BEGINNING STAT.-SOFTWARE+EBOOK ACCESS
BEGINNING STAT.-SOFTWARE+EBOOK ACCESS
2nd Edition
ISBN: 9781941552506
Author: WARREN
Publisher: HAWKES LRN
Question
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Chapter 5.2, Problem 25E
To determine

(a)

To find:

The specified probability for the given scenario.

Expert Solution
Check Mark

Answer to Problem 25E

Solution:

Value of the required probability is 0.9952.

Explanation of Solution

Given information:

Ronnie owns a fireworks stand and knows that in the fireworks business, 1 out of every 13 fireworks is a dud. Suppose that Juanita buys 10 firecrackers at Ronnie’s stand.

Formula used:

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

1P(X>r)=P(Xr)

P(Xr)=P(X=0)+P(X=1)+P(X=2)...+P(X=r)

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 “duds” as a success.

Probability of winning is 1 out of 13.

Probability of success is 113.

p=1130.076

Number of trial is 10.

n=10.

Compute the probability for no more than 3 successes.

Substitute 3 for r, in the formula for no more than r successes.

1P(X>3)=P(X3)

P(X3)=P(X=0)+P(X=1)+P(X=2)+P(X=3)

Substitute 10 for n, 0 for r, and 0.076 for p in the formula of P(X=r) to calculate P(X=0)

P(X=0)=10!0!10!×(0.076)0×(10.076)10=1×1×(0.924)10=0.4536487434

Substitute 10 for n, 1 for r, and 0.076 for p in the formula of P(X=r) to calculate P(X=1)

P(X=1)=10!1!9!×(0.076)1×(10.076)9=10×(0.076)×(0.924)9=0.373131001

Substitute 10 for n, 2 for r, and 0.076 for p in the formula of P(X=r) to calculate P(X=2)

P(X=2)=10!2!8!×(0.076)2×(10.076)8=45×(0.076)2×(0.924)8=0.138106929

Substitute 10 for n, 3 for r, and 0.076 for p in the formula of P(X=r) to calculate P(X=3)

P(X=3)=10!3!7!×(0.076)3×(10.076)7=120×(0.076)3×(0.924)7=0.03029185166

Add the values of P(X=0),P(X=1),P(X=2),andP(X=3).

P(X3)=P(X=0)+P(X=1)+P(X=2)+P(X=3)=0.4536487434+0.373131001+0.138106929+0.030291851660.9952

Conclusion:

Thus, the probability that no more than 3 are duds is 0.9952.

To determine

(b)

To find:

The specified probability for the given scenario.

Expert Solution
Check Mark

Answer to Problem 25E

Solution:

Value of the required probability is 0.45.

Explanation of Solution

Given information:

Ronnie owns a fireworks stand and knows that in the fireworks business, 1 out of every 13 fireworks is a dud. Suppose that Juanita buys 10 firecrackers at Ronnie’s stand.

Formula used:

For a binomial random variable X, the probability of no success 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 “duds” as a success.

Probability of winning is 1 out of 13.

Probability of success is 113.

p=1130.076

Number of trial is 10.

n=10.

Compute the probability for no success.

Substitute 10 for n, 0 for r, and 0.076 for p in the formula of P(X=r) to calculate P(X=0)

P(X=0)=10!0!10!×(0.076)0×(10.076)10=1×1×(0.924)100.45

P(X=0)=0.45

Conclusion:

Thus, the probability of no duds is 0.45.

(c)

To determine

To find:

The specified probability for the given scenario.

(c)

Expert Solution
Check Mark

Answer to Problem 25E

Solution:

Value of the required probability is 0.00003.

Explanation of Solution

Given information:

Ronnie owns a fireworks stand and knows that in the fireworks business, 1 out of every 13 fireworks is a dud. Suppose that Juanita buys 10 firecrackers at Ronnie’s stand.

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 “duds” as a success.

Probability of winning is 1 out of 13.

Probability of success is 113.

p=1130.076

Number of trial is 10.

n=10.

Consider number of more than half of the crackers as a number of success.

r>102=5

Compute the probability for more than 5 successes.

Substitute 5 for r, 10 for n in the formula for more than r successes.

P(X>5)=P(X=6)+P(X=7)+P(X=8)+P(X=9)+P(X=10)

Substitute 10 for n, 6 for r, and 0.076 for p in the formula of P(X=r) to calculate P(X=6)

P(X=6)=10!6!4!×(0.076)6×(10.076)4=210×(0.076)6×(0.924)4=0.0000294977393

Substitute 10 for n, 7 for r, and 0.076 for p in the formula of P(X=r) to calculate P(X=7)

P(X=7)=10!7!3!×(0.076)7×(10.076)3=120×(0.076)7×(0.924)3=0.00000138641199

Substitute 10 for n, 8 for r, and 0.076 for p in the formula of P(X=r) to calculate P(X=8)

P(X=8)=10!8!2!×(0.076)8×(10.076)2=45×(0.076)8×(0.924)20

Substitute 10 for n, 9 for r, and 0.076 for p in the formula of P(X=r) to calculate P(X=9)

P(X=9)=10!9!1!×(0.076)9×(10.076)1=10×(0.076)9×(0.924)10

Substitute 10 for n, 10 for r, and 0.076 for p in the formula of P(X=r) to calculate P(X=10)

P(X=10)=10!10!0!×(0.076)10×(10.076)0=1×(0.076)10×(0.924)00

Add all the values of

P(X=6),P(X=7),P(X=8),P(X=9),andP(X=10)

P(X=6)+P(X=7)+P(X=8)+P(X=9)+P(X=10)=0.0000294977393+0.00000138641199+0+0+00.00003

P(X10)=0.000016

Conclusion:

Thus, the probability of more than half duds is 0.00003.

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Chapter 5 Solutions

BEGINNING STAT.-SOFTWARE+EBOOK ACCESS

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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