To Construct the probability distribution of the given data.

Question

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Answer 1

Question

Chapter 6, Problem 4RE

a.

To determine

To Construct the probability distribution of the given data.

a.

Expert Solution

Explanation of Solution

Given:

Number of Tests	Frequency (in 1000s)
1	953
2	423
3	194
4	80
5	29
6	9
7	3
8	1
Total	1692

To get the probability distribution of no. of tests taken, it is required to divide the respective frequency of no. of test taken by the total frequency.

The probability distribution is given below:

Number of Tests	Frequency (in 1000s)	P (X = x)
1	953	0.563
2	423	0.250
3	194	0.115
4	80	0.047
5	29	0.017
6	9	0.005
7	3	0.002
8	1	0.001
Total	1692	1.000

b.

To determine

To find the probability that student took only one exam.

b.

Expert Solution

Answer to Problem 4RE

Probability that student took only one exam = 0.563

Explanation of Solution

Calculation:

To get the probability distribution of no. of tests taken, it is required to divide the respective frequency of no. of test taken by the total frequency.

The probability distribution is given below:

Number of Tests	Frequency (in 1000s)	P (X = x)
1	953	0.563
2	423	0.250
3	194	0.115
4	80	0.047
5	29	0.017
6	9	0.005
7	3	0.002
8	1	0.001
Total	1692	1.000

Calculation:

P(That student took only one exam)=9531692=0.563

c.

To determine

To find the mean μx

c.

Expert Solution

Answer to Problem 4RE

The mean of the given distribution is 1.73.

Explanation of Solution

Formula Used:

μx=∑x.p(x)

Calculation:

Number of Tests	Frequency (in 1000s)	P (X = x)	x.P(x)
1	953	0.563	0.56
2	423	0.250	0.50
3	194	0.115	0.34
4	80	0.047	0.19
5	29	0.017	0.09
6	9	0.005	0.03
7	3	0.002	0.01
8	1	0.001	0.00
Total	1692	1.000	1.73

The mean μx of a given distribution is 1.73, because the value of ∑x.p(x) = 1.73. Therefore, the mean is 1.73.

d.

To determine

To find the mean μx

d.

Expert Solution

Answer to Problem 4RE

Standard Deviation = 1.05

Explanation of Solution

Formula Used:

μx=∑x.p(x)

Calculation:

Number of Tests	Frequency (in 1000s)	P (X = x)	x.P(x)	x2.P(x)
1	953	0.563	0.56	0.56
2	423	0.250	0.50	1.00
3	194	0.115	0.34	1.03
4	80	0.047	0.19	0.76
5	29	0.017	0.09	0.43
6	9	0.005	0.03	0.19
7	3	0.002	0.01	0.09
8	1	0.001	0.00	0.04
Total	1692	1.000	1.73	4.10

Standard Deviation = ∑x2p(x)− [ xp( x )]2=4.10− 1.732=1.05

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Answer 2

Question

Chapter 6, Problem 4RE

a.

To determine

To Construct the probability distribution of the given data.

a.

Expert Solution

Explanation of Solution

Given:

Number of Tests	Frequency (in 1000s)
1	953
2	423
3	194
4	80
5	29
6	9
7	3
8	1
Total	1692

To get the probability distribution of no. of tests taken, it is required to divide the respective frequency of no. of test taken by the total frequency.

The probability distribution is given below:

Number of Tests	Frequency (in 1000s)	P (X = x)
1	953	0.563
2	423	0.250
3	194	0.115
4	80	0.047
5	29	0.017
6	9	0.005
7	3	0.002
8	1	0.001
Total	1692	1.000

b.

To determine

To find the probability that student took only one exam.

b.

Expert Solution

Answer to Problem 4RE

Probability that student took only one exam = 0.563

Explanation of Solution

Calculation:

To get the probability distribution of no. of tests taken, it is required to divide the respective frequency of no. of test taken by the total frequency.

The probability distribution is given below:

Number of Tests	Frequency (in 1000s)	P (X = x)
1	953	0.563
2	423	0.250
3	194	0.115
4	80	0.047
5	29	0.017
6	9	0.005
7	3	0.002
8	1	0.001
Total	1692	1.000

Calculation:

P(That student took only one exam)=9531692=0.563

c.

