/*! 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 23 Weights of pennies. The distribu... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 4: Problem 23

Weights of pennies. The distribution of weights of United States pennies is approximately normal with a mean of 2.5 grams and a standard deviation of 0.03 grams. (a) What is the probability that a randomly chosen penny weighs less than 2.4 grams? (b) Describe the sampling distribution of the mean weight of 10 randomly chosen pennies. (c) What is the probability that the mean weight of 10 pennies is less than 2.4 grams? (d) Sketch the two distributions (population and sampling) on the same scale. (e) Could you estimate the probabilities from (a) and (c) if the weights of pennies had a skewed distribution?

Short Answer

Expert verified
(a) 0.0004; (b) normal, mean 2.5, SE 0.0095; (c) 0; (d) Sketch shows two curves: population wider, sampling narrower.

Step by step solution

01

Understanding the Standard Normal Distribution

The weights of pennies follow a normal distribution with a mean (\( \mu \)) of 2.5 grams and a standard deviation (\( \sigma \)) of 0.03 grams. Since we are asked for a probability, we will use the standard normal distribution, or \( Z \)-scores, to find the probabilities.
02

Finding Probability for (a)

To find the probability that a penny weighs less than 2.4 grams, convert 2.4 grams to a \( Z \)-score using the formula: \[ Z = \frac{X - \mu}{\sigma} \]. Here, \( X = 2.4 \), \( \mu = 2.5 \), \( \sigma = 0.03 \). Then find \( P(Z < Z_{score}) \) using standard normal distribution tables or software.
03

Calculate Z-score for (a)

\[ Z = \frac{2.4 - 2.5}{0.03} = \frac{-0.1}{0.03} = -3.33 \].
04

Look up Z-score Probability for (a)

Using standard normal distribution tables, find \( P(Z < -3.33) \). This value is approximately 0.0004.
05

Describe Sampling Distribution for (b)

The sampling distribution of the sample mean of \( n = 10 \) pennies is approximately normal. The mean of the sampling distribution is \( \mu_{\bar{X}} = \mu = 2.5 \) grams, and the standard deviation (standard error) is \( \sigma_{\bar{X}} = \frac{\sigma}{\sqrt{n}} = \frac{0.03}{\sqrt{10}} \approx 0.0095 \).
06

Calculate for Mean Probability (c)

To find the probability that the mean weight of 10 pennies is less than 2.4 grams, use the sampling distribution. Calculate the \( Z \)-score: \[ Z = \frac{2.4 - 2.5}{0.0095} \approx -10.53 \].
07

Find Probability for (c)

Using standard normal distribution tables, find \( P(Z < -10.53) \). This probability is essentially 0, as it is extremely rare.
08

Sketch the Two Distributions (d)

On a graph, plot two normal curves on the same scale: one representing the population distribution centered at 2.5 with a SD of 0.03, and a narrower curve for the sampling distribution centered at 2.5 with a SD of 0.0095.
09

Considerations for Skewed Distributions (e)

If the weights were skewed, the accuracy of the probabilities calculated for individual penny weights and sampling distributions would decrease. The Central Limit Theorem helps with sampling distributions, but the normal approximation would not apply well for individual penny weights if n is small.

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.

Standard Deviation
The standard deviation is a crucial concept in statistics, especially when dealing with distributions. It measures the spread or dispersion of a set of data points around the mean. A smaller standard deviation indicates that the data points are closely clustered around the mean, whereas a larger standard deviation suggests more spread-out data values.

In the context of the weights of United States pennies, the standard deviation is 0.03 grams. This means most penny weights are within 0.03 grams of the mean weight of 2.5 grams. The standard deviation provides us with a way to gauge how much individual penny weights can differ from the average, and it's vital for calculating probabilities in a normal distribution.
Sampling Distribution
A sampling distribution is a distribution of a statistic (like the sample mean) that we obtain from many samples drawn from the same population. It tells us how the sample mean is expected to behave.

