/*! 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 90 At Jen and Perry Ice Cream Compa... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 6: Problem 90

At Jen and Perry Ice Cream Company, a machine fills 1-pound cartons of Top Flavor ice cream. The machine can be set to dispense, on average, any amount of ice cream into these cartons. However, the machine does not put exactly the same amount of ice cream into each carton; it varies from carton to carton. It is known that the amount of ice cream put into each such carton has a normal distribution with a standard deviation of 18 ounce. The quality control inspector wants to set the machine such that at least \(90 \%\) of the cartons have more than 16 ounces of ice cream. What should be the mean amount of ice cream put into these cartons by this machine?

Short Answer

Expert verified
The machine should be set to fill the cartons with a mean amount of approximately 2.44 pounds of ice cream.

Step by step solution

01

Calculate the Z-score for 90 % of the Distribution

Since the quality control inspector wants at least 90% of the cartons to have more than 16 ounces of ice cream, this infers the remaining 10% may have less. Consequently, using a Z-table or a calculator which provides the same statistical functionality, you need to find the Z-score for 90%. To be more specific, look for the Z-score, which corresponds to the probability of 0.1 (10 % remaining) being below this Z-score. This yields a Z-score of approximately -1.28.
02

Use Z-score formula to Find the Mean

The standard Z-score formula is \(Z = (X - \mu) / \sigma\), where \(\mu\) is the mean, \(\sigma\) is the standard deviation, and \(X\) is the observed value. In the given problem, \(\sigma = 18\) ounces (converted to pounds), \(X = 16\) ounces (converted to pounds), and \(Z = -1.28\) (from previous step). Solving the Z-score equation for \(\mu\) gives us the formula \(\mu = X - Z \cdot \sigma\). Plug in the known values and solve for the mean.
03

Solve for the Mean

By substituting the values into the equation we get: \(\mu = 1 - (-1.28 * 1.125)\). This gives us a mean of approximately 2.44 pounds.

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

Find the following areas under a normal distribution curve with \(\mu=20\) and \(\sigma=4\). a. Area between \(x=20\) and \(x=27\) b. Area from \(x=23\) to \(x=26\) c. Area between \(x=9.5\) and \(x=17\)

Alpha Corporation is considering two suppliers to secure the large amounts of steel rods that it uses. Company A produces rods with a mean diameter of \(8 \mathrm{~mm}\) and a standard deviation of \(15 \mathrm{~mm}\) and sells 10,000 rods for \(\$ 400\). Company B produces rods with a mean diameter of \(8 \mathrm{~mm}\) and a standard deviation of. \(12 \mathrm{~mm}\) and sells 10,000 rods for \(\$ 460\). A rod is usable only if its diameter is between \(7.8 \mathrm{~mm}\) and \(8.2 \mathrm{~mm}\). Assume that the diameters of the rods produced by each company have a normal distribution. Which of the two companies should Alpha Corporation use as a supplier? Justify your answer with appropriate calculations.

The print on the package of 100 -watt General Electric soft-white lightbulbs states that these bulbs have an average life of 750 hours. Assume that the lives of all such bulbs have a normal distribution with a mean of 750 hours and a standard deviation of 50 hours. a. Let \(x\) be the life of such a lightbulb. Find \(x\) so that only \(2.5 \%\) of such lightbulbs have lives longer than this value. b. Let \(x\) be the life of such a lightbulb. Find \(x\) so that about \(80 \%\) of such lightbulbs have lives shorter than this value.

A charter bus company is advertising a singles outing on a bus that holds 60 passengers. The company has found that, on average, \(10 \%\) of ticket holders do not show up for such trips; hence, the company routinely overbooks such trips. Assume that passengers act independently of one another. a. If the company sells 65 tickets, what is the probability that the bus can hold all the ticket holders who actually show up? In other words, find the probability that 60 or fewer passengers show up. b. What is the largest number of tickets the company can sell and still be at least \(95 \%\) sure that the bus can hold all the ticket holders who actually show up?

Determine the following probabilities for the standard normal distribution. a. \(P(-1.83 \leq z \leq 2.57)\) b. \(P(0 \leq z \leq 2.02)\) c. \(P(-1.99 \leq z \leq 0)\) d. \(P(z \geq 1.48)\)
See all solutions

Recommended explanations on Math Textbooks

Pure Maths

Read Explanation

Logic and Functions

Read Explanation

Geometry

Read Explanation

Theoretical and Mathematical Physics

Read Explanation

Decision Maths

Read Explanation

Discrete Mathematics

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.