/*! 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 30 Consider a population with \(\mu... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Consider a population with \(\mu=43\) and \(\sigma=5.2\). a. Calculate the \(z\) -score for an \(\bar{x}\) of 46.5 from a sample of size 35 b. Could this \(z\) -score be used in calculating probabilities using Table 3 in Appendix B? Why or why not?

Short Answer

Expert verified
a. The \(z\)-score for a sample mean of 46.5 is approximately 3.85. b. This \(z\)-score can't be used for calculating probabilities, assuming we are using a z-table that only goes up to \(z = 3.4\).

Step by step solution

01

Identify given values

Recognize given values: population mean \(\mu = 43\), standard deviation \(\sigma = 5.2\), sample mean \(\bar{x} = 46.5\), and sample size \(n = 35\).
02

Calculate the z-score

Use the z-score formula: \(z = \frac{\bar{x} - \mu}{\sigma / \sqrt{n}}\). Replace \(\bar{x}\) with 46.5, \(\mu\) with 43, \(\sigma\) with 5.2, and \(\sqrt{n}\) with \(\sqrt{35}\). Solving this gives \(z \approx 3.853204\).
03

Evaluate z-score for calculating probabilities

Given the z-score, determine if it can be utilized to figure probabilities. The z-score is, in a standard normal distribution, the number of standard deviations away from the mean a data point is. The z-table records the cumulative probability of the standard normal distribution up to a given z-score. If a z-table only goes up to \(z = 3.4\), using our z-score for probability computations is not possible because our calculated z-score of approximately 3.85 exceeds the z-table limit.

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 91Ó°ÊÓ!

Key Concepts

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

Population Mean
In statistics, the population mean, denoted as \( \mu \), represents the average value of all data points in a complete dataset or entire population. For the given exercise, the population mean is 43. This figure symbolizes what is expected as an average in the population we are surveying or analyzing.
This mean is crucial because it offers a central point to gauge how individual data points deviate from the norm within a dataset. Understanding it helps compare sample data, like a sample mean, to the population's overall trend.
Standard Deviation
Standard deviation, denoted by \( \sigma \), is a measure of how spread out numbers are in a dataset. It gives insight into the variability, or diversity, of data points from the average. In the example, the population standard deviation is 5.2.
Standard deviation is important because it allows us to understand the degree of variation within a dataset. A small standard deviation indicates that data points are close to the mean, whereas a large one suggests significant variation from the mean. This concept helps in calculating the z-score, which is central to the problem at hand.
Sample Size
Sample size, denoted by \( n \), specifies the number of observations or data points selected from a larger population to infer conclusions about that population. In the given example, the sample size is 35.
The sample size is vital because it influences the reliability and precision of statistical conclusions. Larger samples tend to provide more accurate estimates of the population parameters, like the population mean. The sample size is directly involved in the calculation of the z-score through the standard error, defined as \( \frac{\sigma}{\sqrt{n}} \). This standard error adjusts the standard deviation according to the size of the sample, providing a basis to compare the sample mean with the population mean.
Standard Normal Distribution
The standard normal distribution is a special normal distribution where the mean is 0 and the standard deviation is 1. A standard normal distribution is helpful for understanding standard scores, or z-scores.
Z-scores, which are derived from individual data points, indicate how many standard deviations a data point is from the mean. This concept is essential for standardizing different distributions for comparison.
In practice, z-scores are used to help calculate the probability of a data point occurring under the normal distribution. By converting raw scores into z-scores, we can utilize the z-table. This table provides the cumulative probability associated with each z-score. In our analysis, a calculated z-score tells us how significantly a sample mean of 46.5 deviates from the population mean, which helps in making probabilistic introspections about data.

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

According to the June 2004 Readers' Digest article "Only in America," the average amount that a 17 -year old spends on his or her high school prom is \(638\) Assume that the amounts spent are normally distributed with a standard deviation of \(175\) a. Find the probability that the mean cost to attend a high school prom for 36 randomly selected 17-year-olds is between \(550\) and \(700\). b. Find the probability that the mean cost to attend a high school prom for 36 randomly selected 17 -yearolds is greater than \(\$ 750\). c. Do you think the assumption of normality is reasonable? Explain.

a. What is the total measure of the area for any probability distribution? b. Justify the statement " \(\bar{x}\) becomes less variable as \(n\) increases."

The heights of the kindergarten children mentioned in Example 7.6 (p. 328 ) are approximately normally distributed with \(\mu=39\) and \(\sigma=2\). a. If an individual kindergarten child is selected at random, what is the probability that he or she has a height between 38 and 40 inches? b. \(\quad\) A classroom of 30 of these children is used as a sample. What is the probability that the class mean \(\bar{x}\) is between 38 and 40 inches? c. If an individual kindergarten child is selected at random, what is the probability that he or she is taller than 40 inches? d. A classroom of 30 of these kindergarten children is used as a sample. What is the probability that the class mean \(\bar{x}\) is greater than 40 inches?

According to Nielsen's Television Audience Report, in 2009 the average American home had 2.86 television sets (more than the average number of people per household, at \(2.5 \text { people }) .\) If the standard deviation for the number of televisions in a U.S. household is 1.2 and a random sample of 80 American households is selected, the mean of this sample belongs to a sampling distribution. a. What is the shape of this sampling distribution? b. What is the mean of this sampling distribution? c. What is the standard deviation of this sampling distribution?

TIMSS 2007 (Trends in International Mathematics and Science Study) focused on the mathematics and science achievement of eighth-grade students throughout the world. A total of 8 countries (including the United States) participated in the study. The mean math exam score for U.S. students was 509 with a standard deviation of \(88 .\) Assuming the scores are normally distributed, find the following for a sample of 150 students. a. Find the probability that the mean TIMSS score for a randomly selected group of eighth-grade students would be between 495 and 515. b. Find the probability that the mean TIMSS score for a randomly selected group of eighth-grade students would be less than 520. c. Do you think the assumption of normality is reasonable? Explain.

See all solutions

Recommended explanations on Math Textbooks

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.