/*! 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.66 66. In 1940 the average size of ... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

66. In 1940 the average size of a U.S. farm was 174 acres. Let's say that the standard deviation was 55 acres. Suppose we randomly survey 38 farmers from 1940.

a. In words, X=

b. In words, X-=

c. X~-

d. The IQR for X-is from acres to acres.

Short Answer

Expert verified

a. X is the size of the US farm

b. X- is the average size

c.X-~N(174,8.922)

d.168-180acres

Step by step solution

01

Given Information

A random variable is a variable whose potential qualities are mathematical results of an irregular peculiarity. There are two kinds of irregular factors, discrete and nonstop.

02

Explanation Part (1)

a. The size of the US farms is defined as X

b. The average size of the US farm after surveying 38farms

c. Given,

The standard deviation σ=55

Sample size n = 38

and Mean μ=174

From the formula,

=X¯~Nμx,σxn=X¯~N174,5538=X¯~N(174,8.922)

03

Explanation Part (2)

d. Given,

The standard deviation σ=55

Sample size n =38

and Mean = μ=174

X¯~N(174,8.922)

First quartile is25thpercentile

Hence, P(x¯<k)=0.25

The value of k using a calculator is,168acres

The third quartile is 75thpercentile

Hence, P(x¯<k)=0.75

The value of k using a calculator is, invnorm(0.75,174,8.922)=180acres

Thus the IQR is from168-180acres

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

Salaries for teachers in a particular elementary school district are normally distributed with a mean of\(44,000and a standard deviation of \)6,500. We randomly survey ten teachers from that district.

a. Find the90thpercentile for an individual teacher’s salary.

b. Find the 90thpercentile for the average teacher’s salary.

76. The attention span of a two-year-old is exponentially distributed with a mean of about eight minutes. Suppose we randomly survey 60 two-year-olds.

a. In words, X=

b. X~

c. In wordsX-=

d. X-~

e. Before doing any calculations, which do you think will be higher? Explain why.

i. The probability that an individual attention span is less than ten minutes.

ii. The probability that the average attention span for the 60 children in less than ten minutes?

f. Calculate the probabilities in part e.

g. Explain why the distribution for X- is not exponential.

Men have an average weight of 172pounds with a standard deviation of 29pounds.

a. Find the probability that 20randomly selected men will have a sum weight greater than 3600lbs.

b. If 20 men have a sum weight greater than 3500lbs, then their total weight exceeds the safety limits for water taxis. Based on (a), is this a safety concern? Explain.

Use the information in Example \(7.9\), but change the sample size to \(144\).

a. Find \(P(20<\bar{x}<30)\).

b. Find \(P(\sum x\) is at least \(3,000)\).

c. Find the \(75th\) percentile for the sample mean excess time of \(144\) customers.

A manufacturer produces 25-pound lifting weights. The lowest actual weight is 24 pounds, and the highest is 26 pounds. Each weight is equally likely so the distribution of weights is uniform. A sample of 100 weights is taken.

a) What is the distribution for the sum of the weights of 100 25-pound lifting weights?

b) Find P(Σx < 2,450).

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.