To determine

To find the mean μx

c.

Expert Solution

Answer to Problem 4RE

The mean of the given distribution is 1.73.

Explanation of Solution

Formula Used:

μx=∑x.p(x)

Calculation:

Number of Tests	Frequency (in 1000s)	P (X = x)	x.P(x)
1	953	0.563	0.56
2	423	0.250	0.50
3	194	0.115	0.34
4	80	0.047	0.19
5	29	0.017	0.09
6	9	0.005	0.03
7	3	0.002	0.01
8	1	0.001	0.00
Total	1692	1.000	1.73

The mean μx of a given distribution is 1.73, because the value of ∑x.p(x) = 1.73. Therefore, the mean is 1.73.

d.

To determine

To find the mean μx

d.

Expert Solution

Answer to Problem 4RE

Standard Deviation = 1.05

Explanation of Solution

Formula Used:

μx=∑x.p(x)

Calculation:

Number of Tests	Frequency (in 1000s)	P (X = x)	x.P(x)	x2.P(x)
1	953	0.563	0.56	0.56
2	423	0.250	0.50	1.00
3	194	0.115	0.34	1.03
4	80	0.047	0.19	0.76
5	29	0.017	0.09	0.43
6	9	0.005	0.03	0.19
7	3	0.002	0.01	0.09
8	1	0.001	0.00	0.04
Total	1692	1.000	1.73	4.10

Standard Deviation = ∑x2p(x)− [ xp( x )]2=4.10− 1.732=1.05

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Answer 3

Question

Chapter 6, Problem 4RE

a.

To determine

To Construct the probability distribution of the given data.

a.

Expert Solution

Explanation of Solution

Given:

Number of Tests	Frequency (in 1000s)
1	953
2	423
3	194
4	80
5	29
6	9
7	3
8	1
Total	1692

To get the probability distribution of no. of tests taken, it is required to divide the respective frequency of no. of test taken by the total frequency.

The probability distribution is given below:

Number of Tests	Frequency (in 1000s)	P (X = x)
1	953	0.563
2	423	0.250
3	194	0.115
4	80	0.047
5	29	0.017
6	9	0.005
7	3	0.002
8	1	0.001
Total	1692	1.000

b.

To determine

To find the probability that student took only one exam.

b.

Expert Solution

Answer to Problem 4RE

Probability that student took only one exam = 0.563

Explanation of Solution

Calculation:

To get the probability distribution of no. of tests taken, it is required to divide the respective frequency of no. of test taken by the total frequency.

The probability distribution is given below:

Number of Tests	Frequency (in 1000s)	P (X = x)
1	953	0.563
2	423	0.250
3	194	0.115
4	80	0.047
5	29	0.017
6	9	0.005
7	3	0.002
8	1	0.001
Total	1692	1.000

Calculation:

P(That student took only one exam)=9531692=0.563

c.

To determine

To find the mean μx

c.

Expert Solution

Answer to Problem 4RE

The mean of the given distribution is 1.73.

Explanation of Solution

Formula Used:

μx=∑x.p(x)

Calculation:

Number of Tests	Frequency (in 1000s)	P (X = x)	x.P(x)
1	953	0.563	0.56
2	423	0.250	0.50
3	194	0.115	0.34
4	80	0.047	0.19
5	29	0.017	0.09
6	9	0.005	0.03
7	3	0.002	0.01
8	1	0.001	0.00
Total	1692	1.000	1.73

The mean μx of a given distribution is 1.73, because the value of ∑x.p(x) = 1.73. Therefore, the mean is 1.73.

d.

To determine

To find the mean μx

d.

