/*! 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 40 Sugar packaged by a certain mach... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 8: Problem 40

Sugar packaged by a certain machine has a mean weight of \(5 \mathrm{lb}\) and a standard deviation of \(0.02 \mathrm{lb}\). For what values of \(c\) can the manufacturer of the machinery claim that the sugar packaged by this machine has a weight between \(5-c\) and \(5+c \mathrm{lb}\) with probability at least \(96 \%\) ?

Short Answer

Expert verified
The manufacturer can claim that the sugar packaged by this machine has a weight between 4.959 lb and 5.041 lb with a probability of at least 96%.

Step by step solution

01

Identify the normal distribution function and parameters

We know that the given problem follows a normal distribution because the weights of the sugar packages are likely to follow a bell-shaped curve with a mean (\(\mu\)) and a standard deviation (\(\sigma\)). In this case, \(\mu = 5 \mathrm{lb}\) and \(\sigma = 0.02 \mathrm{lb}\). The normal distribution function is: \(f(x) = \frac{1}{\sigma\sqrt{2\pi}}e^{-\frac{(x-\mu)^2}{2\sigma^2}}\)
02

Identify the relevant probability

We need to find the values of 'c' for which the probability of the sugar packaged by the machine has a weight between 5-c and 5+c lb with a probability of at least 96%: \(P(5-c \leq x \leq 5+c) \geq 0.96\)
03

Standardize the normal distribution

In order to solve this problem, we need to standardize this normal distribution by converting it to a standard normal distribution (z-distribution) with \(\mu = 0\) and \(\sigma = 1\). We do this by using the z-score formula: \(z = \frac{x - \mu}{\sigma}\) Let's apply this formula to our bounds: \(z_{lower} = \frac{5 - c - 5}{0.02}\) \(z_{upper} = \frac{5 + c - 5}{0.02}\)
04

Use probability statement and z-table

Now, using the z-scores we calculated above, we can express the probability as follows: \(P(z_{lower} \leq z \leq z_{upper}) \geq 0.96\) To find the values of \(z_{lower}\) and \(z_{upper}\) that satisfy this probability, we can consult a standard normal (z) probability table (or use calculator software like Wolfram Alpha). Looking up the z-score for a probability of \(0.96 + \frac{0.04}{2} = 0.98\), we find that: \(z_{lower} = -2.05\) \(z_{upper} = 2.05\)
05

Solve for the values of "c"

Now that we have the z-scores, we can solve for the values of 'c' by substituting back to the z-score formula: \(z_{lower} = \frac{5 - c - 5}{0.02} = -2.05\) \(z_{upper} = \frac{5 + c - 5}{0.02} = 2.05\) Solving for 'c' in both equations, we get: \(c = 2.05 \times 0.02 = 0.041\) The manufacturer can claim that the sugar packaged by this machine has a weight between \(5-0.041 = 4.959\) lb and \(5 + 0.041 = 5.041\) lb with a probability of at least 96%.

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

The weights, in ounces, of ten packages of potato chips are \(\begin{array}{llllllllll}16.1 & 16 & 15.8 & 16 & 15.9 & 16.1 & 15.9 & 16 & 16 & 16.2\end{array}\) Find the average and the median of these weights.

Find \(C(n, x) p^{x} q^{n-x}\) for the given values of \(n, x\), and \(p\). n=6, x=4, p=\frac{1}{4}

The general manager of the service department of MCA Television has estimated that the time that elapses between the dates of purchase and the dates on which the 50 -in. plasma TVs manufactured by the company first require service is normally distributed with a mean of \(22 \mathrm{mo}\) and a standard deviation of \(4 \mathrm{mo} .\) If the company gives a 1-yr warranty on parts and labor for these TVs, determine the percentage of these TVs manufactured and sold by the company that will require service before the warranty period runs out.

On average, a student takes 100 words/minute midway through an advanced court reporting course at the American Institute of Court Reporting. Assuming that the dictation speeds of the students are normally distributed and that the standard deviation is 20 words/minute, what is the probability that a student randomly selected from the course can take dictation at a speed a. Of more than 120 words/minute? b. Between 80 and 120 words/minute? c. Of less than 80 words/minute?

Suppose \(X\) is a random variable with mean \(\mu\) and standard deviation \(\sigma\). If a large number of trials is observed, at least what percentage of these values is expected to lie between \(\mu-2 \sigma\) and \(\mu+2 \sigma ?\)
See all solutions

Recommended explanations on Math Textbooks

Decision Maths

Read Explanation

Probability and Statistics

Read Explanation

Discrete Mathematics

Read Explanation

Statistics

Read Explanation

Theoretical and Mathematical Physics

Read Explanation

Geometry

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.