/*! 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 10 Suppose that a random sample of ... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Suppose that a random sample of size 64 is to be selected from a population with mean 40 and standard deviation 5. a. What are the mean and standard deviation of the sampling distribution of \(\bar{x}\) ? Describe the shape of the sampling distribution of \(\bar{x}\). b. What is the approximate probability that \(\bar{x}\) will be within 0.5 of the population mean \(\mu\) ? c. What is the approximate probability that \(\bar{x}\) will differ from \(\mu\) by more than \(0.7 ?\)

Short Answer

Expert verified
a. The mean and standard deviation of the sampling distribution of \(\bar{x}\) are 40 and 0.625, respectively. The shape of the sampling distribution of \(\bar{x}\) is approximately normally distributed. b. The approximate probability that \(\bar{x}\) will be within 0.5 of the population mean \(\mu\) is 0.5774. c. The approximate probability that \(\bar{x}\) will differ from \(\mu\) by more than 0.7 is 0.2628.

Step by step solution

01

Find Mean and Standard Deviation of the Sampling Distribution of \(\bar{x}\)#

According to the Central Limit Theorem, the mean of the sampling distribution of \(\bar{x}\) is equal to the population mean (\(\mu\)), and the standard deviation of the sampling distribution of \(\bar{x}\) is equal to the population standard deviation (\(\sigma\)) divided by the square root of the sample size (\(n\)). Mean of the sampling distribution of \(\bar{x}\) (\(\mu_{\bar{x}}\)) = \(\mu\) = 40. Standard deviation of the sampling distribution of \(\bar{x}\) (\(\sigma_{\bar{x}}\)) = \(\frac{\sigma}{\sqrt{n}}\) = \(\frac{5}{\sqrt{64}} = \frac{5}{8} = 0.625\).
02

Describe the Shape of the Sampling Distribution of \(\bar{x}\)#

The Central Limit Theorem states that the shape of the sampling distribution of the sample mean approaches a normal distribution as the sample size increases. Since our sample size is 64, which is large enough, we can assume that the sampling distribution of \(\bar{x}\) will be approximately normally distributed.
03

Calculate the Probability of \(\bar{x}\) Being Within 0.5 of the Population Mean (\(| \bar{x} - \mu | \leq 0.5\))#

To calculate the probability, we will use the z-score, which is defined as the number of standard deviations a value is away from the mean: \(z = \frac{(\bar{x} - \mu)}{\sigma_{\bar{x}}}\). Now, we want to find the probability that the sample mean is within 0.5 of the population mean: \(| \bar{x} - \mu | \leq 0.5\). This can be rewritten as two z-scores: \(z_1 = \frac{(\mu - 0.5 - \mu)}{\sigma_{\bar{x}}}\) and \(z_2 = \frac{(\mu + 0.5 - \mu)}{\sigma_{\bar{x}}}\). Calculating the z-scores: \(z_1 = \frac{-0.5}{0.625} = -0.8\), \(z_2 = \frac{0.5}{0.625} = 0.8\). Now, we can look up these z-scores in the standard normal distribution z-table or use a calculator to find the probability that the z-score is between -0.8 and 0.8, which is approximately \(0.7887 - 0.2113 = 0.5774\).
04

Calculate the Probability of \(\bar{x}\) Differing from \(\mu\) by More than 0.7 (\(|\bar{x} - \mu| > 0.7\))#

To find the probability of the sample mean differing from the population mean by more than 0.7, we need to calculate two z-scores again: \(z_3 = \frac{(\mu - 0.7 - \mu)}{\sigma_{\bar{x}}}\) and \(z_4 = \frac{(\mu + 0.7 - \mu)}{\sigma_{\bar{x}}}\). Calculating the z-scores: \(z_3 = \frac{-0.7}{0.625} = -1.12\), \(z_4 = \frac{0.7}{0.625} = 1.12\). Now, find the probability for z-scores between -1.12 and 1.12 in the standard normal distribution z-table or use a calculator, which is approximately \(0.8686 - 0.1314 = 0.7372\). Finally, to find the probability of the sample mean differing from the population mean by more than 0.7, we need to subtract the probability found above from 1: Probability (\(|\bar{x} - \mu| > 0.7\)) = \(1 - 0.7372 = 0.2628\).

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.