Expert Solution

Answer to Problem 4RE

Standard Deviation = 1.05

Explanation of Solution

Formula Used:

μx=∑x.p(x)

Calculation:

Number of Tests	Frequency (in 1000s)	P (X = x)	x.P(x)	x2.P(x)
1	953	0.563	0.56	0.56
2	423	0.250	0.50	1.00
3	194	0.115	0.34	1.03
4	80	0.047	0.19	0.76
5	29	0.017	0.09	0.43
6	9	0.005	0.03	0.19
7	3	0.002	0.01	0.09
8	1	0.001	0.00	0.04
Total	1692	1.000	1.73	4.10

Standard Deviation = ∑x2p(x)− [ xp( x )]2=4.10− 1.732=1.05

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Answer 4

Question

Chapter 6, Problem 4RE

a.

To determine

To Construct the probability distribution of the given data.

a.

Expert Solution

Explanation of Solution

Given:

Number of Tests	Frequency (in 1000s)
1	953
2	423
3	194
4	80
5	29
6	9
7	3
8	1
Total	1692

To get the probability distribution of no. of tests taken, it is required to divide the respective frequency of no. of test taken by the total frequency.

The probability distribution is given below:

Number of Tests	Frequency (in 1000s)	P (X = x)
1	953	0.563
2	423	0.250
3	194	0.115
4	80	0.047
5	29	0.017
6	9	0.005
7	3	0.002
8	1	0.001
Total	1692	1.000

b.

To determine

To find the probability that student took only one exam.

b.

Expert Solution

Answer to Problem 4RE

Probability that student took only one exam = 0.563

Explanation of Solution

Calculation:

To get the probability distribution of no. of tests taken, it is required to divide the respective frequency of no. of test taken by the total frequency.

The probability distribution is given below:

Number of Tests	Frequency (in 1000s)	P (X = x)
1	953	0.563
2	423	0.250
3	194	0.115
4	80	0.047
5	29	0.017
6	9	0.005
7	3	0.002
8	1	0.001
Total	1692	1.000

Calculation:

P(That student took only one exam)=9531692=0.563

c.

To determine

To find the mean μx

c.

Expert Solution

Answer to Problem 4RE

The mean of the given distribution is 1.73.

Explanation of Solution

Formula Used:

μx=∑x.p(x)

Calculation:

Number of Tests	Frequency (in 1000s)	P (X = x)	x.P(x)
1	953	0.563	0.56
2	423	0.250	0.50
3	194	0.115	0.34
4	80	0.047	0.19
5	29	0.017	0.09
6	9	0.005	0.03
7	3	0.002	0.01
8	1	0.001	0.00
Total	1692	1.000	1.73

The mean μx of a given distribution is 1.73, because the value of ∑x.p(x) = 1.73. Therefore, the mean is 1.73.

d.

To determine

To find the mean μx

d.

Expert Solution

Answer to Problem 4RE

Standard Deviation = 1.05

Explanation of Solution

Formula Used:

μx=∑x.p(x)

Calculation:

Number of Tests	Frequency (in 1000s)	P (X = x)	x.P(x)	x2.P(x)
1	953	0.563	0.56	0.56
2	423	0.250	0.50	1.00
3	194	0.115	0.34	1.03
4	80	0.047	0.19	0.76
5	29	0.017	0.09	0.43
6	9	0.005	0.03	0.19
7	3	0.002	0.01	0.09
8	1	0.001	0.00	0.04
Total	1692	1.000	1.73	4.10

Standard Deviation = ∑x2p(x)− [ xp( x )]2=4.10− 1.732=1.05

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Answer 5

Chapter 6, Problem 4RE

a.

To determine

To Construct the probability distribution of the given data.

a.

Expert Solution

Explanation of Solution

Given:

Number of Tests	Frequency (in 1000s)
1	953
2	423
3	194
4	80
5	29
6	9
7	3
8	1
Total	1692

To get the probability distribution of no. of tests taken, it is required to divide the respective frequency of no. of test taken by the total frequency.

The probability distribution is given below:

Number of Tests	Frequency (in 1000s)	P (X = x)
1	953	0.563
2	423	0.250
3	194	0.115
4	80	0.047
5	29	0.017
6	9	0.005
7	3	0.002
8	1	0.001
Total	1692	1.000

b.

To determine

To find the probability that student took only one exam.

b.

