To find the mean of the given probability distribution.

Question

Want to see more full solutions like this?

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 2RE

a.

To determine

To find the mean of the given probability distribution.

a.

Expert Solution

Answer to Problem 2RE

Mean = 8.37

Explanation of Solution

Given:

x	P ( x )
6	0.21
7	0.12
8	0.29
9	0.11
10	0.01
11	0.26

Concept used:

Mean=∑x.P(x)

x	P ( x )	x.P(x)
6	0.21	1.26
7	0.12	0.84
8	0.29	2.32
9	0.11	0.99
10	0.01	0.1
11	0.26	2.86
Sum	1	8.37

Since the ∑x.P(x) = 8.37, from the above table.

Therefore, the mean of the given distribution is 8.37.

b.

To determine

To find the Variance of the given probability distribution.

b.

Expert Solution

Answer to Problem 2RE

Variance = 3.3131

Explanation of Solution

Given:

x	P ( x )
6	0.21
7	0.12
8	0.29
9	0.11
10	0.01
11	0.26

Concept used:

Variance=∑x2.P(x)−[∑x.P(x)]2

x	P ( x )	x.P(x)	x2.P(x)
6	0.21	1.26	7.56
7	0.12	0.84	5.88
8	0.29	2.32	18.56
9	0.11	0.99	8.91
10	0.01	0.1	1
11	0.26	2.86	31.46
Sum	1	8.37	73.37

From the above table:

∑x.P(x)=8.37∑x2.P(x)=73.37

Therefore, the variance is,

Variance=∑x2.P(x)−[∑x.P(x)]2=73.37−(8.37)2=3.3131

c.

To determine

To find the Standard Deviation of the given probability distribution.

c.

Expert Solution

Answer to Problem 2RE

Standard Deviation = 1.82

Explanation of Solution

Given:

x	P ( x )
6	0.21
7	0.12
8	0.29
9	0.11
10	0.01
11	0.26

Concept used:

Standard Deviation =∑x2.P(x)−[∑x.P( x)]2

x	P ( x )	x.P(x)	x2.P(x)
6	0.21	1.26	7.56
7	0.12	0.84	5.88
8	0.29	2.32	18.56
9	0.11	0.99	8.91
10	0.01	0.1	1
11	0.26	2.86	31.46
Sum	1	8.37	73.37

From the above table:

∑x.P(x)=8.37∑x2.P(x)=73.37

Therefore, the variance is,

Standard Deviation=∑x2.P(x)− [ ∑x.P( x )]2=73.37− ( 8.37 )2=3.3131=1.82

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Students have asked these similar questions

Why the correct answer is letter A? Students in an online course are each randomly assigned to receive either standard practice exercises or adaptivepractice exercises. For the adaptive practice exercises, the next question asked is determined by whether the studentgot the previous question correct. The teacher of the course wants to determine whether there is a differencebetween the two practice exercise types by comparing the proportion of students who pass the course from eachgroup. The teacher plans to test the null hypothesis that versus the alternative hypothesis , whererepresents the proportion of students who would pass the course using standard practice exercises andrepresents the proportion of students who would pass the course using adaptive practice exercises.The teacher knows that the percent confidence interval for the difference in proportion of students passing thecourse for the two practice exercise types (standard minus adaptive) is and the percent…

Carpetland salespersons average $8,000 per week in sales. Steve Contois, the firm's vice president, proposes a compensation plan with new selling incentives. Steve hopes that the results of a trial selling period will enable him to conclude that the compensation plan increases the average sales per salesperson. a. Develop the appropriate null and alternative hypotheses.H 0: H a:

توليد تمرين شامل حول الانحدار الخطي المتعدد بطريقة المربعات الصغرى

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 2RE

a.

To determine

To find the mean of the given probability distribution.

a.

Expert Solution

Answer to Problem 2RE

Mean = 8.37

Explanation of Solution

Given:

x	P ( x )
6	0.21
7	0.12
8	0.29
9	0.11
10	0.01
11	0.26

Concept used:

Mean=∑x.P(x)

x	P ( x )	x.P(x)
6	0.21	1.26
7	0.12	0.84
8	0.29	2.32
9	0.11	0.99
10	0.01	0.1
11	0.26	2.86
Sum	1	8.37

Since the ∑x.P(x) = 8.37, from the above table.

