Beginning Statistics, 2nd Edition
Beginning Statistics, 2nd Edition
2nd Edition
ISBN: 9781932628678
Author: Carolyn Warren; Kimberly Denley; Emily Atchley
Publisher: Hawkes Learning Systems
Question
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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.

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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
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