Beginning Statistics, 2nd Edition
Beginning Statistics, 2nd Edition
2nd Edition
ISBN: 9781932628678
Author: Carolyn Warren; Kimberly Denley; Emily Atchley
Publisher: Hawkes Learning Systems
Question
Book Icon
Chapter 5.2, Problem 21E
To determine

(a)

To find:

The specified probability for the given scenario.

Expert Solution
Check Mark

Answer to Problem 21E

Solution:

Value of the required probability is 1.

Explanation of Solution

Formula used:

For a binomial random variable X, the probability of obtaining at least r successes in n independent trials is given by,

P(Xr)=P(X=r)+P(X=r+1)+...+P(X=n)=1P(Xr1)

P(Xr1)=P(X=0)+P(X=1)+...+P(X=r1)

The formula to calculate P(X=r) is,

P(X=r)=Cnrpr(1p)(nr)

Here, r is the minimum number of successes,

n, is the number of trials, and

p, is the probability of getting a success on any trial.

The formula to calculate Cnr is,

Cnr=n!r!(nr)!

Calculation:

Consider the event “surviving of plant” as a success.

The probability of any plant surviving in Kerry’s garden is 0.8.

Probability of success is 0.8.

p=0.8

Number of trial is 19.

n=19.

Compute the probability for at least 5 successes.

Substitute 5 for r in the formula for at least r successes.

P(X5)=1P(X4)

P(X4)=P(X=0)+P(X=1)+P(X=2)+P(X=3)+P(X=4)

Substitute 19 for n, 0 for r, and 0.8 for p in the formula of P(X=r) to calculate P(X=0).

P(X=0)=19!0!19!×(0.8)0×(10.8)19=1×1×(0.2)190

Substitute 19 for n, 1 for r, and 0.8 for p in the formula of P(X=r) to calculate P(X=1).

P(X=1)=19!1!18!×(0.8)1×(10.8)18=19×(0.8)×(0.2)180

Substitute 19 for n, 2 for r, and 0.8 for p in the formula of P(X=r) to calculate P(X=2).

P(X=2)=19!2!17!×(0.8)2×(10.8)17=19×9×(0.8)2×(0.2)170

Substitute 19 for n, 3 for r, and 0.8 for p in the formula of P(X=r) to calculate P(X=3).

P(X=3)=19!3!16!×(0.8)3×(10.8)16=19×3×17×(0.8)3×(0.2)160

Substitute 19 for n, 4 for r, and 0.8 for p in the formula of P(X=r) to calculate P(X=4).

P(X=4)=19!4!15!×(0.8)4×(10.8)15=19×17×4×(0.8)4×(0.2)150

Add all the values of

P(X=0),P(X=1),P(X=2),P(X=3),andP(X=4)

P(X=0)+P(X=1)+P(X=2)+P(X=3)+P(X=4)=0+0+0+0+0=0

P(X4)=0

Substitute 0 for P(X4) in expansion for P(X5).

P(X5)=10=0

P(X5)=1

Conclusion:

Thus, the probability of surviving at least 5 of them is 1.

To determine

(b)

To find:

The specified probability for the given scenario.

Expert Solution
Check Mark

Answer to Problem 21E

Solution:

Value of the required probability is 0.6733.

Explanation of Solution

Formula used:

For a binomial random variable X, the probability of obtaining more than r successes in n independent trials is given by,

P(X>r)=P(X=r+1)+P(X=r+2)...+P(X=n)

The formula to calculate P(X=r) is,

P(X=r)=Cnrpr(1p)(nr)

Here, r+1 is the minimum number of successes,

n, is the number of trials, and

p, is the probability of getting a success on any trial

The formula to calculate Cnr is,

Cnr=n!r!(nr)!

Calculation:

Consider the event “surviving of plant” as a success.

The probability of any plant surviving in Kerry’s garden is 0.8.

Probability of success is 0.8.

p=0.8

Number of trial is 19.

n=19.

Required number of success is 34of19.

r=34×19=14.25

r is a positive integer.

Take r=14

Compute the probability for more than 14 successes.

Substitute 14 for r, 19 for n in the formula for more than r successes.

P(X>14)=P(X=15)+P(X=16)+P(X=17)+P(X=18)+P(X=19)

Substitute 19 for n, 15 for r, and 0.8 for p in the formula of P(X=r) to calculate P(X=15)

P(X=15)=19!15!4!×(0.8)15×(10.8)4=3876×(0.8)15×(0.2)4=0.2181994019

Substitute 19 for n, 16 for r, and 0.8 for p in the formula of P(X=r) to calculate P(X=16)

P(X=16)=19!16!3!×(0.8)16×(10.8)3=969×(0.8)16×(0.2)3=0.2181994019

Substitute 19 for n, 17 for r, and 0.8 for p in the formula of P(X=r) to calculate P(X=17)

P(X=17)=19!17!2!×(0.8)17×(10.8)2=171×(0.8)17×(0.2)2=0.1540231073

Substitute 19 for n, 18 for r, and 0.8 for p in the formula of P(X=r) to calculate P(X=18)

P(X=18)=19!18!1!×(0.8)18×(10.8)1=19×(0.8)18×(0.2)1=0.06845471434

Substitute 19 for n, 19 for r, and 0.8 for p in the formula of P(X=r) to calculate P(X=19)

P(X=19)=19!19!0!×(0.8)19×(10.8)0=1×(0.8)19×1=0.01441151881

Add all the values of

P(X=15),P(X=16),P(X=17),P(X=18),andP(X=19)

P(X=15)+P(X=16)+P(X=17)+P(X=18)+P(X=19)=0.2181994019+0.2181994019+0.1540231073+0.06845471434+0.014411518810.6733

P(X>14)=0.6733

Conclusion:

Thus, the probability of surviving more than 34 of them is 0.6733.