Therefore, the mean of the given distribution is 8.37.

b.

To determine

To find the Variance of the given probability distribution.

b.

Expert Solution

Answer to Problem 2RE

Variance = 3.3131

Explanation of Solution

Given:

x	P ( x )
6	0.21
7	0.12
8	0.29
9	0.11
10	0.01
11	0.26

Concept used:

Variance=∑x2.P(x)−[∑x.P(x)]2

x	P ( x )	x.P(x)	x2.P(x)
6	0.21	1.26	7.56
7	0.12	0.84	5.88
8	0.29	2.32	18.56
9	0.11	0.99	8.91
10	0.01	0.1	1
11	0.26	2.86	31.46
Sum	1	8.37	73.37

From the above table:

∑x.P(x)=8.37∑x2.P(x)=73.37

Therefore, the variance is,

Variance=∑x2.P(x)−[∑x.P(x)]2=73.37−(8.37)2=3.3131

c.

To determine

To find the Standard Deviation of the given probability distribution.

c.

Expert Solution

Answer to Problem 2RE

Standard Deviation = 1.82

Explanation of Solution

Given:

x	P ( x )
6	0.21
7	0.12
8	0.29
9	0.11
10	0.01
11	0.26

Concept used:

Standard Deviation =∑x2.P(x)−[∑x.P( x)]2

x	P ( x )	x.P(x)	x2.P(x)
6	0.21	1.26	7.56
7	0.12	0.84	5.88
8	0.29	2.32	18.56
9	0.11	0.99	8.91
10	0.01	0.1	1
11	0.26	2.86	31.46
Sum	1	8.37	73.37

From the above table:

∑x.P(x)=8.37∑x2.P(x)=73.37

Therefore, the variance is,

Standard Deviation=∑x2.P(x)− [ ∑x.P( x )]2=73.37− ( 8.37 )2=3.3131=1.82

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

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 2RE

a.

To determine

To find the mean of the given probability distribution.

a.

Expert Solution

Answer to Problem 2RE

Mean = 8.37

Explanation of Solution

Given:

x	P ( x )
6	0.21
7	0.12
8	0.29
9	0.11
10	0.01
11	0.26

Concept used:

Mean=∑x.P(x)

x	P ( x )	x.P(x)
6	0.21	1.26
7	0.12	0.84
8	0.29	2.32
9	0.11	0.99
10	0.01	0.1
11	0.26	2.86
Sum	1	8.37

Since the ∑x.P(x) = 8.37, from the above table.

Therefore, the mean of the given distribution is 8.37.

b.

To determine

To find the Variance of the given probability distribution.

b.

Expert Solution

Answer to Problem 2RE

Variance = 3.3131

Explanation of Solution

Given:

x	P ( x )
6	0.21
7	0.12
8	0.29
9	0.11
10	0.01
11	0.26

Concept used:

Variance=∑x2.P(x)−[∑x.P(x)]2

x	P ( x )	x.P(x)	x2.P(x)
6	0.21	1.26	7.56
7	0.12	0.84	5.88
8	0.29	2.32	18.56
9	0.11	0.99	8.91
10	0.01	0.1	1
11	0.26	2.86	31.46
Sum	1	8.37	73.37

From the above table:

∑x.P(x)=8.37∑x2.P(x)=73.37

Therefore, the variance is,

Variance=∑x2.P(x)−[∑x.P(x)]2=73.37−(8.37)2=3.3131

c.

To determine

To find the Standard Deviation of the given probability distribution.

c.