For example, if we repeatedly draw samples of 10 pennies and calculate the average weight for each sample, these sample means will form a sampling distribution. According to the Central Limit Theorem, this distribution will be approximately normal, even if the underlying distribution is not. The mean of this sampling distribution is the same as the population mean (2.5 grams), but the standard deviation, known as the standard error, is reduced. Here, it is calculated as \( \frac{0.03}{\sqrt{10}} \), which roughly equals 0.0095 grams. This smaller standard deviation makes the sampling distribution narrower and more concentrated around the mean.
Mean Weight
Mean weight is the average weight of a set of data points. In the context of the penny example, it refers to the average weight of U.S. pennies.

The mean weight of the pennies is given as 2.5 grams. This number is a crucial reference point when discussing the standard deviation and sampling distributions. Understanding the mean allows us to evaluate the probability of a penny weighing less than a specific amount, such as 2.4 grams, by comparing it to the standard deviation and using it to calculate Z-scores.
Z-score
A Z-score is a statistical measurement that explains how many standard deviations a data point is from the mean. It's an essential tool in statistics for determining probabilities and evaluating how typical or atypical a certain observation is within a given distribution.

To find the Z-score, we use the formula \( Z = \frac{X - \mu}{\sigma} \), where \( X \) is the value being examined, \( \mu \) is the mean, and \( \sigma \) is the standard deviation. For a penny weighing 2.4 grams, we calculated the Z-score as -3.33. This tells us that 2.4 grams is 3.33 standard deviations below the mean of 2.5 grams. Using Z-scores with standard normal distribution tables or software helps us to determine the probability of an observation or a sample mean being less than a particular value.

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

Area under the curve, Part II. What percent of a standard normal distribution \(N(\mu=0, \sigma=1)\) is found in each region? Be sure to draw a graph. (a) \(Z>-1.13\) (b) \(Z<0.18\) (c) \(Z>8\) (d) \(|Z|<0.5\)

Find the SD. Cholesterol levels for women aged 20 to 34 follow an approximately normal distribution with mean 185 milligrams per deciliter \((\mathrm{mg} / \mathrm{dl})\). Women with cholesterol levels above \(220 \mathrm{mg} / \mathrm{dl}\) are considered to have high cholesterol and about \(18.5 \%\) of women fall into this category. Find the standard deviation of this distribution.

Spray paint, Part 1. Suppose the area that can be painted using a single can of spray paint is slightly variable and follows a nearly normal distribution with a mean of 25 square feet and a standard deviation of 3 square feet. Suppose also that you buy three cans of spray paint. (a) How much area would you expect to cover with these three cans of spray paint? (b) What is the standard deviation of the area you expect to cover with these three eans of spray paint? (c) The area you wanted to cover is 80 square feet. What is the probability that you will be able to cover this entire area with these three cans of spray paint?

Bernoulli, the mean. Use the probability rules from Section 3.5 to derive the mean of a Bernoulli random variable, i.e. a random variable \(X\) that takes value 1 with probability \(p\) and value 0 with probability \(1-p .\) That is, compute the expected value of a generic Bernoulli random variable.

Defective rate. A machine that produces a special type of transistor (a component of computers) has a \(2 \%\) defective rate. The production is considered a random process where each transistor is independent of the others. (a) What is the probability that the \(10^{t h}\) transistor produced is the first. with a defect? (b) What is the probability that the machine produces no defective transistors in a batch of \(100 ?\) (c) On average, how many transistors would you expect to be produced before the first with a defect? What is the standard deviation? (d) Another machine that also produces transistors has a \(5 \%\) defective rate where each transistor is produced independent of the others. On average how many transistors would you expect to be produced with this machine before the first with a defect? What is the standard deviation? (e) Based on your answers to parts (c) and (d), how does increasing the probability of an event affect the mean and standard deviation of the wait time until success?
See all solutions

Recommended explanations on Math Textbooks

Logic and Functions

Read Explanation

Decision Maths

Read Explanation

Theoretical and Mathematical Physics

Read Explanation

Statistics

Read Explanation

Applied Mathematics

Read Explanation

Probability and Statistics

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.