Introduction to Probability and Statistics
Introduction to Probability and Statistics
14th Edition
ISBN: 9781133103752
Author: Mendenhall, William
Publisher: Cengage Learning
bartleby

Concept explainers

bartleby

Videos

Textbook Question
Book Icon
Chapter 6.4, Problem 6.38E

Let x be a binomial random variable with n = 15 and p = .5 .

a. Is the normal approximation appropriate?

b. Find P ( x 6 ) using the normal approximation.

c. Find P ( x > 6 ) using the normal approximation.

d. Find the exact probabilities for parts b and c, and compare these with your approximations.

a.

Expert Solution
Check Mark
To determine

To find: Whether the normal approximation will be appropriate.

Answer to Problem 6.38E

The normal approximation will be appropriate.

Explanation of Solution

Given information: The values of n is 15 and value of p is 0.5 .

Calculation:

The value of np is,

  np=15×0.5=7.5

The value of nq is,

  nq=15×(10.5)=7.5

Since, the value np and nq are greater than five.

Thus, the normal approximation will be appropriate.

b.

Expert Solution
Check Mark
To determine

To find: The value of P(x6) .

Answer to Problem 6.38E

The value of P(x6) is 0.8485 .

Explanation of Solution

Given information: The values of n is 15 and value of p is 0.5 .

Calculation:

The mean is,

  μ=np=15×0.5=7.5

The standard deviation is,

  σ=npq=15×0.5×0.5=1.94

The z score is,

  z=xμσ=5.57.51.94=1.03

The probability is,

  P(x6)=P(z1.03)=1P(z<1.03)=10.1515=0.8485

Thus, the value of P(x6) is 0.8485 .

c.

Expert Solution
Check Mark
To determine

To find: The value of P(x>6) .

Answer to Problem 6.38E

The value of P(x>6) is 0.6985 .

Explanation of Solution

Given information: The values of n is 15 and value of p is 0.5 .

Calculation:

The z score is,

  z=xμσ=6.57.51.94=0.52

The probability is,

  P(x>6)=P(z>0.52)=1P(z0.52)=10.3015=0.6985

Thus, the value of P(x>6) is 0.6985 .

d.

Expert Solution
Check Mark
To determine

To find: The exact value of P(x6) and P(x>6) .

Answer to Problem 6.38E

The exact value of P(x6) and P(x>6) is 0.849 and 0.696 .

Explanation of Solution

Given information: The values of n is 15 and value of p is 0.5 .

Calculation:

The probability using the table is,

  P(x6)=1P(x<6)=10.151=0.849

The probability using the table is,

  P(x>6)=1P(x6)=10.304=0.696

Thus, the exact value of P(x6) and P(x>6) is 0.849 and 0.696 .

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
Twenty-eight applicants interested in working for the Food Stamp program took an examination designed to measure their aptitude for social work. A stem-and-leaf plot of the 28 scores appears below, where the first column is the count per branch, the second column is the stem value, and the remaining digits are the leaves. a) List all the values. Count 1 Stems Leaves 4 6 1 4 6 567 9 3688 026799 9 8 145667788 7 9 1234788 b) Calculate the first quartile (Q1) and the third Quartile (Q3). c) Calculate the interquartile range. d) Construct a boxplot for this data.
Pam, Rob and Sam get a cake that is one-third chocolate, one-third vanilla, and one-third strawberry as shown below. They wish to fairly divide the cake using the lone chooser method.  Pam likes strawberry twice as much as chocolate or vanilla.   Rob only likes chocolate.  Sam, the chooser, likes vanilla and strawberry twice as much as chocolate.  In the first division, Pam cuts the strawberry piece off and lets Rob choose his favorite piece.  Based on that, Rob chooses the chocolate and vanilla parts. Note: All cuts made to the cake shown below are vertical.Which is a second division that Rob would make of his share of the cake?
Three players (one divider and two choosers) are going to divide a cake fairly using the lone divider method. The divider cuts the cake into three slices (s1, s2, and s3). If the choosers' declarations are Chooser 1: {s1 , s2} and Chooser 2: {s2 , s3}. Using the lone-divider method, how many different fair divisions of this cake are possible?