Expert Solution

Answer to Problem 2RE

Standard Deviation = 1.82

Explanation of Solution

Given:

x	P ( x )
6	0.21
7	0.12
8	0.29
9	0.11
10	0.01
11	0.26

Concept used:

Standard Deviation =∑x2.P(x)−[∑x.P( x)]2

x	P ( x )	x.P(x)	x2.P(x)
6	0.21	1.26	7.56
7	0.12	0.84	5.88
8	0.29	2.32	18.56
9	0.11	0.99	8.91
10	0.01	0.1	1
11	0.26	2.86	31.46
Sum	1	8.37	73.37

From the above table:

∑x.P(x)=8.37∑x2.P(x)=73.37

Therefore, the variance is,

Standard Deviation=∑x2.P(x)− [ ∑x.P( x )]2=73.37− ( 8.37 )2=3.3131=1.82

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

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 2RE

a.

To determine

To find the mean of the given probability distribution.

a.

Expert Solution

Answer to Problem 2RE

Mean = 8.37

Explanation of Solution

Given:

x	P ( x )
6	0.21
7	0.12
8	0.29
9	0.11
10	0.01
11	0.26

Concept used:

Mean=∑x.P(x)

x	P ( x )	x.P(x)
6	0.21	1.26
7	0.12	0.84
8	0.29	2.32
9	0.11	0.99
10	0.01	0.1
11	0.26	2.86
Sum	1	8.37

Since the ∑x.P(x) = 8.37, from the above table.

Therefore, the mean of the given distribution is 8.37.

b.

To determine

To find the Variance of the given probability distribution.

b.

Expert Solution

Answer to Problem 2RE

Variance = 3.3131

Explanation of Solution

Given:

x	P ( x )
6	0.21
7	0.12
8	0.29
9	0.11
10	0.01
11	0.26

Concept used:

Variance=∑x2.P(x)−[∑x.P(x)]2

x	P ( x )	x.P(x)	x2.P(x)
6	0.21	1.26	7.56
7	0.12	0.84	5.88
8	0.29	2.32	18.56
9	0.11	0.99	8.91
10	0.01	0.1	1
11	0.26	2.86	31.46
Sum	1	8.37	73.37

From the above table:

∑x.P(x)=8.37∑x2.P(x)=73.37

Therefore, the variance is,

Variance=∑x2.P(x)−[∑x.P(x)]2=73.37−(8.37)2=3.3131

c.

To determine

To find the Standard Deviation of the given probability distribution.

c.

Expert Solution

Answer to Problem 2RE

Standard Deviation = 1.82

Explanation of Solution

Given:

x	P ( x )
6	0.21
7	0.12
8	0.29
9	0.11
10	0.01
11	0.26

Concept used:

Standard Deviation =∑x2.P(x)−[∑x.P( x)]2

x	P ( x )	x.P(x)	x2.P(x)
6	0.21	1.26	7.56
7	0.12	0.84	5.88
8	0.29	2.32	18.56
9	0.11	0.99	8.91
10	0.01	0.1	1
11	0.26	2.86	31.46
Sum	1	8.37	73.37

From the above table:

∑x.P(x)=8.37∑x2.P(x)=73.37

Therefore, the variance is,

Standard Deviation=∑x2.P(x)− [ ∑x.P( x )]2=73.37− ( 8.37 )2=3.3131=1.82

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Answer 5

Chapter 6, Problem 2RE

a.

To determine

To find the mean of the given probability distribution.

a.

Expert Solution

Answer to Problem 2RE

Mean = 8.37

Explanation of Solution

Given:

x	P ( x )
6	0.21
7	0.12
8	0.29
9	0.11
10	0.01
11	0.26

Concept used:

Mean=∑x.P(x)

x	P ( x )	x.P(x)
6	0.21	1.26
7	0.12	0.84
8	0.29	2.32
9	0.11	0.99
10	0.01	0.1
11	0.26	2.86
Sum	1	8.37

Since the ∑x.P(x) = 8.37, from the above table.

Therefore, the mean of the given distribution is 8.37.

b.

To determine

To find the Variance of the given probability distribution.

b.

