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 5.3, Problem 69E

(a)

To determine

The values of n and p for X, which is binomially distributed.

(a)

Expert Solution
Check Mark

Answer to Problem 69E

Solution: The required distribution of X is p=0.62 and n=4_.

Explanation of Solution

It is provided that 62% of the selected people admitted that they had received nasty messages on social sites. So, the probability of success for this case is p=0.62. The total number of people, which is 4, is randomly selected, so the total number of trails n=4. Assume X of them admitted that they had received nasty messages. Here X is binomially distributed with parameters p=0.62 and n=4. So, the required distribution of X is p=0.62 and n=4.

(b)

To determine

To find: The probability of every possible value of X.

(b)

Expert Solution
Check Mark

Answer to Problem 69E

Solution: Probability distribution for each value of X is:

X

P(X)

0

0.02085

1

0.1360

2

0.333

3

0.3622

4

0.1477

Explanation of Solution

Calculation: In the provided problem the distribution can be calculated as:

P(X)=(nX)P(1P)X1

For X=0.

P(0)=(40)(0.62)(10.62)4=(40)(0.62)(0.38)4=0.02085

For X=1.

P(1)=(41)(0.62)(10.62)3=(41)(0.62)(0.38)3=0.1360

For X=2.

P(2)=(42)(0.62)(10.62)2=(42)(0.62)(0.38)2=0.333

For X=3.

P(3)=(43)(0.62)(10.62)1=(43)(0.62)(0.38)1=0.3622

For X=4.

P(4)=(44)(0.62)(10.62)0=(44)(0.62)(0.38)0=0.1477

Hence, the table representing every possible value of X and the probability parallel to X is shown below:

X

P(X)

0

0.02085

1

0.1360

2

0.333

3

0.3622

4

0.1477

To determine

To graph: A probability histogram for the obtained distribution.

Expert Solution
Check Mark

Explanation of Solution

Graph: In the provided problem, the distribution of X is obtained in the part above. Now, the probability histogram can be obtained by using Minitab. The steps to be followed are enlisted below:

Step 1: Input the values of X and the probability in Minitab worksheet.

Step 2: Go to Graph > Histogram.

Step 3: Choose ‘View simple’ and click OK.

Step 4: Insert the value of X and the probability p in the column of graph variables.

Step 5: Click on Ok.

The Histogram is obtained as:

Introduction to the Practice of Statistics, Chapter 5.3, Problem 69E , additional homework tip  1

The probability histogram is obtained for the required distribution of X.

(c)

To determine

To Find: The average number of responders who said yes.

(c)

Expert Solution
Check Mark

Answer to Problem 69E

Solution: The required average is 2.48.

Explanation of Solution

Calculation: Binomial mean can be calculated using this formula,

μx=np

Substitute the values p=0.62 and n=4 in the formula above to obtain the mean. Mean can be calculated as:

μx=np=4(0.62)=2.48

The required average is 2.48.

To determine

To graph: The average value in the obtained histogram.

Expert Solution
Check Mark

Explanation of Solution

Graph: The obtained average value is 2.48. To mark this value in the obtained histogram, the steps are as follows:

Step 1: Go to the histogram which is obtained above.

Step 2: Double click on the X axis.

Step 3: Select ‘Position of ticks’ and enter the obtained mean value 2.48.

Step 4: Click on Ok.

The histogram obtained is:

Introduction to the Practice of Statistics, Chapter 5.3, Problem 69E , additional homework tip  2

The histogram with the mark of mean value 2.48 is obtained.

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Chapter 5 Solutions

Introduction to the Practice of Statistics

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. 5.2 - Prob. 17UYKCh. 5.2 - Prob. 18UYKCh. 5.2 - Prob. 19UYKCh. 5.2 - Prob. 20UYKCh. 5.2 - Prob. 21UYKCh. 5.2 - Prob. 22UYKCh. 5.2 - Prob. 23UYKCh. 5.2 - Prob. 24UYKCh. 5.2 - Prob. 25ECh. 5.2 - Prob. 26ECh. 5.2 - Prob. 27ECh. 5.2 - Prob. 28ECh. 5.2 - Prob. 29ECh. 5.2 - Prob. 30ECh. 5.2 - Prob. 31ECh. 5.2 - Prob. 32ECh. 5.2 - Prob. 33ECh. 5.2 - Prob. 34ECh. 5.2 - Prob. 35ECh. 5.2 - Prob. 36ECh. 5.2 - Prob. 37ECh. 5.2 - Prob. 38ECh. 5.2 - Prob. 39ECh. 5.2 - Prob. 40ECh. 5.2 - Prob. 41ECh. 5.2 - Prob. 42ECh. 5.3 - Prob. 43UYKCh. 5.3 - Prob. 44UYKCh. 5.3 - Prob. 45UYKCh. 5.3 - Prob. 46UYKCh. 5.3 - Prob. 47UYKCh. 5.3 - Prob. 48UYKCh. 5.3 - Prob. 49UYKCh. 5.3 - Prob. 50UYKCh. 5.3 - Prob. 51UYKCh. 5.3 - Prob. 52UYKCh. 5.3 - Prob. 53UYKCh. 5.3 - Prob. 54UYKCh. 5.3 - Prob. 55UYKCh. 5.3 - Prob. 56UYKCh. 5.3 - Prob. 57ECh. 5.3 - Prob. 58ECh. 5.3 - Prob. 59ECh. 5.3 - Prob. 60ECh. 5.3 - Prob. 61ECh. 5.3 - Prob. 62ECh. 5.3 - Prob. 63ECh. 5.3 - Prob. 64ECh. 5.3 - Prob. 65ECh. 5.3 - Prob. 66ECh. 5.3 - Prob. 67ECh. 5.3 - Prob. 68ECh. 5.3 - Prob. 69ECh. 5.3 - Prob. 70ECh. 5.3 - Prob. 71ECh. 5.3 - Prob. 72ECh. 5.3 - Prob. 73ECh. 5.3 - Prob. 74ECh. 5.3 - Prob. 75ECh. 5.3 - Prob. 76ECh. 5.3 - Prob. 77ECh. 5.3 - Prob. 78ECh. 5.3 - Prob. 79ECh. 5.3 - Prob. 80ECh. 5.3 - Prob. 81ECh. 5.3 - Prob. 82ECh. 5.3 - Prob. 83ECh. 5 - Prob. 84ECh. 5 - Prob. 85ECh. 5 - Prob. 86ECh. 5 - Prob. 87ECh. 5 - Prob. 88ECh. 5 - Prob. 89ECh. 5 - Prob. 90ECh. 5 - Prob. 91ECh. 5 - Prob. 92ECh. 5 - Prob. 93ECh. 5 - Prob. 94ECh. 5 - Prob. 95ECh. 5 - Prob. 96ECh. 5 - Prob. 97ECh. 5 - Prob. 98ECh. 5 - Prob. 99ECh. 5 - Prob. 100ECh. 5 - Prob. 101ECh. 5 - Prob. 102E
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