Expert Solution

Answer to Problem 4RE

Probability that student took only one exam = 0.563

Explanation of Solution

Calculation:

To get the probability distribution of no. of tests taken, it is required to divide the respective frequency of no. of test taken by the total frequency.

The probability distribution is given below:

Number of Tests	Frequency (in 1000s)	P (X = x)
1	953	0.563
2	423	0.250
3	194	0.115
4	80	0.047
5	29	0.017
6	9	0.005
7	3	0.002
8	1	0.001
Total	1692	1.000

Calculation:

P(That student took only one exam)=9531692=0.563

c.

To determine

To find the mean μx

c.

Expert Solution

Answer to Problem 4RE

The mean of the given distribution is 1.73.

Explanation of Solution

Formula Used:

μx=∑x.p(x)

Calculation:

Number of Tests	Frequency (in 1000s)	P (X = x)	x.P(x)
1	953	0.563	0.56
2	423	0.250	0.50
3	194	0.115	0.34
4	80	0.047	0.19
5	29	0.017	0.09
6	9	0.005	0.03
7	3	0.002	0.01
8	1	0.001	0.00
Total	1692	1.000	1.73

The mean μx of a given distribution is 1.73, because the value of ∑x.p(x) = 1.73. Therefore, the mean is 1.73.

d.

To determine

To find the mean μx

d.

Expert Solution

Answer to Problem 4RE

Standard Deviation = 1.05

Explanation of Solution

Formula Used:

μx=∑x.p(x)

Calculation:

Number of Tests	Frequency (in 1000s)	P (X = x)	x.P(x)	x2.P(x)
1	953	0.563	0.56	0.56
2	423	0.250	0.50	1.00
3	194	0.115	0.34	1.03
4	80	0.047	0.19	0.76
5	29	0.017	0.09	0.43
6	9	0.005	0.03	0.19
7	3	0.002	0.01	0.09
8	1	0.001	0.00	0.04
Total	1692	1.000	1.73	4.10

Standard Deviation = ∑x2p(x)− [ xp( x )]2=4.10− 1.732=1.05

Answer 6

Chapter 6, Problem 4RE

a.

To determine

To Construct the probability distribution of the given data.

a.

Expert Solution

Explanation of Solution

Given:

Number of Tests	Frequency (in 1000s)
1	953
2	423
3	194
4	80
5	29
6	9
7	3
8	1
Total	1692

To get the probability distribution of no. of tests taken, it is required to divide the respective frequency of no. of test taken by the total frequency.

The probability distribution is given below:

Number of Tests	Frequency (in 1000s)	P (X = x)
1	953	0.563
2	423	0.250
3	194	0.115
4	80	0.047
5	29	0.017
6	9	0.005
7	3	0.002
8	1	0.001
Total	1692	1.000

Answer 7

Chapter 6, Problem 4RE

a.

To determine

To Construct the probability distribution of the given data.

Answer 8

a.

Expert Solution

Explanation of Solution

Given:

Number of Tests	Frequency (in 1000s)
1	953
2	423
3	194
4	80
5	29
6	9
7	3
8	1
Total	1692

To get the probability distribution of no. of tests taken, it is required to divide the respective frequency of no. of test taken by the total frequency.

The probability distribution is given below:

Number of Tests	Frequency (in 1000s)	P (X = x)
1	953	0.563
2	423	0.250
3	194	0.115
4	80	0.047
5	29	0.017
6	9	0.005
7	3	0.002
8	1	0.001
Total	1692	1.000

Answer 9

a.

Expert Solution

Answer 10

a.

Expert Solution

Answer 11

Expert Solution

Answer 12

b.

To determine

To find the probability that student took only one exam.

b.

Expert Solution

Answer to Problem 4RE

Probability that student took only one exam = 0.563

Explanation of Solution

Calculation:

To get the probability distribution of no. of tests taken, it is required to divide the respective frequency of no. of test taken by the total frequency.

The probability distribution is given below:

Number of Tests	Frequency (in 1000s)	P (X = x)
1	953	0.563
2	423	0.250
3	194	0.115
4	80	0.047
5	29	0.017
6	9	0.005
7	3	0.002
8	1	0.001
Total	1692	1.000

Calculation:

P(That student took only one exam)=9531692=0.563

Answer 13

b.

