/*! 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 42 a. Is the normal approximation t... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 7: Problem 42

a. Is the normal approximation to the sampling distribution of \(\hat{p}\) appropriate when \(n=400\) and \(p=.8 ?\) b. Use the results of part a to find the probability that \(\hat{p}\) is greater than \(.83 .\) c. Use the results of part a to find the probability that \(\hat{p}\) lies between .76 and .84

Short Answer

Expert verified
If yes, calculate the probabilities that \(\hat{p}\) is greater than 0.83 and that \(\hat{p}\) lies between 0.76 and 0.84. Answer: Yes, the normal approximation can be used for the given sample size (n=400) and true proportion value (p=0.8). The probability that \(\hat{p}\) is greater than 0.83 is approximately 0.067, and the probability that \(\hat{p}\) lies between 0.76 and 0.84 is approximately 0.954.

Step by step solution

01

Check the normal approximation conditions

In order to use the normal approximation, we must check if the following conditions are met: \(np \geq 10\), and \(n(1-p) \geq 10\). Let's calculate these values: \(np = 400 \times 0.8 = 320\) \(n(1-p) = 400 \times (1-0.8) = 400 \times 0.2 = 80\) Since both values are greater than or equal to 10, the normal approximation can be used.
02

Find the standard deviation of the sampling distribution of \(\hat{p}\)

We must now find the standard deviation of the sampling distribution of \(\hat{p}\) using the formula: \(\sigma_{\hat{p}} = \sqrt{\frac{p(1-p)}{n}}\) Substitute the given values and calculate the standard deviation: \(\sigma_{\hat{p}} = \sqrt{\frac{0.8(1-0.8)}{400}} = \sqrt{\frac{0.8\times0.2}{400}} = \sqrt{\frac{0.16}{400}} \approx 0.02\)
03

Find the Z-score for the given probabilities in parts b and c

For part b, we need to find the probability that \(\hat{p}\) is greater than 0.83. For part c, we need to find the probability that \(\hat{p}\) lies between 0.76 and 0.84. To find these probabilities, we'll first calculate the corresponding Z-scores using the formula: \(Z = \frac{\hat{p}-p}{\sigma_{\hat{p}}}\) For part b: \(Z_{0.83} = \frac{0.83-0.8}{0.02} = \frac{0.03}{0.02} = 1.5\) For part c: \(Z_{0.76} = \frac{0.76-0.8}{0.02} = \frac{-0.04}{0.02} = -2\) \(Z_{0.84} = \frac{0.84-0.8}{0.02} = \frac{0.04}{0.02} = 2\)
04

Calculate the probabilities for parts b and c

Using the Z-scores, we can now find the probabilities for parts b and c using the standard normal distribution table or software. For part b: \(P(\hat{p} > 0.83) = P(Z > 1.5) = 1 - P(Z \leq 1.5) \approx 1 - 0.933 = 0.067\) For part c: \(P(0.76 < \hat{p} < 0.84) = P(-2 < Z < 2) = P(Z < 2) - P(Z < -2) \approx 0.977 - 0.023 = 0.954\) The probability that \(\hat{p}\) is greater than 0.83 is approximately 0.067. The probability that \(\hat{p}\) lies between 0.76 and 0.84 is approximately 0.954.

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!

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

A manufacturer of paper used for packaging requires a minimum strength of 20 pounds per square inch. To check on the quality of the paper, a random sample of 10 pieces of paper is selected each hour from the previous hour's production and a strength measurement is recorded for each. The standard deviation \(\sigma\) of the strength measurements, computed by pooling the sum of squares of deviations of many samples, is known to equal 2 pounds per square inch, and the strength measurements are normally distributed. a. What is the approximate sampling distribution of the sample mean of \(n=10\) test pieces of paper? b. If the mean of the population of strength measurements is 21 pounds per square inch, what is the approximate probability that, for a random sample of \(n=10\) test pieces of paper, \(\bar{x}<20 ?\) c. What value would you select for the mean paper strength \(\mu\) in order that \(P(\bar{x}<20)\) be equal to \(.001 ?\)

A questionnaire was mailed to 1000 registered municipal voters selected at random. Only 500 questionnaires were returned, and of the 500 returned, 360 respondents were strongly opposed to a surcharge proposed to support the city Parks and Recreation Department. Are you willing to accept the \(72 \%\) figure as a valid estimate of the percentage in the city who are opposed to the surcharge? Why or why not?

A bottler of soft drinks packages cans in six-packs. Suppose that the fill per can has an approximate normal distribution with a mean of 12 fluid ounces and a standard deviation of 0.2 fluid ounces. a. What is the distribution of the total fill for a case of 24 cans? b. What is the probability that the total fill for a case is less than 286 fluid ounces? c. If a six-pack of soda can be considered a random sample of size \(n=6\) from the population, what is the probability that the average fill per can for a six-pack of soda is less than 11.8 fluid ounces?

During long production runs of canned tomatoes, the average weights (in ounces) of samples of five cans of standard-grade tomatoes in pureed form were taken at 30 control points during an 11 -day period. These results are shown in the table. \({ }^{20}\) When the machine is performing normally, the average weight per can is 21 ounces with a standard deviation of 1.20 ounces. a. Compute the upper and lower control limits and the centerline for the \(\bar{x}\) chart. b. Plot the sample data on the \(\bar{x}\) chart and determine whether the performance of the machine is in control.

Suppose that college faculty with the rank of professor at two-year institutions earn an average of \(\$ 64,571\) per year \(^{7}\) with a standard deviation of \(\$ 4,000 .\) In an attempt to verify this salary level, a random sample of 60 professors was selected from a personnel database for all two- year institutions in the United States. a. Describe the sampling distribution of the sample \(\operatorname{mean} \bar{x}\) b. Within what limits would you expect the sample average to lie, with probability \(.95 ?\) c. Calculate the probability that the sample mean \(\bar{x}\) is greater than \(\$ 66,000 ?\) d. If your random sample actually produced a sample mean of \(\$ 66,000,\) would you consider this unusual? What conclusion might you draw?
See all solutions

Recommended explanations on Math Textbooks

Decision Maths

Read Explanation

Mechanics Maths

Read Explanation

Pure Maths

Read Explanation

Probability and Statistics

Read Explanation

Theoretical and Mathematical Physics

Read Explanation

Logic and Functions

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.