/*! 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 73 The number of wiring packages th... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 7: Problem 73

The number of wiring packages that can be assembled by a company's employees has a normal distribution, with a mean equal to 16.4 per hour and a standard deviation of 1.3 per hour. a. What are the mean and standard deviation of the number \(x\) of packages produced per worker in an 8-hour day? b. Do you expect the probability distribution for \(x\) to be mound-shaped and approximately normal? Explain. c. What is the probability that a worker will produce at least 135 packages per 8 -hour day?

Short Answer

Expert verified
Answer: The probability that a worker will produce at least 135 packages in an 8-hour day is approximately 0.356 or 35.6%.

Step by step solution

01

Finding mean and standard deviation in an 8-hour day

To find the mean and the standard deviation of the number of packages produced per worker in an 8-hour day, we will multiply the given mean and standard deviation by the number of hours (8) in a day. The updated mean and standard deviation will be: Mean: \(16.4 \times 8 = 131.2\) Standard deviation: \(1.3 \times 8 = 10.4\) Thus, the mean is 131.2 packages in an 8-hour day, and the standard deviation is 10.4.
02

Determine if the distribution remains mound-shaped and approximately normal

Since the number of packages produced per hour follows a normal distribution, when we change the time unit from hour to 8-hour day, it will still be a normal distribution. The reason is that the normal distribution is based on continuous random variables, and regardless of the units of measurement, as long as the relationship between the variables is linear, the distribution will be normal. Hence, the probability distribution for \(x\) (packages produced per 8-hour day) remains mound-shaped and approximately normal.
03

Calculate the probability of producing at least 135 packages in an 8-hour day

To calculate this probability, we will first find the z-score corresponding to 135 packages produced in an 8-hour day by using the z-score formula: \(z = \frac{x - \mu}{\sigma} = \frac{135 - 131.2}{10.4} = \frac{3.8}{10.4} \approx 0.37\) Next, we will use a standard normal table or a calculator to find the probability of producing at least 135 packages in an 8-hour day. The probability of producing less than 135 packages is: \(P(x < 135) = P(z < 0.37) \approx 0.644\) Since we are looking for the probability of producing at least 135 packages in an 8-hour day, we will subtract the above result from 1: \(P(x \ge 135) = 1 - P(x < 135) = 1 - 0.644 = 0.356\) Thus, the probability that a worker will produce at least 135 packages in an 8-hour day is approximately 0.356 or 35.6%.

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

Samples of \(n=200\) items were selected hourly over a 100 -hour period, and the sample proportion of defectives was calculated each hour. The mean of the 100 sample proportions was \(.041 .\) a. Use the data to find the upper and lower control limits for a \(p\) chart. b. Construct a \(p\) chart for the process and explain how it can be used.

The "Cheeseburger Bill" In the spring of \(2004,\) the U.S. Congress considered a bill that would prevent Americans from suing fast-food giants like McDonald's for making them overweight. Although the fast-food industry may not be to blame, a study by Children's Hospital in Boston reports that about twothirds of adult Americans and about \(15 \%\) of children and adolescents are overweight. \({ }^{13}\) A random sample of 100 children is selected. a. What is the probability that the sample percentage of overweight children exceeds \(25 \%?\) b. What is the probability that the sample percentage of overweight children is less than \(12 \% ?\) c. Would it be unusual to find that \(30 \%\) of the sampled children were overweight? Explain.

The normal daily human potassium requirement is in the range of 2000 to 6000 milligrams (mg), with larger amounts required during hot summer weather. The amount of potassium in food varies, depending on the food. For example, there are approximately \(7 \mathrm{mg}\) in a cola drink, \(46 \mathrm{mg}\) in a beer, \(630 \mathrm{mg}\) in a banana, \(300 \mathrm{mg}\) in a carrot, and \(440 \mathrm{mg}\) in a glass of orange juice. Suppose the distribution of potassium in a banana is normally distributed, with mean equal to \(630 \mathrm{mg}\) and standard deviation equal to \(40 \mathrm{mg}\) per banana. You eat \(n=3\) bananas per day, and \(T\) is the total number of milligrams of potassium you receive from them. a. Find the mean and standard deviation of \(T\). b. Find the probability that your total daily intake of potassium from the three bananas will exceed \(2000 \mathrm{mg}\). (HINT: Note that \(T\) is the sum of three random variables, \(x_{1}, x_{2}\), and \(x_{3}\), where \(x_{1}\) is the amount of potassium in banana number 1 , etc.)

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

Are you a chocolate "purist," or do you like other ingredients in your chocolate? American Demographics reports that almost \(75 \%\) of consumers like traditional ingredients such as nuts or caramel in their chocolate. They are less enthusiastic about the taste of mint or coffee that provide more distinctive flavors. \({ }^{14}\) A random sample of 200 consumers is selected and the number who like nuts or caramel in their chocolate is recorded. a. What is the approximate sampling distribution for the sample proportion \(\hat{p}\) ? What are the mean and standard deviation for this distribution? b. What is the probability that the sample percentage is greater than \(80 \% ?\) c. Within what limits would you expect the sample proportion to lie about \(95 \%\) of the time?
See all solutions

Recommended explanations on Math Textbooks

Statistics

Read Explanation

Logic and Functions

Read Explanation

Geometry

Read Explanation

Decision Maths

Read Explanation

Probability and Statistics

Read Explanation

Calculus

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.