PRACTICE OF STATISTICS F/AP EXAM
PRACTICE OF STATISTICS F/AP EXAM
6th Edition
ISBN: 9781319113339
Author: Starnes
Publisher: MAC HIGHER
bartleby

Concept explainers

bartleby

Videos

Question
Book Icon
Chapter 7.3, Problem 65E

(a)

To determine

To Calculate: the probability that the mean loss from fire is greater than $300 for an SRS of 1000 homeowners.

(a)

Expert Solution
Check Mark

Answer to Problem 65E

37.45%

Explanation of Solution

Given:

  μ=250σ=5000n=1000x¯=300

Formula used:

  z=xμx¯σx¯

Calculation:

Since the population distribution is normal, the sampling distribution of the sample mean x¯ is also normal.

The Z-score is

  z=xμx¯σx¯=x¯μσ/n=3002505000/1000=0.32

The associating probability using the normal probability

  P(Z<0.32) is given in the row starting with 0.3 and in the column starting with 0.2 of the standard normal probability

  P(X¯300)=P(Z>0.32)=1P(Z<0.32)=10.6255=0.3745=37.45%

(b)

To determine

To find: the company charge.

(b)

Expert Solution
Check Mark

Answer to Problem 65E

$452.39

Explanation of Solution

Given:

  μ=250σ=5000n=1000P(X¯x¯)=90%

Formula used:

  z=xμx¯σx¯

Calculation:

Finding the z-score that associates with a probability of 90% or 0.90 in the normal probability and it is observed that the closest probability is 0.8997 which lies in the row 1.2 and in the column .08 of the normal probability table and therefore the corresponding z-score is then 1.2+.08=1.28

  z=1.28

The z-score is

  z=xμx¯σx¯=x¯μσ/n=x¯2505000/1000

The two found expressions of the z-score than have to be equal:

  x¯2505000/1000=1.28

  x¯250=1.28(5000/1000)

Add 250 to each side:

  x¯250+1.28(5000/1000)

Evaluate:

  x=452.39

Therefore the company should charge $ 452.39 .