Expert Solution

Answer to Problem 2RE

Variance = 3.3131

Explanation of Solution

Given:

x	P ( x )
6	0.21
7	0.12
8	0.29
9	0.11
10	0.01
11	0.26

Concept used:

Variance=∑x2.P(x)−[∑x.P(x)]2

x	P ( x )	x.P(x)	x2.P(x)
6	0.21	1.26	7.56
7	0.12	0.84	5.88
8	0.29	2.32	18.56
9	0.11	0.99	8.91
10	0.01	0.1	1
11	0.26	2.86	31.46
Sum	1	8.37	73.37

From the above table:

∑x.P(x)=8.37∑x2.P(x)=73.37

Therefore, the variance is,

Variance=∑x2.P(x)−[∑x.P(x)]2=73.37−(8.37)2=3.3131

c.

To determine

To find the Standard Deviation of the given probability distribution.

c.

Expert Solution

Answer to Problem 2RE

Standard Deviation = 1.82

Explanation of Solution

Given:

x	P ( x )
6	0.21
7	0.12
8	0.29
9	0.11
10	0.01
11	0.26

Concept used:

Standard Deviation =∑x2.P(x)−[∑x.P( x)]2

x	P ( x )	x.P(x)	x2.P(x)
6	0.21	1.26	7.56
7	0.12	0.84	5.88
8	0.29	2.32	18.56
9	0.11	0.99	8.91
10	0.01	0.1	1
11	0.26	2.86	31.46
Sum	1	8.37	73.37

From the above table:

∑x.P(x)=8.37∑x2.P(x)=73.37

Therefore, the variance is,

Standard Deviation=∑x2.P(x)− [ ∑x.P( x )]2=73.37− ( 8.37 )2=3.3131=1.82

Answer 6

Chapter 6, Problem 2RE

a.

To determine

To find the mean of the given probability distribution.

a.

Expert Solution

Answer to Problem 2RE

Mean = 8.37

Explanation of Solution

Given:

x	P ( x )
6	0.21
7	0.12
8	0.29
9	0.11
10	0.01
11	0.26

Concept used:

Mean=∑x.P(x)

x	P ( x )	x.P(x)
6	0.21	1.26
7	0.12	0.84
8	0.29	2.32
9	0.11	0.99
10	0.01	0.1
11	0.26	2.86
Sum	1	8.37

Since the ∑x.P(x) = 8.37, from the above table.

Therefore, the mean of the given distribution is 8.37.

Answer 7

Chapter 6, Problem 2RE

a.

To determine

To find the mean of the given probability distribution.

Answer 8

a.

Expert Solution

Answer to Problem 2RE

Mean = 8.37

Answer 9

a.

Expert Solution

Answer 10

a.

Expert Solution

Answer 11

Expert Solution

Answer 12

Explanation of Solution

Given:

x	P ( x )
6	0.21
7	0.12
8	0.29
9	0.11
10	0.01
11	0.26

Concept used:

Mean=∑x.P(x)

x	P ( x )	x.P(x)
6	0.21	1.26
7	0.12	0.84
8	0.29	2.32
9	0.11	0.99
10	0.01	0.1
11	0.26	2.86
Sum	1	8.37

Since the ∑x.P(x) = 8.37, from the above table.

Therefore, the mean of the given distribution is 8.37.

Answer 13

b.

To determine

To find the Variance of the given probability distribution.

b.

Expert Solution

Answer to Problem 2RE

Variance = 3.3131

Explanation of Solution

Given:

x	P ( x )
6	0.21
7	0.12
8	0.29
9	0.11
10	0.01
11	0.26

Concept used:

Variance=∑x2.P(x)−[∑x.P(x)]2

x	P ( x )	x.P(x)	x2.P(x)
6	0.21	1.26	7.56
7	0.12	0.84	5.88
8	0.29	2.32	18.56
9	0.11	0.99	8.91
10	0.01	0.1	1
11	0.26	2.86	31.46
Sum	1	8.37	73.37

From the above table:

∑x.P(x)=8.37∑x2.P(x)=73.37

Therefore, the variance is,

Variance=∑x2.P(x)−[∑x.P(x)]2=73.37−(8.37)2=3.3131

Answer 14

b.

To determine

To find the Variance of the given probability distribution.

Answer 15

b.

