Introduction to the Practice of Statistics
Introduction to the Practice of Statistics
9th Edition
ISBN: 9781319013387
Author: David S. Moore, George P. McCabe, Bruce A. Craig
Publisher: W. H. Freeman
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Chapter 4.4, Problem 83E

(a)

Section 1:

To determine

The mean of the number of heads.

(a)

Section 1:

Expert Solution
Check Mark

Answer to Problem 83E

Solution: The mean of the number of heads (μX) is 0.5.

Explanation of Solution

The chance of the occurrence of the number of heads in the tossing of single coin is either 0 or 1, which means each has probability of 12. So, the formula for mean is equal to the expectation of chance. Thus, the mean can be calculated as shown below:

Mean(μX)=i=12xi×p(xi)=0×0.5+1×0.5=0.5

Section 2:

To determine

The standard deviation of number of heads.

Section 2:

Expert Solution
Check Mark

Answer to Problem 83E

Solution: The standard deviation (σX) is 0.5.

Explanation of Solution

The chance of the occurrence of number of heads in the tossing of single coin is 0 or 1, which means each has probability of 12. So, the formula for variance is as follows:

Variance(σX2)=(xiμX)2×pi

And it is calculated as

Variance(σX2)=(xiμX)2×pi=(00.5)2×12+(10.5)2×12=0.25

So, the standard deviation is calculated as

Standard deviation (σ)=Variance =0.25=0.5

(b)

Section 1:

To determine

The mean of the number of heads.

(b)

Section 1:

Expert Solution
Check Mark

Answer to Problem 83E

Solution: The mean is 2.

Explanation of Solution

The coin is tossed four times, so there will be five possible numbers of heads (0, 1, 2, 3, 4). So, the number of arrangements for the number of heads is as follows:

For 0 head, there will be 4C0=1 arrangement that has no heads.

For 1 head, there will be 4C1=4 arrangements that have 1 head.

For 2 heads, there will be 4C2=6 arrangements that have 2 heads.

For 3 heads, there will be 4C3=4 arrangements that have 3 heads.

For 4 heads, there will be 4C4=1 arrangement that has 4 heads.

So, there will be total 16(1+4+6+4+1) arrangements. Thus, the corresponding probabilities are as follows:

Number of heads (xi)

Favorable arrangements

Probability (pi)

0

1

1/16

1

4

4/16

2

6

6/16

3

4

4/16

4

1

1/16

Now, the formula of mean is as follows:

Mean(μX)=i=15xi×pi

And it is calculated as

Mean(μX)=i=15xi×pi=(0×116)+(1×416)+(2×616)+(3×416)+(4×116)=2

Section 2:

To determine

The standard deviation.

Section 2:

Expert Solution
Check Mark

Answer to Problem 83E

Solution: The standard deviation is 1.

Explanation of Solution

The formula for standard deviation is as follows:

Standarddeviation (σX)=i=15(xiμX)2×pi

And it is calculated as

Standarddeviation (σX)=i=15(xiμX)2×pi=((02)2×116+(12)2×416+(22)2×616+(32)2×416+(42)2×116)=1

(c)

Section 1:

To determine

The mean using the distribution provided in Example 4.23.

(c)

Section 1:

Expert Solution
Check Mark

Answer to Problem 83E

Solution: The mean is 2.

Explanation of Solution

The mean for the referred distribution is calculated by using the formula:

MeanμX=i=15xi×pi=0×0.0625+1×0.25+2×0.375+3×0.25+4×0.0625=2

Section 2:

To determine

The standard deviation using the distribution provided in Example 4.23.

Section 2:

Expert Solution
Check Mark

Answer to Problem 83E

Solution: The standard deviation is 1.

Explanation of Solution

The standard deviation for the referred distribution is calculated by using the formula:

Standard deviation (σX)=i=15(xiμX)2×pi

And it is calculated as

Standard deviation (σX)=i=15(xiμX)2×pi=((02)2×0.0625+(12)2×0.25+(22)2×0.375+(32)2×0.25+(42)2×0.0625)=1

Section 3:

To determine

Whether the results obtained in part (b) and part (c) are the same.

Section 3:

Expert Solution
Check Mark

Answer to Problem 83E

Solution: Yes, the results obtained in part (b) and parts (c) are same.

Explanation of Solution

The mean and standard deviation obtained in part (b) are 2 and 1, respectively. Similarly, the mean and standard deviation obtained in part (c) are 2 and 1, respectively. Hence, the values obtained for the mean and standard deviation is the same in both the parts.

