The 40 t h percentile of the diameter.

Question

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Answer 1

Question

Features Features Normal distribution is characterized by two parameters, mean (µ) and standard deviation (σ). When graphed, the mean represents the center of the bell curve and the graph is perfectly symmetric about the center. The mean, median, and mode are all equal for a normal distribution. The standard deviation measures the data's spread from the center. The higher the standard deviation, the more the data is spread out and the flatter the bell curve looks. Variance is another commonly used measure of the spread of the distribution and is equal to the square of the standard deviation.

Chapter 7.2, Problem 50E

(a)

To determine

The 40^t^h percentile of the diameter.

(a)

Expert Solution

Answer to Problem 50E

The 40^t^h percentile of the diameter is 72.97

Explanation of Solution

It is given in the question that,

μ=75σ=8

Probability density function of normal distribution is f(x,σ,μ)=12πσe−12( x−μ)2:−∞<x<∞

Now, for determining 60^t^h percentile of the diameter,

∫−∞x1 2πσe −12 ( x−μ )2=0.4

Using calculator, x=72.97

(b)

To determine

The 65^t^h percentile of the diameter.

(b)

Expert Solution

Answer to Problem 50E

The 65^t^h percentile of the diameter is 78.08

Explanation of Solution

It is given in the question that,

μ=75σ=8

Probability density function of normal distribution is f(x,σ,μ)=12πσe−12( x−μ)2:−∞<x<∞

Now, for determining 32^t^h percentile of the diameter,

∫−∞x1 2πσe −12 ( x−μ )2=0.65

Using calculator, x=78.08

(c)

To determine

The minimum score above which top 10% of students will lie

(c)

Expert Solution

Answer to Problem 50E

The minimum score above which top 10% of students will lie is 85.25

Explanation of Solution

It is given in the question that,

μ=75σ=8

Probability density function of normal distribution is f(x,σ,μ)=12πσe−12( x−μ)2:−∞<x<∞

For determining the minimum score above which top 10% of students will lie, one has to find 90^t^h percentile of the score of statistics exam.

Now,

∫−∞x1 2πσe −12 ( x−μ )2=0.90

Using calculator, x=85.25

The minimum score above which top 10% of students will lie is 85.25

(d)

To determine

The range in which middle 60% of the diameters will lie.

(d)

Expert Solution

Answer to Problem 50E

Between 68.26 25.04 and 81.73 middle 60% of the scores of statistics exam will lie.

Explanation of Solution

It is given in the question that,

μ=25.1σ=0.08

Probability density function of normal distribution is f(x,σ,μ)=12πσe−12( x−μ)2:−∞<x<∞

For determining the range in which middle 60% of the scores will lie, one has to find 20^t^hpercentile from left side and 20^t^h percentile from right side.

For left 25^t^h percentile,

∫−∞x1 2πσe −12 ( x−μ )2=0.2

Using calculator, x=68.26

For right 25^t^h percentile,

∫x∞1 2πσe −12 ( x−μ )2=0.2

Using calculator, x=81.73

Now, between 68.26 and 81.73 middle 60% of the scores of statistics exam will lie

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Answer 2

Question

Features Features Normal distribution is characterized by two parameters, mean (µ) and standard deviation (σ). When graphed, the mean represents the center of the bell curve and the graph is perfectly symmetric about the center. The mean, median, and mode are all equal for a normal distribution. The standard deviation measures the data's spread from the center. The higher the standard deviation, the more the data is spread out and the flatter the bell curve looks. Variance is another commonly used measure of the spread of the distribution and is equal to the square of the standard deviation.

Chapter 7.2, Problem 50E

(a)

To determine

The 40^t^h percentile of the diameter.

(a)

Expert Solution

Answer to Problem 50E

The 40^t^h percentile of the diameter is 72.97

Explanation of Solution

It is given in the question that,

μ=75σ=8

Probability density function of normal distribution is f(x,σ,μ)=12πσe−12( x−μ)2:−∞<x<∞

Now, for determining 60^t^h percentile of the diameter,

∫−∞x1 2πσe −12 ( x−μ )2=0.4

Using calculator, x=72.97

(b)

To determine

The 65^t^h percentile of the diameter.