Expert Solution

Answer to Problem 2RE

Variance = 3.3131

Answer 16

b.

Expert Solution

Answer 17

b.

Expert Solution

Answer 18

Expert Solution

Answer 19

Explanation of Solution

Given:

x	P ( x )
6	0.21
7	0.12
8	0.29
9	0.11
10	0.01
11	0.26

Concept used:

Variance=∑x2.P(x)−[∑x.P(x)]2

x	P ( x )	x.P(x)	x2.P(x)
6	0.21	1.26	7.56
7	0.12	0.84	5.88
8	0.29	2.32	18.56
9	0.11	0.99	8.91
10	0.01	0.1	1
11	0.26	2.86	31.46
Sum	1	8.37	73.37

From the above table:

∑x.P(x)=8.37∑x2.P(x)=73.37

Therefore, the variance is,

Variance=∑x2.P(x)−[∑x.P(x)]2=73.37−(8.37)2=3.3131

Answer 20

c.

To determine

To find the Standard Deviation of the given probability distribution.

c.

Expert Solution

Answer to Problem 2RE

Standard Deviation = 1.82

Explanation of Solution

Given:

x	P ( x )
6	0.21
7	0.12
8	0.29
9	0.11
10	0.01
11	0.26

Concept used:

Standard Deviation =∑x2.P(x)−[∑x.P( x)]2

x	P ( x )	x.P(x)	x2.P(x)
6	0.21	1.26	7.56
7	0.12	0.84	5.88
8	0.29	2.32	18.56
9	0.11	0.99	8.91
10	0.01	0.1	1
11	0.26	2.86	31.46
Sum	1	8.37	73.37

From the above table:

∑x.P(x)=8.37∑x2.P(x)=73.37

Therefore, the variance is,

Standard Deviation=∑x2.P(x)− [ ∑x.P( x )]2=73.37− ( 8.37 )2=3.3131=1.82

Answer 21

c.

To determine

To find the Standard Deviation of the given probability distribution.

Answer 22

c.

Expert Solution

Answer to Problem 2RE

Standard Deviation = 1.82

Answer 23

c.

Expert Solution

Answer 24

c.

Expert Solution

Answer 25

Expert Solution

Answer 26

Explanation of Solution

Given:

x	P ( x )
6	0.21
7	0.12
8	0.29
9	0.11
10	0.01
11	0.26

Concept used:

Standard Deviation =∑x2.P(x)−[∑x.P( x)]2

x	P ( x )	x.P(x)	x2.P(x)
6	0.21	1.26	7.56
7	0.12	0.84	5.88
8	0.29	2.32	18.56
9	0.11	0.99	8.91
10	0.01	0.1	1
11	0.26	2.86	31.46
Sum	1	8.37	73.37

From the above table:

∑x.P(x)=8.37∑x2.P(x)=73.37

Therefore, the variance is,

Standard Deviation=∑x2.P(x)− [ ∑x.P( x )]2=73.37− ( 8.37 )2=3.3131=1.82

Answer 27

Ch. 6.1 - Exercises 9-12, fill in each blank with the...Ch. 6.1 - Exercises 9-12, fill in each blank with the...Ch. 6.1 - Exercises 9-12, fill in each blank with the...Ch. 6.1 - Exercises 9-12, fill in each blank with the...Ch. 6.1 - In Exercises 13-16, determine whether the...Ch. 6.1 - In Exercises 13-16, determine whether the...Ch. 6.1 - In Exercises 13-16, determine whether the...Ch. 6.1 - In Exercises 13-16, determine whether the...Ch. 6.1 - In Exercises 17-26, determine whether the random...Ch. 6.1 - In Exercises 17-26, determine whether the random...

To find the mean of the given probability distribution.

Concept explainers

Videos

To find the mean of the given probability distribution.

Answer to Problem 2RE

Explanation of Solution

To find the Variance of the given probability distribution.

Answer to Problem 2RE

Explanation of Solution

To find the Standard Deviation of the given probability distribution.

Answer to Problem 2RE

Explanation of Solution

Want to see more full solutions like this?

Chapter 6 Solutions