To determine

(c)

To find:

The specified probability for the given scenario.

Expert Solution
Check Mark

Answer to Problem 21E

Solution:

Value of the required probability is 0

Explanation of Solution

Formula used:

For a binomial random variable X, the probability of obtaining no successes in n independent trials is given by,

P(X=0)=Cn0p0(1p)(n0)

n, is the number of trials, and

p, is the probability of getting a success on any trial

The formula to calculate Cnr is,

Cnr=n!r!(nr)!

Calculation:

Consider the event “surviving of plant” as a success.

The probability of any plant surviving in Kerry’s garden is 0.8.

Probability of success is 0.8.

p=0.8

Number of trial is 19.

n=19.

Compute the probability for no successes.

Substitute 19 for n, 0 for r, and 0.8 for p in the formula of P(X=0)

P(X=0)=19!0!19!×(0.8)0×(10.8)19=1×1×(0.2)190

Conclusion:

Thus, the probability that no plant survive is 0.

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
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?
Theorem 2.6 (The Minkowski inequality) Let p≥1. Suppose that X and Y are random variables, such that E|X|P <∞ and E|Y P <00. Then X+YpX+Yp

Chapter 5 Solutions

Beginning Statistics, 2nd Edition

Ch. 5.1 - Prob. 11ECh. 5.1 - Prob. 12ECh. 5.1 - Prob. 13ECh. 5.1 - Prob. 14ECh. 5.1 - Prob. 15ECh. 5.1 - Prob. 16ECh. 5.1 - Prob. 17ECh. 5.1 - Prob. 18ECh. 5.1 - Prob. 19ECh. 5.1 - Prob. 20ECh. 5.1 - Prob. 21ECh. 5.1 - Prob. 22ECh. 5.1 - Prob. 23ECh. 5.2 - Prob. 1ECh. 5.2 - Prob. 2ECh. 5.2 - Prob. 3ECh. 5.2 - Prob. 4ECh. 5.2 - Prob. 5ECh. 5.2 - Prob. 6ECh. 5.2 - Prob. 7ECh. 5.2 - Prob. 8ECh. 5.2 - Prob. 9ECh. 5.2 - Prob. 10ECh. 5.2 - Prob. 11ECh. 5.2 - Prob. 12ECh. 5.2 - Prob. 13ECh. 5.2 - Prob. 14ECh. 5.2 - Prob. 15ECh. 5.2 - Prob. 16ECh. 5.2 - Prob. 17ECh. 5.2 - Prob. 18ECh. 5.2 - Prob. 19ECh. 5.2 - Prob. 20ECh. 5.2 - Prob. 21ECh. 5.2 - Prob. 22ECh. 5.2 - Prob. 23ECh. 5.2 - Prob. 24ECh. 5.2 - Prob. 25ECh. 5.3 - Prob. 1ECh. 5.3 - Prob. 2ECh. 5.3 - Prob. 3ECh. 5.3 - Prob. 4ECh. 5.3 - Prob. 5ECh. 5.3 - Prob. 6ECh. 5.3 - Prob. 7ECh. 5.3 - Prob. 8ECh. 5.3 - Prob. 9ECh. 5.3 - Prob. 10ECh. 5.3 - Prob. 11ECh. 5.3 - Prob. 12ECh. 5.3 - Prob. 13ECh. 5.3 - Prob. 14ECh. 5.3 - Prob. 15ECh. 5.3 - Prob. 16ECh. 5.3 - Prob. 17ECh. 5.3 - Prob. 18ECh. 5.3 - Prob. 19ECh. 5.3 - Prob. 20ECh. 5.3 - Prob. 21ECh. 5.3 - Prob. 22ECh. 5.3 - Prob. 23ECh. 5.3 - Prob. 24ECh. 5.4 - Prob. 1ECh. 5.4 - Prob. 2ECh. 5.4 - Prob. 3ECh. 5.4 - Prob. 4ECh. 5.4 - Prob. 5ECh. 5.4 - Prob. 6ECh. 5.4 - Prob. 7ECh. 5.4 - Prob. 8ECh. 5.4 - Prob. 9ECh. 5.4 - Prob. 10ECh. 5.4 - Prob. 11ECh. 5.4 - Prob. 12ECh. 5.4 - Prob. 13ECh. 5.4 - Prob. 14ECh. 5.4 - Prob. 15ECh. 5.4 - Prob. 16ECh. 5.4 - Prob. 17ECh. 5.4 - Prob. 18ECh. 5.4 - Prob. 19ECh. 5.CR - Prob. 1CRCh. 5.CR - Prob. 2CRCh. 5.CR - Prob. 3CRCh. 5.CR - Prob. 4CRCh. 5.CR - Prob. 5CRCh. 5.CR - Prob. 6CRCh. 5.CR - Prob. 7CRCh. 5.CR - Prob. 8CRCh. 5.CR - Prob. 9CRCh. 5.CR - Prob. 10CRCh. 5.CR - Prob. 11CRCh. 5.CR - Prob. 12CRCh. 5.CR - Prob. 13CRCh. 5.P - Prob. 1PCh. 5.P - Prob. 2PCh. 5.P - Prob. 3PCh. 5.P - Prob. 4PCh. 5.P - Prob. 5PCh. 5.P - Prob. 6PCh. 5.P - Prob. 7PCh. 5.P - Prob. 8PCh. 5.P - Prob. 9PCh. 5.P - Prob. 10PCh. 5.P - Prob. 11PCh. 5.P - Prob. 12PCh. 5.P - Prob. 13PCh. 5.P - Prob. 14PCh. 5.P - Prob. 15PCh. 5.P - Prob. 16PCh. 5.P - Prob. 17P
Knowledge Booster
Background pattern image
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