are around the mean. In simpler terms, it indicates the typical distance of each data point from the mean.

For a normal distribution, the standard deviation allows us to gauge the spread of the data. In our case, the crew members' hourly wages have a standard deviation of \\(3.50 from the mean of \\)20.50. This information is crucial because it shows how much the wages typically deviate from the average, helping us not only to understand the dataset but also to calculate probabilities using Z-scores. It acts like a measuring stick, showing whether data points are typical or unusual.

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Most popular questions from this chapter

According to the South Dakota Department of Health the mean number of hours of TV viewing per week is higher among adult women than men. A recent study showed women spent an average of 34 hours per week watching TV and men 29 hours per week. Assume that the distribution of hours watched follows the normal distribution for both groups, and that the standard deviation among the women is 4.5 hours and is 5.1 hours for the men. a. What percent of the women watch TV less than 40 hours per week? b. What percent of the men watch TV more than 25 hours per week? c. How many hours of TV do the one percent of women who watch the most TV per week watch? Find the comparable value for the men.

A uniform distribution is defined over the interval from 2 to \(5 .\) a. What are the values for \(a\) and \(b\) ? b. What is the mean of this uniform distribution? c. What is the standard deviation? d. Show that the total area is 1.00 . e. Find the probability of a value more than 2.6 f. Find the probability of a value between 2.9 and 3.7 .

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

A recent report in USA Today indicated a typical family of four spends \(\$ 490\) per month on food. Assume the distribution of food expenditures for a family of four follows the normal distribution, with a mean of \(\$ 490\) and a standard deviation of \(\$ 90\). a. What percent of the families spend more than \(\$ 30\) but less than \(\$ 490\) per month on food? b. What percent of the families spend less than \(\$ 430\) per month on food? c. What percent spend between \(\$ 430\) and \(\$ 600\) per month on food? d. What percent spend between \(\$ 500\) and \(\$ 600\) per month on food?

According to the Insurance Institute of America, a family of four spends between \(\$ 400\) and \(\$ 3,800\) per year on all types of insurance. Suppose the money spent is uniformly distributed between these amounts. a. What is the mean amount spent on insurance? b. What is the standard deviation of the amount spent? c. If we select a family at random, what is the probability they spend less than \(\$ 2,000\) per year on insurance per year? d. What is the probability a family spends more than \(\$ 3,000\) per year?

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