Introduction to the Practice of Statistics: w/CrunchIt/EESEE Access Card
Introduction to the Practice of Statistics: w/CrunchIt/EESEE Access Card
8th Edition
ISBN: 9781464158933
Author: David S. Moore, George P. McCabe, Bruce A. Craig
Publisher: W. H. Freeman
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Chapter 5.2, Problem 70E

(a)

To determine

Whether the provided expression is true or not.

(a)

Expert Solution
Check Mark

Answer to Problem 70E

Solution: The provided expression is true.

Explanation of Solution

Calculation: Binomial coefficient is defined as the number of ways of ordering k successes among n observations. The formula is

Cnk=n!k!(nk)!   , k=0,1,2,,n

Where,

n!=n×(n1)×(n2)××3×2×1

Now, by using binomial coefficient, the provided expression can be calculated as:

Cnn=n!n!(nn)! =n!n!×0!=n!n!×1=1

Hence, the provided expression is true.

To determine

To explain: The provided expression in words.

Expert Solution
Check Mark

Answer to Problem 70E

Solution: From the above solution, it can be seen that there is only one way of distributing ‘n’ successes among ‘n’ trials.

Explanation of Solution

By using binomial coefficient, the provided expression shows the number of ways of ordering ‘n’ successes among ‘n’ observations. Hence, from the above solution, it can be seen that there is only one way of ordering ‘n’ successes among ‘n’ observations.

(b)

To determine

To test: Whether the provided expression is true or not.

(b)

Expert Solution
Check Mark

Answer to Problem 70E

Solution: The provided expression is true.

Explanation of Solution

Calculation: By using binomial coefficient, the provided expression can be calculated as:

Cnn1=n!(n1)!{n(n1)}! =n!(n1)!×1!=n×(n1)!(n1)!×1=n

Conclusion: Hence, the provided expression is true.

To determine

To explain: The provided expression in words.

Expert Solution
Check Mark

Answer to Problem 70E

Solution: From the above solution, it can be seen that there are only n ways of distributing n1 successes among ‘n’ trials.

Explanation of Solution

By using binomial coefficient, the provided expression is shows the number of ways of ordering n1 successes among ‘n’ observations. Hence, from the above solution, it can be seen that there are n ways of ordering n1 successes among n observations.

(c)

To determine

To test: Whether the provided expression is true or not.

(c)

Expert Solution
Check Mark

Answer to Problem 70E

Solution: The provided expression is true.

Explanation of Solution

Calculation: By using binomial coefficient, the provided expression can be calculated as:

L.H.S,

Cnk=n!k!(nk)!

R.H.S,

Cnnk=n!(nk)!{n(nk)}!=n!(nk)!{nn+k}!=n!(nk)!k!

From the above solution, it can be seen that L.H.S = R.H.S. Hence, the provided expression is true.

To determine

To explain: The provided expression in words.

Expert Solution
Check Mark

Answer to Problem 70E

Solution: From the above solution, it can be seen that the number of ways of distributing ‘k’ successes to ‘n’ trials is exactly equal to the number of ways of distributing nk successes to ‘n’ trials.

Explanation of Solution

By using binomial coefficient, the provided expression Cnk is shows the number of ways of ordering ‘k’ successes among ‘n’ observations and the expression Cnnk is defined as the number of ways of ordering nk successes among ‘n’ observations. Hence, from the above solution, it can be seen that the number of ways of ordering ‘k’ successes among ‘n’ observations is equal to the number of ways of ordering nk observations among ‘n’ observations, that is, Cnk=Cnnk.

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Chapter 5 Solutions

Introduction to the Practice of Statistics: w/CrunchIt/EESEE Access Card

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. 5.1 - Prob. 24ECh. 5.1 - Prob. 25ECh. 5.1 - Prob. 26ECh. 5.1 - Prob. 27ECh. 5.1 - Prob. 28ECh. 5.1 - Prob. 29ECh. 5.1 - Prob. 30ECh. 5.1 - Prob. 31ECh. 5.1 - Prob. 32ECh. 5.1 - Prob. 33ECh. 5.1 - Prob. 34ECh. 5.2 - Prob. 35UYKCh. 5.2 - Prob. 36UYKCh. 5.2 - Prob. 37UYKCh. 5.2 - Prob. 38UYKCh. 5.2 - Prob. 39UYKCh. 5.2 - Prob. 40UYKCh. 5.2 - Prob. 41UYKCh. 5.2 - Prob. 42UYKCh. 5.2 - Prob. 43UYKCh. 5.2 - Prob. 44UYKCh. 5.2 - Prob. 45UYKCh. 5.2 - Prob. 46ECh. 5.2 - Prob. 47ECh. 5.2 - Prob. 48ECh. 5.2 - Prob. 49ECh. 5.2 - Prob. 50ECh. 5.2 - Prob. 51ECh. 5.2 - Prob. 52ECh. 5.2 - Prob. 53ECh. 5.2 - Prob. 54ECh. 5.2 - Prob. 55ECh. 5.2 - Prob. 56ECh. 5.2 - Prob. 57ECh. 5.2 - Prob. 58ECh. 5.2 - Prob. 59ECh. 5.2 - Prob. 60ECh. 5.2 - Prob. 61ECh. 5.2 - Prob. 62ECh. 5.2 - Prob. 63ECh. 5.2 - Prob. 64ECh. 5.2 - Prob. 65ECh. 5.2 - Prob. 66ECh. 5.2 - Prob. 67ECh. 5.2 - Prob. 68ECh. 5.2 - Prob. 69ECh. 5.2 - Prob. 70ECh. 5.2 - Prob. 71ECh. 5.2 - Prob. 72ECh. 5.2 - Prob. 73ECh. 5.2 - Prob. 74ECh. 5.2 - Prob. 75ECh. 5 - Prob. 76ECh. 5 - Prob. 77ECh. 5 - Prob. 78ECh. 5 - Prob. 79ECh. 5 - Prob. 80ECh. 5 - Prob. 81ECh. 5 - Prob. 82ECh. 5 - Prob. 83ECh. 5 - Prob. 84ECh. 5 - Prob. 85ECh. 5 - Prob. 86ECh. 5 - Prob. 87ECh. 5 - Prob. 88ECh. 5 - Prob. 89ECh. 5 - Prob. 90E
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