To determine

To find the probability that student took only one exam.

Answer 14

b.

Expert Solution

Answer to Problem 4RE

Probability that student took only one exam = 0.563

Answer 15

b.

Expert Solution

Answer 16

b.

Expert Solution

Answer 17

Expert Solution

Answer 18

Explanation of Solution

Calculation:

To get the probability distribution of no. of tests taken, it is required to divide the respective frequency of no. of test taken by the total frequency.

The probability distribution is given below:

Number of Tests	Frequency (in 1000s)	P (X = x)
1	953	0.563
2	423	0.250
3	194	0.115
4	80	0.047
5	29	0.017
6	9	0.005
7	3	0.002
8	1	0.001
Total	1692	1.000

Calculation:

P(That student took only one exam)=9531692=0.563

Answer 19

c.

To determine

To find the mean μx

c.

Expert Solution

Answer to Problem 4RE

The mean of the given distribution is 1.73.

Explanation of Solution

Formula Used:

μx=∑x.p(x)

Calculation:

Number of Tests	Frequency (in 1000s)	P (X = x)	x.P(x)
1	953	0.563	0.56
2	423	0.250	0.50
3	194	0.115	0.34
4	80	0.047	0.19
5	29	0.017	0.09
6	9	0.005	0.03
7	3	0.002	0.01
8	1	0.001	0.00
Total	1692	1.000	1.73

The mean μx of a given distribution is 1.73, because the value of ∑x.p(x) = 1.73. Therefore, the mean is 1.73.

Answer 20

c.

To determine

To find the mean μx

Answer 21

c.

Expert Solution

Answer to Problem 4RE

The mean of the given distribution is 1.73.

Answer 22

c.

Expert Solution

Answer 23

c.

Expert Solution

Answer 24

Expert Solution

Answer 25

Explanation of Solution

Formula Used:

μx=∑x.p(x)

Calculation:

Number of Tests	Frequency (in 1000s)	P (X = x)	x.P(x)
1	953	0.563	0.56
2	423	0.250	0.50
3	194	0.115	0.34
4	80	0.047	0.19
5	29	0.017	0.09
6	9	0.005	0.03
7	3	0.002	0.01
8	1	0.001	0.00
Total	1692	1.000	1.73

The mean μx of a given distribution is 1.73, because the value of ∑x.p(x) = 1.73. Therefore, the mean is 1.73.

Answer 26

d.

To determine

To find the mean μx

d.

Expert Solution

Answer to Problem 4RE

Standard Deviation = 1.05

Explanation of Solution

Formula Used:

μx=∑x.p(x)

Calculation:

Number of Tests	Frequency (in 1000s)	P (X = x)	x.P(x)	x2.P(x)
1	953	0.563	0.56	0.56
2	423	0.250	0.50	1.00
3	194	0.115	0.34	1.03
4	80	0.047	0.19	0.76
5	29	0.017	0.09	0.43
6	9	0.005	0.03	0.19
7	3	0.002	0.01	0.09
8	1	0.001	0.00	0.04
Total	1692	1.000	1.73	4.10

Standard Deviation = ∑x2p(x)− [ xp( x )]2=4.10− 1.732=1.05

Answer 27

d.

To determine

To find the mean μx

Answer 28

d.

Expert Solution

Answer to Problem 4RE

Standard Deviation = 1.05

Answer 29

d.

Expert Solution

Answer 30

d.

Expert Solution

Answer 31

Expert Solution

Answer 32

Explanation of Solution

Formula Used:

μx=∑x.p(x)

Calculation:

Number of Tests	Frequency (in 1000s)	P (X = x)	x.P(x)	x2.P(x)
1	953	0.563	0.56	0.56
2	423	0.250	0.50	1.00
3	194	0.115	0.34	1.03
4	80	0.047	0.19	0.76
5	29	0.017	0.09	0.43
6	9	0.005	0.03	0.19
7	3	0.002	0.01	0.09
8	1	0.001	0.00	0.04
Total	1692	1.000	1.73	4.10