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Chapter 4 Solutions

Introduction to the Practice of Statistics

Ch. 4.2 - Prob. 11UYKCh. 4.2 - Prob. 12UYKCh. 4.2 - Prob. 13UYKCh. 4.2 - Prob. 14UYKCh. 4.2 - Prob. 15UYKCh. 4.2 - Prob. 16UYKCh. 4.2 - Prob. 17ECh. 4.2 - Prob. 18ECh. 4.2 - Prob. 19ECh. 4.2 - Prob. 20ECh. 4.2 - Prob. 21ECh. 4.2 - Prob. 22ECh. 4.2 - Prob. 23ECh. 4.2 - Prob. 24ECh. 4.2 - Prob. 25ECh. 4.2 - Prob. 26ECh. 4.2 - Prob. 27ECh. 4.2 - Prob. 28ECh. 4.2 - Prob. 29ECh. 4.2 - Prob. 30ECh. 4.2 - Prob. 31ECh. 4.2 - Prob. 32ECh. 4.2 - Prob. 33ECh. 4.2 - Prob. 34ECh. 4.2 - Prob. 35ECh. 4.2 - Prob. 36ECh. 4.2 - Prob. 37ECh. 4.2 - Prob. 38ECh. 4.2 - Prob. 39ECh. 4.2 - Prob. 40ECh. 4.2 - Prob. 41ECh. 4.3 - Prob. 42UYKCh. 4.3 - Prob. 43UYKCh. 4.3 - Prob. 44UYKCh. 4.3 - Prob. 45ECh. 4.3 - Prob. 46ECh. 4.3 - Prob. 47ECh. 4.3 - Prob. 48ECh. 4.3 - Prob. 49ECh. 4.3 - Prob. 50ECh. 4.3 - Prob. 51ECh. 4.3 - Prob. 52ECh. 4.3 - Prob. 53ECh. 4.3 - Prob. 54ECh. 4.3 - Prob. 55ECh. 4.3 - Prob. 56ECh. 4.3 - Prob. 57ECh. 4.3 - Prob. 58ECh. 4.3 - Prob. 59ECh. 4.3 - Prob. 60ECh. 4.3 - Prob. 61ECh. 4.3 - Prob. 62ECh. 4.4 - Prob. 63UYKCh. 4.4 - Prob. 64UYKCh. 4.4 - Prob. 65UYKCh. 4.4 - Prob. 66UYKCh. 4.4 - Prob. 67UYKCh. 4.4 - Prob. 68ECh. 4.4 - Prob. 69ECh. 4.4 - Prob. 70ECh. 4.4 - Prob. 71ECh. 4.4 - Prob. 72ECh. 4.4 - Prob. 73ECh. 4.4 - Prob. 74ECh. 4.4 - Prob. 75ECh. 4.4 - Prob. 76ECh. 4.4 - Prob. 77ECh. 4.4 - Prob. 78ECh. 4.4 - Prob. 79ECh. 4.4 - Prob. 80ECh. 4.4 - Prob. 81ECh. 4.4 - Prob. 82ECh. 4.4 - Prob. 83ECh. 4.4 - Prob. 84ECh. 4.4 - Prob. 85ECh. 4.4 - Prob. 86ECh. 4.4 - Prob. 87ECh. 4.4 - Prob. 88ECh. 4.5 - Prob. 89UYKCh. 4.5 - Prob. 90UYKCh. 4.5 - Prob. 91UYKCh. 4.5 - Prob. 92UYKCh. 4.5 - Prob. 93UYKCh. 4.5 - Prob. 94UYKCh. 4.5 - Prob. 95UYKCh. 4.5 - Prob. 96ECh. 4.5 - Prob. 97ECh. 4.5 - Prob. 98ECh. 4.5 - Prob. 99ECh. 4.5 - Prob. 100ECh. 4.5 - Prob. 101ECh. 4.5 - Prob. 102ECh. 4.5 - Prob. 103ECh. 4.5 - Prob. 104ECh. 4.5 - Prob. 105ECh. 4.5 - Prob. 106ECh. 4.5 - Prob. 107ECh. 4.5 - Prob. 108ECh. 4.5 - Prob. 109ECh. 4.5 - Prob. 110ECh. 4.5 - Prob. 111ECh. 4.5 - Prob. 112ECh. 4.5 - Prob. 113ECh. 4.5 - Prob. 114ECh. 4.5 - Prob. 115ECh. 4.5 - Prob. 116ECh. 4.5 - Prob. 117ECh. 4.5 - Prob. 118ECh. 4.5 - Prob. 119ECh. 4.5 - Prob. 120ECh. 4.5 - Prob. 121ECh. 4.5 - Prob. 122ECh. 4.5 - Prob. 123ECh. 4 - Prob. 124ECh. 4 - Prob. 125ECh. 4 - Prob. 126ECh. 4 - Prob. 127ECh. 4 - Prob. 128ECh. 4 - Prob. 129ECh. 4 - Prob. 130ECh. 4 - Prob. 131ECh. 4 - Prob. 132ECh. 4 - Prob. 133ECh. 4 - Prob. 134ECh. 4 - Prob. 135ECh. 4 - Prob. 136ECh. 4 - Prob. 137ECh. 4 - Prob. 138ECh. 4 - Prob. 139E
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