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Chapter 7: Problem 22

The mean starting salary for college graduates in the spring of 2005 was \(\$ 36,280 .\) Assume that the distribution of starting salaries follows the normal distribution with a standard deviation of \(\$ 3,300 .\) What percent of the graduates have starting salaries: a. Between \(\$ 35,000\) and \(\$ 40,000 ?\) b. More than \(\$ 45,000 ?\) c. Between \(\$ 40,000\) and \(\$ 45,000 ?\)

Short Answer

Expert verified
(a) 52.25%, (b) 0.41%, (c) 12.51%

Step by step solution

01

Identify the Given Information

We are given that the mean starting salary is \( \mu = 36,280 \) dollars and the standard deviation is \( \sigma = 3,300 \) dollars. This suggests we need to use the normal distribution to find percentages related to salaries.
02

Understanding Normal Distribution

Since salary distribution is normally distributed, we can use the standard normal distribution to find the required probabilities by converting given salaries to z-scores using the formula \( z = \frac{X - \mu}{\sigma} \).
03

Calculate Z-scores for Part (a)

For salaries between \\(35,000 and \\)40,000, calculate z-scores. \[ z_{1} = \frac{35,000 - 36,280}{3,300} = -0.39 \] \[ z_{2} = \frac{40,000 - 36,280}{3,300} = 1.13 \]
04

Find Probability for Part (a)

Using a standard normal distribution table or calculator, find the probabilities: - Probability(\(z<1.13\)) \( \approx 0.8708 \)- Probability(\(z<-0.39\)) \( \approx 0.3483 \)The probability between \\(35,000 and \\)40,000 is \(0.8708 - 0.3483 = 0.5225\) or 52.25%.
05

Calculate Z-score for Part (b)

For salaries more than \$45,000, calculate the z-score.\[ z = \frac{45,000 - 36,280}{3,300} = 2.64 \]
06

Find Probability for Part (b)

Using the standard normal distribution table or calculator, find the probability: - Probability(\(z>2.64\)) is \(1 - 0.9959 = 0.0041\) or 0.41%.
07

Calculate Z-scores for Part (c)

For salaries between \\(40,000 and \\)45,000, calculate z-scores.\[ z_{1} = \frac{40,000 - 36,280}{3,300} = 1.13 \] \[ z_{2} = \frac{45,000 - 36,280}{3,300} = 2.64 \]
08

Find Probability for Part (c)

Using a standard normal distribution table or calculator, find the probabilities: - Probability(\(z<2.64\)) \( \approx 0.9959 \)- Probability(\(z<1.13\)) \( \approx 0.8708 \)The probability between \\(40,000 and \\)45,000 is \(0.9959 - 0.8708 = 0.1251\) or 12.51%.

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Key Concepts

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

Z-score
Imagine that a Z-score is a translation tool, allowing us to turn actual data points within our dataset into valuable positional information.
  • The Z-score tells us how many standard deviations a particular data value is from the mean of the dataset.
  • If a Z-score is 0, the data point is at the mean.
  • A positive Z-score indicates that the data point is above the mean, while a negative Z-score means it's below.
To compute a Z-score, use the formula \( z = \frac{X - \mu}{\sigma} \), where \( X \) is your data point, \( \mu \) represents the mean, and \( \sigma \), the standard deviation. This gives a standardized way to evaluate how unusual or typical a particular income is, relative to the overall salary distribution.
Standard Deviation
Standard deviation is a crucial statistic that measures the amount of variation or dispersion in a dataset.
  • It tells us how much individual data points tend to deviate from the mean.
  • A small standard deviation implies that data points are generally close to the mean, creating a narrow distribution.
  • Conversely, a large standard deviation signifies more spread-out data points.
In the context of our exercise, the standard deviation of \( \\(3,300 \) indicates how typical salaries deviate from the average salary of \( \\)36,280 \).Visualizing this can help understand the variations in the salary distribution and how different or similar they are to the average.
Probability Calculation
Calculating probability in a normal distribution allows us to answer questions like: How likely is a salary to fall within a certain range?
  • Once Z-scores are determined, we can use them to find probabilities through a Z-table or a calculator, which shows the cumulative probability associated with each Z-score.
  • For example, to find the probability of a salary between two values, subtract the smaller Z-score's probability from the larger one.
  • To find the probability of salaries more than a certain amount, subtract the Z-score's cumulative probability from 1.
Applying this in our exercise, these calculations give insight into how salaries are distributed around the mean, showing how likely someone is to earn within specific salary brackets.

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Most popular questions from this chapter

A study of long distance phone calls made from the corporate offices of the Pepsi Bottling Group, Inc., in Somers, New York, showed the calls follow the normal distribution. The mean length of time per call was 4.2 minutes and the standard deviation was 0.60 minutes. a. What fraction of the calls last between 4.2 and 5 minutes? b. What fraction of the calls last more than 5 minutes? c. What fraction of the calls last between 5 and 6 minutes? d. What fraction of the calls last between 4 and 6 minutes? e. As part of her report to the president, the Director of Communications would like to report the length of the longest (in duration) 4 percent of the calls. What is this time?

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.

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

The number of passengers on the Carnival Sensation during one-week cruises in the Caribbean follows the normal distribution. The mean number of passengers per cruise is 1,820 and the standard deviation is \(120 .\) a. What percent of the cruises will have between 1,820 and 1,970 passengers? b. What percent of the cruises will have 1,970 passengers or more? c. What percent of the cruises will have 1,600 or fewer passengers? d. How many passengers are on the cruises with the fewest 25 percent of passengers?

A tube of Listerine Tartar Control toothpaste contains 4.2 ounces. As people use the toothpaste, the amount remaining in any tube is random. Assume the: amount of toothpaste left in the tube follows a uniform distribution. From this information, we can determine the following information about the amount remaining in a toothpaste tube without invading anyone's privacy. a. How much toothpaste would you expect to be remaining in the tube? b. What is the standard deviation of the amount remaining in the tube? c. What is the likelihood there is less than 3.0 ounces remaining in the tube? d. What is the probability there is more than 1.5 ounces remaining in the tube?
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