/*! 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 16 A simple random sample of size \... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

A simple random sample of size \(n=36\) is obtained from a population with \(\mu=64\) and \(\sigma=18\). (a) Describe the sampling distribution of \(\bar{x}\). (b) What is \(P(\bar{x}<62.6) ?\) (c) What is \(P(\bar{x} \geq 68.7) ?\) (d) What is \(P(59.8<\bar{x}<65.9) ?\)

Short Answer

Expert verified
a) Normal distribution, mean=64, standard error=3. b) 0.3192 c) 0.0571 d) 0.6549

Step by step solution

01

- Describe the sampling distribution of \(\bar{x}\)

The sampling distribution of the sample mean \(\bar{x}\) can be described using the Central Limit Theorem, which states that if we have a sufficiently large sample size, the sampling distribution of \(\bar{x}\) will be approximately normally distributed. In this case, the population mean \(\text{μ} = 64\) and the population standard deviation \(\text{σ} = 18\). The mean of the sampling distribution \(\text{μ}_{\bar{x}}\) is equal to the population mean, \(\text{μ}_{\bar{x}} = 64\), and the standard deviation of the sampling distribution (standard error) \(\text{σ}_{\bar{x}}\) is given by \(\frac{\text{σ}}{\text{√n}} = \frac{18}{\text{√36}} = 3\).
02

- Calculate \(P(\bar{x}

To find \(P(\bar{x}<62.6)\), convert the sample mean to a z-score using the formula \(z = \frac{x - μ_{\bar{x}}}{σ_{\bar{x}}}\): \[ z = \frac{62.6 - 64}{3} = \frac{-1.4}{3} = -0.467\]. Then, use a z-table or standard normal distribution calculator to find the probability corresponding to z = -0.47. \(P(\bar{x}<62.6)\) is approximately 0.3192.
03

- Calculate \(P(\bar{x} \geq\ 68.7)\)

To find \(P(\bar{x} \geq\ 68.7)\), convert the sample mean to a z-score: \[z = \frac{68.7 - 64}{3} = \frac{4.7}{3} = 1.567\]. Using a z-table or a standard normal distribution calculator, find the probability corresponding to z = 1.57 and subtract it from 1. The probability of z being 1.57 is approximately 0.9429, so \(P(\bar{x} \geq\ 68.7) = 1 - 0.9429 = 0.0571\).
04

- Calculate \(P(59.8

To find \(P(59.8<\bar{x}<65.9)\), convert both sample means to z-scores: \[z_1 = \frac{59.8 - 64}{3} = \frac{-4.2}{3} = -1.4\] and \[z_2 = \frac{65.9 - 64}{3} = \frac{1.9}{3} = 0.633\]. Use the z-table to find the probabilities corresponding to z = -1.4 and z = 0.633. \(P(z < -1.4)\) is approximately 0.0808, and \(P(z < 0.63)\) is approximately 0.7357. Thus, \(P(59.8<\bar{x}<65.9) = 0.7357 - 0.0808 = 0.6549\).

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.

Sampling Distribution
The sampling distribution of \( \bar{x} \) refers to the probability distribution of the sample mean from a set of samples taken from the same population. According to the Central Limit Theorem, the sampling distribution of \( \bar{x} \) tends to be normal if the sample size is sufficiently large, even if the population distribution is not normal. For our case, where \( n = 36 \), the theorem applies, and we can state that the sampling distribution of \( \bar{x} \) is approximately normal.

The mean of the sampling distribution (\( \text{μ}_{\bar{x}} \)) is equal to the population mean (\( \text{μ}=64 \)), and the standard deviation of the sampling distribution (also known as the standard error) is calculated using the formula: \(\text{σ}_{\bar{x}} = \frac{\text{σ}}{\text{√n}} \) which in our example simplifies to \( \frac{18}{6} = 3 \). Therefore, \( \text{μ}_{\bar{x}} = 64 \) and \(\text{σ}_{\bar{x}} = 3 \).
Z-Score
A Z-score quantifies the number of standard deviations a data point is from the mean of the sampling distribution. It helps in standardizing different data points so we can compare them.