Central Limit Theorem
The Central Limit Theorem (CLT) is a statistical principle with profound importance in probability theory. When dealing with the mean of independent and identically distributed random variables, the CLT tells us that as the sample size grows, the distribution of the sample means will approximate a normal distribution (also known as a Gaussian distribution), irrespective of the original distribution of the data. This allows statisticians to make inferences about population parameters using the normal distribution, even when the sample comes from a non-normal distribution. This is key to understanding why, in our exercise, with a large sample size of 64, we can assume that the sampling distribution of \(\bar{x}\) is approximately normally distributed.
standard deviation
Standard deviation is a measure that quantifies the amount of variation or dispersion of a set of data values. A low standard deviation indicates that the values tend to be close to the mean (also called the expected value) of the set, while a high standard deviation indicates that the values are spread out over a wider range. In the exercise, we calculated the standard deviation of the sampling distribution of the sample mean, \(\sigma_{\bar{x}}\), which is the population standard deviation \(\sigma\) divided by the square root of the sample size \(n\). This calculation of the standard deviation of the sampling distribution is crucial because it allows us to measure the spread of all possible sample means about the population mean and, consequently, to calculate probabilities using the z-score.
z-score
A z-score, also known as a standard score, measures the number of standard deviations an individual data point or sample mean is from the mean of a distribution. It is a dimensionless quantity that provides a way to compare different data points from different normal distributions. In our context, calculating the z-score for the sample mean allows us to determine where it falls within the normal distribution of sample means. We can then use this score to calculate probabilities associated with that sample mean. For example, to find the probability that the sample mean is within a certain range of the population mean, as the exercise requested, we calculate corresponding z-scores and use the standard normal distribution to find these probabilities. The z-score transforms the problem into one that can be easily handled by referring to standard normal distribution tables or software calculations.
normal distribution
The normal distribution is a continuous probability distribution that is symmetrical around its mean, representing a bell-shaped curve where most of the observations cluster around the central peak and probabilities for values further from the mean taper off equally in both directions. It is paramount in many statistical analyses because of its unique properties and the CLT, which allows us to assume a normal distribution for the sampling distribution of the sample mean, given a large enough sample size. In the exercise, the normal distribution is the basis for comparing the sample mean to the population mean and calculating the associated probabilities of divergence between the two. The normal distribution is the reference point for z-scores, allowing us to determine the likelihood of observing sample means within specific ranges.

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 authors of the paper "Driving Performance While Using a Mobile Phone: A Simulation Study of Greek Professional Drivers" (Transportation Research Part \(F\) [2016]: 164-170) describe a study to evaluate the effect of mobile phone use by taxi drivers in Greece. Fifty taxi drivers drove in a driving simulator where they were following a lead car. The drivers were asked to carry on a conversation on a mobile phone while driving, and the following distance (the distance between the taxi and the lead car) was recorded. The sample mean following distance was 3.20 meters and the sample standard deviation was 1.11 meters. a. Construct and interpret a \(95 \%\) confidence interval for \(\mu,\) the population mean following distance while talking on a mobile phone for the population of taxi drivers. b. What assumption must be made in order to generalize this confidence interval to the population of all taxi drivers in Greece?

Suppose that the population mean value of interpupillary distance (the distance between the pupils of the left and right eyes) for adult males is \(65 \mathrm{~mm}\) and that the population standard deviation is \(5 \mathrm{~mm}\). a. If the distribution of interpupillary distance is normal and a random sample of \(n=25\) adult males is to be selected, what is the probability that the sample mean distance \(\bar{x}\) for these 25 will be between 64 and \(67 \mathrm{~mm}\) ? At least \(68 \mathrm{~mm}\) ? b. Suppose that a random sample of 100 adult males is to be selected. Without assuming that interpupillary distance is normally distributed, what is the approximate probability that the sample mean distance will be between 64 and 67 \(\mathrm{mm}\) ? At least \(68 \mathrm{~mm} ?\)

A sign in the elevator of a college library indicates a limit of 16 persons. In addition, there is a weight limit of 2500 pounds. Assume that the average weight of students, faculty, and staff at this college is 150 pounds, that the standard deviation is 27 pounds, and that the distribution of weights of individuals on campus is approximately normal. A random sample of 16 persons from the campus will be selected. a. What is the mean of the sampling distribution of \(\bar{x} ?\) b. What is the standard deviation of the sampling distribution of \(\bar{x} ?\) c. What average weights for a sample of 16 people will result in the total weight exceeding the weight limit of 2500 pounds? d. What is the probability that a random sample of 16 people will exceed the weight limit?

Medical research has shown that repeated wrist extension beyond 20 degrees increases the risk of wrist and hand injuries. Each of 24 students at Cornell University used a proposed new computer mouse design, and while using the mouse, each student's wrist extension was recorded. Data consistent with summary values given in the paper "Comparative Study of Two Computer Mouse Designs" (Cornell Human Factors Laboratory Technical Report RP7992) are given. Use these data to test the hypothesis that the mean wrist extension for people using this new mouse design is greater than 20 degrees. Are any assumptions required in order for it to be appropriate to generalize the results of your test to the population of all Cornell students? To the population of all university students? $$ \begin{array}{lllllll} 27 & 28 & 24 & 26 & 27 & 25 & 25 \\ 24 & 24 & 24 & 25 & 28 & 22 & 25 \\ 24 & 28 & 27 & 26 & 31 & 25 & 28 \\ 27 & 27 & 25 & & & & \end{array} $$

Give as much information as you can about the \(P\) -value of a \(t\) test in each of the following situations. (Hint: See discussion on page \(594 .\) ) a. Upper-tailed test, \(\mathrm{df}=8, t=2.0\) b. Lower-tailed test, \(\mathrm{df}=10, t=-2.4\) c. Lower-tailed test, \(n=22, t=-4.2\) d. Two-tailed test, \(\mathrm{df}=15, t=-1.6\)

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.