To Construct the probability distribution of the given data.

Question

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Answer 1

Question

Definition Definition Measure of central tendency that is the average of a given data set. The mean value is evaluated as the quotient of the sum of all observations by the sample size. The mean, in contrast to a median, is affected by extreme values. Very large or very small values can distract the mean from the center of the data. Arithmetic mean: The most common type of mean is the arithmetic mean. It is evaluated using the formula: μ = 1 N ∑ i = 1 N x i Other types of means are the geometric mean, logarithmic mean, and harmonic mean. Geometric mean: The nth root of the product of n observations from a data set is defined as the geometric mean of the set: G = x 1 x 2 ... x n n Logarithmic mean: The difference of the natural logarithms of the two numbers, divided by the difference between the numbers is the logarithmic mean of the two numbers. The logarithmic mean is used particularly in heat transfer and mass transfer. ln x 2 − ln x 1 x 2 − x 1 Harmonic mean: The inverse of the arithmetic mean of the inverses of all the numbers in a data set is the harmonic mean of the data. 1 1 x 1 + 1 x 2 + ...

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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Suppose you are gambling on a roulette wheel. Each time the wheel is spun, the result is one of the outcomes 0, 1, and so on through 36. Of these outcomes, 18 are red, 18 are black, and 1 is green. On each spin you bet $5 that a red outcome will occur and $1 that the green outcome will occur. If red occurs, you win a net $4. (You win $10 from red and nothing from green.) If green occurs, you win a net $24. (You win $30 from green and nothing from red.) If black occurs, you lose everything you bet for a loss of $6. a. Use simulation to generate 1,000 plays from this strategy. Each play should indicate the net amount won or lost. Then, based on these outcomes, calculate a 95% confidence interval for the total net amount won or lost from 1,000 plays of the game. (Round your answers to two decimal places and if your answer is negative value, enter "minus" sign.) I worked out the Upper Limit, but I can't seem to arrive at the correct answer for the Lower Limit. What is the Lower Limit?…

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Consider the hypothesis test Ho: 0 s² = = 4.5; s² = 2.3. Use a = 0.01. = σ against H₁: 6 > σ2. Suppose that the sample sizes are n₁ = 20 and 2 = 8, and that (a) Test the hypothesis. Round your answers to two decimal places (e.g. 98.76). The test statistic is fo = 1.96 The critical value is f = 6.18 Conclusion: fail to reject the null hypothesis at a = 0.01. (b) Construct the confidence interval on 02/2/622 which can be used to test the hypothesis: (Round your answer to two decimal places (e.g. 98.76).) 035

Answer 2

Question

Definition Definition Measure of central tendency that is the average of a given data set. The mean value is evaluated as the quotient of the sum of all observations by the sample size. The mean, in contrast to a median, is affected by extreme values. Very large or very small values can distract the mean from the center of the data. Arithmetic mean: The most common type of mean is the arithmetic mean. It is evaluated using the formula: μ = 1 N ∑ i = 1 N x i Other types of means are the geometric mean, logarithmic mean, and harmonic mean. Geometric mean: The nth root of the product of n observations from a data set is defined as the geometric mean of the set: G = x 1 x 2 ... x n n Logarithmic mean: The difference of the natural logarithms of the two numbers, divided by the difference between the numbers is the logarithmic mean of the two numbers. The logarithmic mean is used particularly in heat transfer and mass transfer. ln x 2 − ln x 1 x 2 − x 1 Harmonic mean: The inverse of the arithmetic mean of the inverses of all the numbers in a data set is the harmonic mean of the data. 1 1 x 1 + 1 x 2 + ...

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

Definition Definition Measure of central tendency that is the average of a given data set. The mean value is evaluated as the quotient of the sum of all observations by the sample size. The mean, in contrast to a median, is affected by extreme values. Very large or very small values can distract the mean from the center of the data. Arithmetic mean: The most common type of mean is the arithmetic mean. It is evaluated using the formula: μ = 1 N ∑ i = 1 N x i Other types of means are the geometric mean, logarithmic mean, and harmonic mean. Geometric mean: The nth root of the product of n observations from a data set is defined as the geometric mean of the set: G = x 1 x 2 ... x n n Logarithmic mean: The difference of the natural logarithms of the two numbers, divided by the difference between the numbers is the logarithmic mean of the two numbers. The logarithmic mean is used particularly in heat transfer and mass transfer. ln x 2 − ln x 1 x 2 − x 1 Harmonic mean: The inverse of the arithmetic mean of the inverses of all the numbers in a data set is the harmonic mean of the data. 1 1 x 1 + 1 x 2 + ...

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

Definition Definition Measure of central tendency that is the average of a given data set. The mean value is evaluated as the quotient of the sum of all observations by the sample size. The mean, in contrast to a median, is affected by extreme values. Very large or very small values can distract the mean from the center of the data. Arithmetic mean: The most common type of mean is the arithmetic mean. It is evaluated using the formula: μ = 1 N ∑ i = 1 N x i Other types of means are the geometric mean, logarithmic mean, and harmonic mean. Geometric mean: The nth root of the product of n observations from a data set is defined as the geometric mean of the set: G = x 1 x 2 ... x n n Logarithmic mean: The difference of the natural logarithms of the two numbers, divided by the difference between the numbers is the logarithmic mean of the two numbers. The logarithmic mean is used particularly in heat transfer and mass transfer. ln x 2 − ln x 1 x 2 − x 1 Harmonic mean: The inverse of the arithmetic mean of the inverses of all the numbers in a data set is the harmonic mean of the data. 1 1 x 1 + 1 x 2 + ...

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