Pearson eText for Basic Business Statistics -- Instant Access (Pearson+)
Pearson eText for Basic Business Statistics -- Instant Access (Pearson+)
14th Edition
ISBN: 9780137400119
Author: MARK BERENSON, David Levine
Publisher: PEARSON+
bartleby

Concept explainers

bartleby

Videos

Textbook Question
Book Icon
Chapter 6, Problem 1PS

Given a standardized normal distribution (with a mean of 0 and standard deviation of 1, as in Table E.2), What is probability that

a . Z is less than 1.57 ? b . Z is greater than 1.84 ? c . Z is between 1.57 and 1.84 ? d . Z is less than 1.57 greater than 1.84 ?

a.

Expert Solution
Check Mark
To determine

The probability that Z is less than 1.57.

Answer to Problem 1PS

The required probability is 0.9417.

Explanation of Solution

The Table E.2 given in the Appendix of the book provides the area under the cumulative standardized normal distribution from to Z .

From the table, the cumulative probability corresponding to the value 1.57 is 0.9418, which can be written as:

PZ<1.57=0.9417

Thus, the probability is 0.9417.

b.

Expert Solution
Check Mark
To determine

The probability that Z is greater than 1.84.

Answer to Problem 1PS

The required probability is 0.0329.

Explanation of Solution

The probability that Z greater than 1.84 can be written as:

PZ>1.84=1PZ<1.84

From the table E.2, the cumulative probability corresponding to the value 1.84 is 0.9671, which can be written as, PZ<1.84=0.9671 .

Substituting the value in the above equation, the required probability is obtained as:

PZ>1.84=10.9671=0.0329

Thus, the probability is 0.0329.

c.

Expert Solution
Check Mark
To determine

The probability that Z is between 1.57 and 1.84.

Answer to Problem 1PS

The required probability is 0.0254.

Explanation of Solution

The probability that Z is between 1.57 and 1.84 can be written as:

P1.57<Z<1.84=PZ<1.84PZ<1.57

From the table E.2, PZ<1.57=0.9417 and PZ<1.84=0.9671 . Substituting these values in the above equation, the required probability is obtained as:

P1.54<Z<1.84=0.96710.9417=0.0254

Thus, the probability is 0.0254.

d.

Expert Solution
Check Mark
To determine

The probability that Z is less than 1.57 or greater than 1.84.

Answer to Problem 1PS

The required probability is 0.9746.

Explanation of Solution

The probability that Z is less than 1.57 or greater than 1.84 can be written as;

PZ<1.57 or Z>1.84=PZ<1.57+PZ>1.84

From part (a) and (b), it is obtained that PZ<1.57=0.9417 and PZ>1.84=0.0329 . Substituting these values in the above equation, the required probability is obtained as:

PZ<1.57 or Z>1.84=0.9417+0.0329=0.9746

Thus, the probability is 0.9746.

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!
Students have asked these similar questions
Customers experiencing technical difficulty with their Internet cable service may call an 800 number for technical support. It takes the technician between 30 seconds and 11 minutes to resolve the problem. The distribution of this support time follows the uniform distribution. Required: a. What are the values for a and b in minutes? Note: Do not round your intermediate calculations. Round your answers to 1 decimal place. b-1. What is the mean time to resolve the problem? b-2. What is the standard deviation of the time? c. What percent of the problems take more than 5 minutes to resolve? d. Suppose we wish to find the middle 50% of the problem-solving times. What are the end points of these two times?
Exercise 6-6 (Algo) (LO6-3)   The director of admissions at Kinzua University in Nova Scotia estimated the distribution of student admissions for the fall semester on the basis of past experience. Admissions Probability 1,100 0.5 1,400 0.4 1,300 0.1     Click here for the Excel Data File   Required: What is the expected number of admissions for the fall semester?       Compute the variance and the standard deviation of the number of admissions. Note: Round your standard deviation to 2 decimal places.
1. Find the mean of the x-values (x-bar) and the mean of the y-values (y-bar) and write/label each here: 2. Label the second row in the table using proper notation; then, complete the table. In the fifth and sixth columns, show the 'products' of what you're multiplying, as well as the answers. X y x minus x-bar y minus y-bar (x minus x-bar)(y minus y-bar) (x minus x-bar)^2 xy 16 20 34 4-2 5 2 3. Write the sums that represents Sxx and Sxy in the table, at the bottom of their respective columns. 4. Find the slope of the Regression line: bi = (simplify your answer) 5. Find the y-intercept of the Regression line, and then write the equation of the Regression line. Show your work. Then, BOX your final answer. Express your line as "y-hat equals...

Chapter 6 Solutions

Pearson eText for Basic Business Statistics -- Instant Access (Pearson+)

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
Glencoe Algebra 1, Student Edition, 9780079039897...
Algebra
ISBN:9780079039897
Author:Carter
Publisher:McGraw Hill
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