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Chapter 6: Problem 10

The average teacher's salary in Connecticut (ranked first among states) is \(\$ 57,337\). Suppose that the distribution of salaries is normal with a standard deviation of \(\$ 7500 .\) a. What is the probability that a randomly selected teacher makes less than \(\$ 52,000\) per year? b. If we sample 100 teachers' salaries, what is the probability that the sample mean is less than \(\$ 56,000 ?\)

Short Answer

Expert verified
a) 23.8% chance for a single teacher; b) 3.7% chance for a sample mean.

Step by step solution

01

Understand the Problem

We need to calculate two probabilities using the normal distribution. First, we need the probability that a single teacher earns less than \(52,000, given that teacher salaries are normally distributed with a mean (\)\mu\() of \)57,337 and a standard deviation (\(\sigma\)) of \(7,500. Second, we need the probability that the sample mean of 100 teachers is less than \)56,000.
02

Calculate Z-Score for Part (a)

To find the probability of a single teacher earning less than $52,000, we first calculate the Z-score using the formula \( Z = \frac{X - \mu}{\sigma} \). Here, \( X = 52,000 \), \( \mu = 57,337 \), and \( \sigma = 7,500 \). Substitute these values to get \( Z = \frac{52,000 - 57,337}{7,500} \approx -0.712 \).
03

Find Probability for Part (a)

Using the Z-score of -0.712 from standard normal distribution tables or a calculator, we determine the probability \( P(Z < -0.712) \approx 0.238 \). This means there's about a 23.8% chance that a randomly selected teacher earns less than $52,000.
04

Calculate Standard Error for Part (b)

For the sample of 100 teachers, we need to find the standard error (SE) of the mean. The formula is \( SE = \frac{\sigma}{\sqrt{n}} \), where \( n = 100 \). So, \( SE = \frac{7,500}{\sqrt{100}} = 750 \).
05

Calculate Z-Score for Part (b)

We calculate the Z-score for the sample mean to be below $56,000 using \( Z = \frac{X - \mu}{SE} \). Here, \( X = 56,000 \), \( \mu = 57,337 \), and \( SE = 750 \). Substitute these to find \( Z = \frac{56,000 - 57,337}{750} \approx -1.7827 \).
06

Find Probability for Part (b)

Using the Z-score of -1.7827, determine the probability from standard normal distribution tables or a calculator: \( P(Z < -1.7827) \approx 0.037 \). This means there's about a 3.7% chance that the sample mean salary of 100 teachers is less than $56,000.

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Key Concepts

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

Z-score
A Z-score tells us how many standard deviations away from the mean a particular value is in a normal distribution. It is a way to standardize scores on different scales to compare them more easily. In our exercise, the goal was to find out how likely it is for a teacher's salary to fall under a specific threshold.
To calculate the Z-score, use the formula:
  • \[Z = \frac{X - \mu}{\sigma}\]
Where:
  • \(X\) is the salary we are analyzing.
  • \(\mu\) is the average salary, in this case, \(57,337.
  • \(\sigma\) is the standard deviation, which is \)7,500 for this example.
By placing these numbers into the formula, we can determine the Z-score for a teacher making $52,000. Here, the Z-score calculated was approximately -0.712. This suggests the salary is below average, though not drastically so.
Standard Error
The standard error (SE) measures how much variability or "spread" can be expected in the sample mean compared to the actual population mean. It helps in understanding how well a sample represents the full population.
For a sample size of \(n\) teachers, the formula to calculate the standard error is:
  • \[SE = \frac{\sigma}{\sqrt{n}} \]
Where:
  • \(\sigma\) is the population standard deviation, \(7,500 in this case.
  • \(n\) is the number of sampled teachers, here 100.
By using the formula, the standard error is computed as \)750. This means if we sample 100 teachers' salaries, the average鈥攐r mean鈥攕alary can typically be expected to vary by about $750 up or down from the true population mean.
Teacher Salaries
Teacher salaries are often analyzed to understand economic conditions and disparities among different areas or states. Here, we consider salaries in Connecticut, which ranks first among states with an average of $57,337.
This information is useful when you want to compare teacher salaries across different regions. Understanding the average alongside the standard deviation, $7,500 in this case, gives a view of how much individual salaries tend to stray from the mean. It provides insight into income distribution for teachers
and highlights how a substantial or subtle variation鈥攚hether teachers in Connecticut generally earn around or mostly above this average.
Probability Calculation
Probability calculations in a normal distribution context give us the likelihood of a specific event occurring. For instance, determining how probable it is for a teacher to earn less than a certain amount, like $52,000.
By calculating the Z-score first, we can determine probabilities using standard normal distribution tables or software.
  • In our problem, a Z-score of 鈭�0.712 was used to find out that there's roughly a 23.8% chance a teacher makes less than $52,000.
  • For the sample mean of 100 teachers earning less than $56,000, the Z-score was 鈭�1.7827, leading us to deduce only about 3.7% probability of the sample mean being that low.
These calculations provide vital data for decision-making processes, forecasting, and understanding economic trends. It can aid policymakers in aligning salary structures with economic goals, ensuring educator retention, or adjusting budgets adequately.

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Most popular questions from this chapter

Use the normal approximation to find the probabilities for the specific value(s) of X. a. \(n=10, p=0.5, X \geq 7\) b. \(n=20, p=0.7, X \leq 12\) c. \(n=50, p=0.6, X \leq 40\)

The average commute to work (one way) is 25 minutes according to the 2005 American Community Survey. If we assume that commuting times are normally distributed and that the standard deviation is 6.1 minutes, what is the probability that a randomly selected commuter spends more than 30 minutes commuting one way? Less than 18 minutes?

Find the z value to the right of the mean so that a. 54.78% of the area under the distribution curve lies to the left of it. b. 69.85% of the area under the distribution curve lies to the left of it. c. 88.10% of the area under the distribution curve lies to the left of it.

An instructor gives a 100-point examination in which the grades are normally distributed. The mean is 60 and the standard deviation is 10. If there are 5% A鈥檚 and 5% F鈥檚, 15% B鈥檚 and 15% D鈥檚, and 60% C鈥檚, find the scores that divide the distribution into those categories.

Find the probabilities for each, using the standard normal distribution. \(P(z< 1.42)\)
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