/*! 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} Q 51. Potato chips Refer to Exercise 4... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Potato chips Refer to Exercise 47 Use the 68–95–99.7 rule to answer the

following questions.

a. About what percent of bags weigh less than 9.02 ounces? Show your method clearly.

b. A bag that weighs9.07 ounces is at about what percentile in this distribution? Justify your answer.

Short Answer

Expert verified

Part (a) 2.5% of the bags weigh less than 9.02 ounces.

Part (b) The bag is at 16th percentile.

Step by step solution

01

Part (a) Step 1: Given information

Mean,μ=9.12

Standard deviation,σ=0.05

02

Part (a) Step 2: Concept

Graph of normal probability- We can claim that the distribution is roughly Normal if the Normal probability plot has a linear structure.

03

Part (a) Step 3: Calculation

According to 68−95−99.7 rule:

In a normal distribution, 68 percent of the data lies within 1 standard deviation of the mean.

A normal distribution has 95 percent of its data within two standard deviations of the mean.

A normal distribution has 99.7% of its data inside 1 standard deviation of the mean.

Then

The general Normal density graph is represented as:

Note that

9.02 lies 2σ below the mean.

μ−2σ=9.12−2(0.05)=9.02

According to 68−95−99.7 rule:

95% of the data values are within 2 standard deviations of the mean. 68−95−99.7n.

Although,

Data values in total are 100%

Then

100%−95%=5%

5% of the data values are greater than 2 standard deviations off the mean.

We also know that

The normal distribution is symmetric around the mean.

That implies

2.5 percent of the data points are more than 2 standard deviations above the mean.

And

2.5 percent of the data points are more than 2 standard deviations below the mean.

Therefore,

2.5% of the bags weigh less than 9.02 ounces.

04

Part (b) Step 1: Calculation

According to 68−95−99.7 rule:

In a normal distribution, 68 percent of the data lies within 1 standard deviation of the mean.

In a normal distribution, 95% of the data lies within 2 standard deviations of the mean.

A normal distribution has 99.7% of its data inside 1 standard deviation of the mean.

Then

The general Normal density graph is represented as:

Note that

9.02 lies σ below the mean.

According to 68−95−99.7 rule:

68% of the data values lie within σ (1 standard deviation) of the mean.

Although,

Data values in total are 100%

Then

μ−σ=9.12−0.05=9.07

According to 68−95−99.7 rule:

68% of the data values lie within σ (1 standard deviation) of the mean.

Although,

Data values in total are 100%

Then

100%−68%=32%

32% of the data values lie more than σ (1 standard deviation) from the mean.

We also know that

The normal distribution is symmetric about the mean.

That implies

16% of the data values are more than σ (1 standard deviation) above the mean.

We also know that

The normal distribution is symmetric about the mean.

That implies

16% of the data values are more than σ (1 standard deviation) above the mean.

And

16% of the data values are more than σ (1 standard deviation) below the mean.

The data value represented by the Xth percentile includes x% of the data values below it.

That implies

Bag that weighs 9.07 ounces has about 16% of the other weighs below it.

Thus,

The bag is at 16th percentile.

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 91Ó°ÊÓ!

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

Watch the salt! Refer to Exercise 55About what percent of the meals

ordered contained between 1200 mg and 1800 mg of sodium?

Batter up! Refer to Exercise 48Use the 68–95–99.7rule to answer the following questions.

a. About what percent of Major League Baseball players with 100 plate appearances had batting averages of 0.363or higher? Show your method clearly.

b. A player with a batting average of 0.227 is at about what percentile in this distribution? Justify your answer.

Making money (2.1) The parallel dot plots show the total family income of randomly chosen individuals from Indiana (38 individuals) and New Jersey (44 individuals). Means and standard deviations are given below the dot plots.

Consider individuals in each state with total family incomes of $95,000 Which individual has a higher income, relative to others in his or her state? Use percentiles and z-scores to support your answer.

Mean and median The figure displays two density curves that model

different distributions of quantitative data. Identify the location of the mean and median by letter for each graph. Justify your answers.

The number of absences during the fall semester was recorded for each student in a large elementary school. The distribution of absences is displayed in the following cumulative relative frequency graph.

If the distribution of absences was displayed in a histogram, what would be the best description of the histogram’s shape?

a. Symmetric

b. Uniform

c. Skewed left

d. Skewed right

e. Cannot be determined

See all solutions

Recommended explanations on Math Textbooks

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.