Problem 22
Question
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 \(\$ 45,000 ?\)
Step-by-Step Solution
Verified Answer
(a) 52.25%, (b) 0.41%, (c) 12.51%
1Step 1: 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.
2Step 2: 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 \( z = \frac{X - \mu}{\sigma} \).
3Step 3: Calculate Z-scores for Part (a)
For salaries between \\(35,000 and \\)40,000, calculate z-scores. \[ z_{1} = \frac{35,000 - 36,280}{3,300} = -0.39 \] \[ z_{2} = \frac{40,000 - 36,280}{3,300} = 1.13 \]
4Step 4: 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%.
5Step 5: Calculate Z-score for Part (b)
For salaries more than \$45,000, calculate the z-score.\[ z = \frac{45,000 - 36,280}{3,300} = 2.64 \]
6Step 6: 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%.
7Step 7: Calculate Z-scores for Part (c)
For salaries between \\(40,000 and \\)45,000, calculate z-scores.\[ z_{1} = \frac{40,000 - 36,280}{3,300} = 1.13 \] \[ z_{2} = \frac{45,000 - 36,280}{3,300} = 2.64 \]
8Step 8: 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%.
Key Concepts
Z-scoreStandard DeviationProbability Calculation
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.
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.
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.
Other exercises in this chapter
Problem 20
The amounts of money requested on home loan applications at Down River Federal Savings follow the normal distribution, with a mean of \(\$ 70,000\) and a standa
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A normal distribution has a mean of 50 and a standard deviation of \(4 .\) Determine the value below which 95 percent of the observations will occur.
View solution Problem 24
A normal distribution has a mean of 80 and a standard deviation of \(14 .\) Determine the value above which 80 percent of the values will occur.
View solution