Chapter 7 Solutions

PRACTICE OF STATISTICS F/AP EXAM

Ch. 7.1 - Prob. 11ECh. 7.1 - Prob. 12ECh. 7.1 - Prob. 13ECh. 7.1 - Prob. 14ECh. 7.1 - Prob. 15ECh. 7.1 - Prob. 16ECh. 7.1 - Prob. 17ECh. 7.1 - Prob. 18ECh. 7.1 - Prob. 19ECh. 7.1 - Prob. 20ECh. 7.1 - Prob. 21ECh. 7.1 - Prob. 22ECh. 7.1 - Prob. 23ECh. 7.1 - Prob. 24ECh. 7.1 - Prob. 25ECh. 7.1 - Prob. 26ECh. 7.1 - Prob. 27ECh. 7.1 - Prob. 28ECh. 7.1 - Prob. 29ECh. 7.1 - Prob. 30ECh. 7.1 - Prob. 31ECh. 7.1 - Prob. 32ECh. 7.2 - Prob. 33ECh. 7.2 - Prob. 34ECh. 7.2 - Prob. 35ECh. 7.2 - Prob. 36ECh. 7.2 - Prob. 37ECh. 7.2 - Prob. 38ECh. 7.2 - Prob. 39ECh. 7.2 - Prob. 40ECh. 7.2 - Prob. 41ECh. 7.2 - Prob. 42ECh. 7.2 - Prob. 43ECh. 7.2 - Prob. 44ECh. 7.2 - Prob. 45ECh. 7.2 - Prob. 46ECh. 7.2 - Prob. 47ECh. 7.2 - Prob. 48ECh. 7.2 - Prob. 49ECh. 7.2 - Prob. 50ECh. 7.2 - Prob. 51ECh. 7.2 - Prob. 52ECh. 7.3 - Prob. 53ECh. 7.3 - Prob. 54ECh. 7.3 - Prob. 55ECh. 7.3 - Prob. 56ECh. 7.3 - Prob. 57ECh. 7.3 - Prob. 58ECh. 7.3 - Prob. 59ECh. 7.3 - Prob. 60ECh. 7.3 - Prob. 61ECh. 7.3 - Prob. 62ECh. 7.3 - Prob. 63ECh. 7.3 - Prob. 64ECh. 7.3 - Prob. 65ECh. 7.3 - Prob. 66ECh. 7.3 - Prob. 67ECh. 7.3 - Prob. 68ECh. 7.3 - Prob. 69ECh. 7.3 - Prob. 70ECh. 7.3 - Prob. 71ECh. 7.3 - Prob. 72ECh. 7.3 - Prob. 73ECh. 7.3 - Prob. 74ECh. 7.3 - Prob. 75ECh. 7.3 - Prob. 76ECh. 7.3 - Prob. 77ECh. 7.3 - Prob. 78ECh. 7 - Prob. R7.1RECh. 7 - Prob. R7.2RECh. 7 - Prob. R7.3RECh. 7 - Prob. R7.4RECh. 7 - Prob. R7.5RECh. 7 - Prob. R7.6RECh. 7 - Prob. R7.7RECh. 7 - Prob. T7.1SPTCh. 7 - Prob. T7.2SPTCh. 7 - Prob. T7.3SPTCh. 7 - Prob. T7.4SPTCh. 7 - Prob. T7.5SPTCh. 7 - Prob. T7.6SPTCh. 7 - Prob. T7.7SPTCh. 7 - Prob. T7.8SPTCh. 7 - Prob. T7.9SPTCh. 7 - Prob. T7.10SPTCh. 7 - Prob. T7.11SPTCh. 7 - Prob. T7.12SPTCh. 7 - Prob. T7.13SPTCh. 7 - Prob. AP2.1CPTCh. 7 - Prob. AP2.2CPTCh. 7 - Prob. AP2.3CPTCh. 7 - Prob. AP2.4CPTCh. 7 - Prob. AP2.5CPTCh. 7 - Prob. AP2.6CPTCh. 7 - Prob. AP2.7CPTCh. 7 - Prob. AP2.8CPTCh. 7 - Prob. AP2.9CPTCh. 7 - Prob. AP2.10CPTCh. 7 - Prob. AP2.11CPTCh. 7 - Prob. AP2.12CPTCh. 7 - Prob. AP2.13CPTCh. 7 - Prob. AP2.14CPTCh. 7 - Prob. AP2.15CPTCh. 7 - Prob. AP2.16CPTCh. 7 - Prob. AP2.17CPTCh. 7 - Prob. AP2.18CPTCh. 7 - Prob. AP2.19CPTCh. 7 - Prob. AP2.20CPTCh. 7 - Prob. AP2.21CPTCh. 7 - Prob. AP2.22CPTCh. 7 - Prob. AP2.23CPTCh. 7 - Prob. AP2.24CPTCh. 7 - Prob. AP2.25CPT
Knowledge Booster
Background pattern image
Statistics
Learn more about
Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, statistics and related others by exploring similar questions and additional content below.
Similar questions
SEE MORE QUESTIONS
Recommended textbooks for you
Text book image
MATLAB: An Introduction with Applications
Statistics
ISBN:9781119256830
Author:Amos Gilat
Publisher:John Wiley & Sons Inc
Text book image
Probability and Statistics for Engineering and th...
Statistics
ISBN:9781305251809
Author:Jay L. Devore
Publisher:Cengage Learning
Text book image
Statistics for The Behavioral Sciences (MindTap C...
Statistics
ISBN:9781305504912
Author:Frederick J Gravetter, Larry B. Wallnau
Publisher:Cengage Learning
Text book image
Elementary Statistics: Picturing the World (7th E...
Statistics
ISBN:9780134683416
Author:Ron Larson, Betsy Farber
Publisher:PEARSON
Text book image
The Basic Practice of Statistics
Statistics
ISBN:9781319042578
Author:David S. Moore, William I. Notz, Michael A. Fligner
Publisher:W. H. Freeman
Text book image
Introduction to the Practice of Statistics
Statistics
ISBN:9781319013387
Author:David S. Moore, George P. McCabe, Bruce A. Craig
Publisher:W. H. Freeman
Statistics 4.1 Point Estimators; Author: Dr. Jack L. Jackson II;https://www.youtube.com/watch?v=2MrI0J8XCEE;License: Standard YouTube License, CC-BY
Statistics 101: Point Estimators; Author: Brandon Foltz;https://www.youtube.com/watch?v=4v41z3HwLaM;License: Standard YouTube License, CC-BY
Central limit theorem; Author: 365 Data Science;https://www.youtube.com/watch?v=b5xQmk9veZ4;License: Standard YouTube License, CC-BY
Point Estimate Definition & Example; Author: Prof. Essa;https://www.youtube.com/watch?v=OTVwtvQmSn0;License: Standard Youtube License
Point Estimation; Author: Vamsidhar Ambatipudi;https://www.youtube.com/watch?v=flqhlM2bZWc;License: Standard Youtube License