/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 12 The average teacher's salary in ... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 6: Problem 12

The average teacher's salary in North Dakota is \(\$ 37,764\). Assume a normal distribution with \(\sigma=\$ 5100 .\) a. What is the probability that a randomly selected teacher's salary is greater than \(\$ 45,000 ?\) b. For a sample of 75 teachers, what is the probability that the sample mean is greater than \(\$ 38,000 ?\)

Short Answer

Expert verified
a. The probability is 0.0778. b. The probability is 0.3446.

Step by step solution

01

Understand the Problem

We need to find the probabilities related to a normal distribution of teacher salaries in North Dakota. First, we'll find the probability that a single teacher's salary is greater than $45,000. Then, we'll find the probability that the average salary of a sample of 75 teachers exceeds $38,000.
02

Calculate Z-Score for Individual Salary

To determine the probability of a salary being greater than $45,000, calculate the z-score using the formula: \( z = \frac{x - \mu}{\sigma} \) where \( x = 45,000 \), \( \mu = 37,764 \), and \( \sigma = 5,100 \). Substitute the values to get \( z = \frac{45,000 - 37,764}{5,100} \approx 1.42 \).
03

Find Probability for Individual Salary

Using the z-score of 1.42, find the probability that corresponds to this z-value from the standard normal distribution table. The probability of \( Z < 1.42 \) is about 0.9222. Hence, the probability that \( Z > 1.42 \) is \( 1 - 0.9222 = 0.0778 \).
04

Calculate Standard Error for Sample Mean

When considering a sample of 75 teachers, calculate the standard error of the mean using \( \text{SE} = \frac{\sigma}{\sqrt{n}} \) where \( \sigma = 5,100 \) and \( n = 75 \). This gives \( \text{SE} = \frac{5,100}{\sqrt{75}} \approx 589 \).
05

Calculate Z-Score for Sample Mean

Find the z-score for the sample mean salary equaling $38,000 using \( z = \frac{x - \mu}{\text{SE}} \) where \( x = 38,000 \). Thus, \( z = \frac{38,000 - 37,764}{589} \approx 0.40 \).
06

Find Probability for Sample Mean

Using the z-score of 0.40, find the corresponding probability from the standard normal distribution table. The probability of \( Z < 0.40 \) is about 0.6554. Therefore, the probability that \( Z > 0.40 \) is \( 1 - 0.6554 = 0.3446 \).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

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

Z-Score Calculation
When working with a normal distribution, z-scores help us understand the position of a specific value within the distribution. A z-score indicates how many standard deviations an element is from the mean. For instance, if we're using the mean teacher salary of \(37,764 in North Dakota and want to see the probability of earning more than \)45,000, we calculate a z-score. The formula is:\[ z = \frac{x - \mu}{\sigma} \]Where:
  • \( x \) is the value of interest, in this case, \( \\( 45,000 \).
  • \( \mu \) is the mean, which is \( \\) 37,764 \).
  • \( \sigma \) is the standard deviation, \( \\( 5,100 \).
Substitute these values to obtain:\[ z = \frac{45,000 - 37,764}{5,100} \approx 1.42 \]This tells us that \)45,000 is 1.42 standard deviations above the mean salary.
Standard Error
Understanding standard error is essential when dealing with sample data. It shows how much the sample mean (average) is expected to vary from the true population mean. When you take a sample, say 75 teachers' salaries, you compute the standard error (SE) with:\[ \text{SE} = \frac{\sigma}{\sqrt{n}} \]Where:
  • \( \sigma \) is the population standard deviation, \( \$ 5,100 \).
  • \( n \) is the sample size, here it's 75.
Plugging in these numbers, we find:\[ \text{SE} = \frac{5,100}{\sqrt{75}} \approx 589 \]This SE value helps us understand the variability we could expect if we were to take many samples of 75 teachers. It's like zooming in on a smaller snapshot within the bigger population picture.
Probability
Probability helps us determine the likelihood of a certain outcome. In our context, it's used for events like whether a teacher earns more than a particular amount. After finding the z-score, we look up this value in a standard normal distribution table to find probabilities.For example, for a teacher earning more than \(45,000:
  • The z-score: 1.42.
  • Probability \( Z < 1.42 \) from the table is about 0.9222.
  • Thus, \( P(Z > 1.42) = 1 - 0.9222 = 0.0778 \).
This tells us there's a 7.78% chance a randomly selected teacher earns more than \)45,000.
Teacher Salaries
When considering salaries, it's important to understand how averages can provide insight. The mean salary of teachers in North Dakota is an example of a population mean. Exploring how individual salaries deviate from this mean reveals several insights.With the normal distribution of teacher salaries:
  • The mean (average) is \( \\( 37,764 \).
  • The standard deviation represents typical salary variability, \( \\) 5,100 \).
These measures help us analyze individual and sample salary probabilities. For instance, if a sample of 75 teachers has an average salary of more than \( \$ 38,000 \), we calculate the relevant probabilities using the normal distribution, as demonstrated. This approach offers useful predictions and assessments regarding salaries in the educational sector.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

The average lifetime of smoke detectors that a company manufactures is 5 years, or 60 months, and the standard deviation is 8 months. Find the probability that a random sample of 30 smoke detectors will have a mean lifetime between 58 and 63 months.

In a recent year, Delaware had the highest per capita annual income with \(\$ 51,803\). If \(\sigma=\$ 4850\), what is the probability that a random sample of 34 state residents had a mean income greater than \(\$ 50,000 ?\) Less than \(\$ 48,000 ?\)

The average length of a hospital stay for all diagnoses is 4.8 days. If we assume that the lengths of hospital stays are normally distributed with a variance of 2.1, then 10% of hospital stays are longer than how many days? Thirty percent of stays are less than how many days?

College students often make up a substantial portion of the population of college cities and towns. State College, Pennsylvania, ranks first with 71.1% of its population made up of college students. What is the probability that in a random sample of 150 people from State College, more than 50 are not college students?

To qualify for a medical study, an applicant must have a systolic blood pressure in the 50% of the middle range. If the systolic blood pressure is normally distributed with a mean of 120 and a standard deviation of 4, find the upper and lower limits of blood pressure a person must have to qualify for the study.
See all solutions

Recommended explanations on Math Textbooks

Decision Maths

Read Explanation

Pure Maths

Read Explanation

Applied Mathematics

Read Explanation

Probability and Statistics

Read Explanation

Logic and Functions

Read Explanation

Mechanics Maths

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.