Chapter 6 Solutions

Introduction to Probability and Statistics

Ch. 6.3 - Find the following percentiles for the standard...Ch. 6.3 - A normal random variable x has mean =10 and...Ch. 6.3 - Prob. 6.13ECh. 6.3 - Prob. 6.14ECh. 6.3 - A normal random variable x has mean 35 and...Ch. 6.3 - Prob. 6.16ECh. 6.3 - A normal random variable x has an unknown mean and...Ch. 6.3 - Hamburger Meat The meat department at a local...Ch. 6.3 - Human Heights Human heights are one of many...Ch. 6.3 - Christmas Trees The diameters of Douglas firs...Ch. 6.3 - Cerebral Blood Flow Cerebral blood flow (CBF) in...Ch. 6.3 - Braking Distances For a car traveling 30 miles per...Ch. 6.3 - Elevator Capacities Suppose that you must...Ch. 6.3 - Sunflowers An experimenter publishing in the...Ch. 6.3 - Prob. 6.26ECh. 6.3 - Prob. 6.27ECh. 6.3 - Prob. 6.28ECh. 6.3 - Prob. 6.30ECh. 6.3 - How Many Words? A publisher has discovered that...Ch. 6.3 - Prob. 6.32ECh. 6.4 - Consider a binomial random variable x with n=25...Ch. 6.4 - Prob. 6.36ECh. 6.4 - Prob. 6.37ECh. 6.4 - Let x be a binomial random variable with n=15 and...Ch. 6.4 - Let x be a binomial random variable with n=100 and...Ch. 6.4 - Prob. 6.40ECh. 6.4 - Prob. 6.41ECh. 6.4 - Prob. 6.42ECh. 6.4 - Prob. 6.43ECh. 6.4 - Prob. 6.44ECh. 6.4 - Prob. 6.45ECh. 6.4 - Prob. 6.46ECh. 6.4 - No Shows Airlines and hotels often grant...Ch. 6.4 - Prob. 6.49ECh. 6.4 - Pepsi’s Market Share Two of the biggest soft drink...Ch. 6.4 - Prob. 6.52ECh. 6.4 - Prob. 6.53ECh. 6 - Prob. 6.54SECh. 6 - Prob. 6.55SECh. 6 - Prob. 6.56SECh. 6 - Prob. 6.57SECh. 6 - a. Find a z0 such that P(z0zz0)=.95 . b. Find a z0...Ch. 6 - Prob. 6.59SECh. 6 - Let x be a binomial random variable with n=36 and...Ch. 6 - Prob. 6.61SECh. 6 - Prob. 6.62SECh. 6 - Prob. 6.63SECh. 6 - Prob. 6.64SECh. 6 - Prob. 6.65SECh. 6 - Prob. 6.66SECh. 6 - Prob. 6.67SECh. 6 - Prob. 6.68SECh. 6 - Prob. 6.69SECh. 6 - Used Cars A used-car dealership has found that the...Ch. 6 - Washers The life span of a type of automatic...Ch. 6 - Prob. 6.74SECh. 6 - Prob. 6.75SECh. 6 - Prob. 6.77SECh. 6 - Prob. 6.78SECh. 6 - Prob. 6.79SECh. 6 - Prob. 6.80SECh. 6 - No Shows An airline finds that 5% of the persons...Ch. 6 - Prob. 6.82SECh. 6 - Plant Genetics In Exercise 5.75, a cross between...Ch. 6 - Prob. 6.84SECh. 6 - Prob. 6.86SECh. 6 - Prob. 6.87SECh. 6 - Introvert or Extrovert? A psychological...Ch. 6 - Normal Distribution? The chest measurements for...Ch. 6 - Prob. 6.91SECh. 6 - Prob. 6.92SECh. 6 - Prob. 6.93SECh. 6 - Prob. 6.94SECh. 6 - Prob. 6.95SECh. 6 - Prob. 6.96SECh. 6 - Prob. 6.97SE
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
Algebra & Trigonometry with Analytic Geometry
Algebra
ISBN:9781133382119
Author:Swokowski
Publisher:Cengage
Text book image
College Algebra (MindTap Course List)
Algebra
ISBN:9781305652231
Author:R. David Gustafson, Jeff Hughes
Publisher:Cengage Learning
Text book image
College Algebra
Algebra
ISBN:9781305115545
Author:James Stewart, Lothar Redlin, Saleem Watson
Publisher:Cengage Learning
Text book image
Holt Mcdougal Larson Pre-algebra: Student Edition...
Algebra
ISBN:9780547587776
Author:HOLT MCDOUGAL
Publisher:HOLT MCDOUGAL
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