For instance, to find the probability \( P(\bar{x}<62.6) \), we compute the Z-score using the formula: \( z = \frac{x - μ_{\bar{x}}}{σ_{\bar{x}}} \). Substituting into the equation, \( z = \frac{62.6 - 64}{3} = -0.467 \). This Z-score tells us how far away 62.6 is from the average sample mean in standard deviation units.
Standard Error
The standard error (SE) measures an estimate of the standard deviation of the sample mean's sampling distribution. It provides an understanding of how much variability we can expect in the sample mean from sample to sample.

From our example, with \( n = 36 \) and a population standard deviation of \( \text{σ} = 18 \), the SE is calculated as \( \frac{\text{σ}}{\text{√n}} = \frac{18}{6} = 3 \). The smaller the SE, the more accurate our sample mean is as an estimate of the population mean.

Probability Calculations
Probability calculations allow us to find the likelihood of a certain sample mean occurring within the sampling distribution.

For instance, to find \( P(\bar{x} \text{ < } 62.6) \), we compute the Z-score and then find the corresponding probability using a Z-table. Given \( z = -0.467 \), we find the probability of \( P(z < -0.467) \) which is approximately 0.3192.

Similarly, for \( P(\bar{x} \text{ ≥ } 68.7) \), the process involves computing \( z = 1.567 \), finding the area to the left from Z-tables and subtracting it from 1 to get the result: \( P(\bar{x} \text{ ≥ } 68.7) = 0.0571 \).
Population Mean
The population mean (\( \text{μ} \)) is the average of all values in the population. It is a fixed value and doesn't change. In our example, the population mean is given as 64.

In sampling, we use the population mean to compare with sample means. The goal is often to understand how sample results relate to the overall population.

When computing probabilities or Z-scores, we use this population mean as our benchmark value.
Standard Deviation
The standard deviation (\text{σ}) measures how spread out the values in a population are. It gives insights into data variability.

In our case, the population standard deviation is 18. When we take samples, we often calculate the standard error (SE), which is the standard deviation of the sampling distribution. For our given sample size, we use \( \text{σ}_{\bar{x}} = \frac{\text{σ}}{\text{√n}} = \frac{18}{6} = 3 \).

Standard deviation helps in understanding the dispersion of data and is a key component in calculating Z-scores and probabilities in a sampling distribution.

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

Determine \(\mu_{\bar{x}}\) and \(\sigma_{\bar{x}}\) from the given parameters of the population and the sample size. \(\mu=27, \sigma=6, n=15\)

Suppose you want to study the number of hours of sleep full-time college students at your college get each evening. To do so, you obtain a list of full-time students at your college, obtain a simple random sample of ten students, and ask each of them to disclose how many hours of sleep they obtained the most recent Monday. (a) What is the population of interest in this study? What is the sample? (b) Explain why number of hours of sleep in this study is a random variable. (c) After you obtain your ten observations, you compute the mean number of hours of sleep. Is this a statistic or a parameter? Why? (d) Is the mean number of hours computed in part (c) a random variable? Why? If it is a random variable, what is the source of variation? How does the source of variation in this study differ from that of Problem \(40 ?\)

The shape of the distribution of the time required to get an oil change at a 10 -minute oil-change facility is unknown. However, records indicate that the mean time for an oil change is 11.4 minutes, and the standard deviation for oilchange time is 3.2 minutes. (a) To compute probabilities regarding the sample mean using the normal model, what size sample would be required? (b) What is the probability that a random sample of \(n=40\) oil changes results in a sample mean time of less than 10 minutes? (c) Suppose the manager agrees to pay each employee a \(\$ 50\) bonus if they meet a certain goal. On a typical Saturday, the oil-change facility will perform 40 oil changes between 10 A.M. and 12 P.M. Treating this as a random sample, what mean oil-change time would there be a \(10 \%\) chance of being at or below? This will be the goal established by the manager.

Determine \(\mu_{\bar{x}}\) and \(\sigma_{\bar{x}}\) from the given parameters of the population and the sample size. \(\mu=80, \sigma=14, n=49\)

Afraid to Fly According to a study conducted by the Gallup organization, the proportion of Americans who are afraid to fly is \(0.10 .\) A random sample of 1100 Americans results in 121 indicating that they are afraid to fly. Explain why this is not necessarily evidence that the proportion of Americans who are afraid to fly has increased since the time of the Gallup study.

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.