(b)

Expert Solution

Answer to Problem 50E

The 65^t^h percentile of the diameter is 78.08

Explanation of Solution

It is given in the question that,

μ=75σ=8

Probability density function of normal distribution is f(x,σ,μ)=12πσe−12( x−μ)2:−∞<x<∞

Now, for determining 32^t^h percentile of the diameter,

∫−∞x1 2πσe −12 ( x−μ )2=0.65

Using calculator, x=78.08

(c)

To determine

The minimum score above which top 10% of students will lie

(c)

Expert Solution

Answer to Problem 50E

The minimum score above which top 10% of students will lie is 85.25

Explanation of Solution

It is given in the question that,

μ=75σ=8

Probability density function of normal distribution is f(x,σ,μ)=12πσe−12( x−μ)2:−∞<x<∞

For determining the minimum score above which top 10% of students will lie, one has to find 90^t^h percentile of the score of statistics exam.

Now,

∫−∞x1 2πσe −12 ( x−μ )2=0.90

Using calculator, x=85.25

The minimum score above which top 10% of students will lie is 85.25

(d)

To determine

The range in which middle 60% of the diameters will lie.

(d)

Expert Solution

Answer to Problem 50E

Between 68.26 25.04 and 81.73 middle 60% of the scores of statistics exam will lie.

Explanation of Solution

It is given in the question that,

μ=25.1σ=0.08

Probability density function of normal distribution is f(x,σ,μ)=12πσe−12( x−μ)2:−∞<x<∞

For determining the range in which middle 60% of the scores will lie, one has to find 20^t^hpercentile from left side and 20^t^h percentile from right side.

For left 25^t^h percentile,

∫−∞x1 2πσe −12 ( x−μ )2=0.2

Using calculator, x=68.26

For right 25^t^h percentile,

∫x∞1 2πσe −12 ( x−μ )2=0.2

Using calculator, x=81.73

Now, between 68.26 and 81.73 middle 60% of the scores of statistics exam will lie

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

Features Features Normal distribution is characterized by two parameters, mean (µ) and standard deviation (σ). When graphed, the mean represents the center of the bell curve and the graph is perfectly symmetric about the center. The mean, median, and mode are all equal for a normal distribution. The standard deviation measures the data's spread from the center. The higher the standard deviation, the more the data is spread out and the flatter the bell curve looks. Variance is another commonly used measure of the spread of the distribution and is equal to the square of the standard deviation.

Chapter 7.2, Problem 50E

(a)

To determine

The 40^t^h percentile of the diameter.

(a)

Expert Solution

Answer to Problem 50E

The 40^t^h percentile of the diameter is 72.97

Explanation of Solution

It is given in the question that,

μ=75σ=8

Probability density function of normal distribution is f(x,σ,μ)=12πσe−12( x−μ)2:−∞<x<∞

Now, for determining 60^t^h percentile of the diameter,

∫−∞x1 2πσe −12 ( x−μ )2=0.4

Using calculator, x=72.97

(b)

To determine

The 65^t^h percentile of the diameter.

(b)

Expert Solution

Answer to Problem 50E

The 65^t^h percentile of the diameter is 78.08

Explanation of Solution

It is given in the question that,

μ=75σ=8

Probability density function of normal distribution is f(x,σ,μ)=12πσe−12( x−μ)2:−∞<x<∞

Now, for determining 32^t^h percentile of the diameter,

∫−∞x1 2πσe −12 ( x−μ )2=0.65

Using calculator, x=78.08

(c)

To determine

The minimum score above which top 10% of students will lie

(c)

Expert Solution

Answer to Problem 50E

The minimum score above which top 10% of students will lie is 85.25

Explanation of Solution

It is given in the question that,

μ=75σ=8

Probability density function of normal distribution is f(x,σ,μ)=12πσe−12( x−μ)2:−∞<x<∞

For determining the minimum score above which top 10% of students will lie, one has to find 90^t^h percentile of the score of statistics exam.

Now,

∫−∞x1 2πσe −12 ( x−μ )2=0.90

Using calculator, x=85.25

The minimum score above which top 10% of students will lie is 85.25

(d)

To determine

The range in which middle 60% of the diameters will lie.

(d)

Expert Solution

Answer to Problem 50E

Between 68.26 25.04 and 81.73 middle 60% of the scores of statistics exam will lie.

Explanation of Solution

It is given in the question that,

μ=25.1σ=0.08

Probability density function of normal distribution is f(x,σ,μ)=12πσe−12( x−μ)2:−∞<x<∞

For determining the range in which middle 60% of the scores will lie, one has to find 20^t^hpercentile from left side and 20^t^h percentile from right side.

For left 25^t^h percentile,

∫−∞x1 2πσe −12 ( x−μ )2=0.2

Using calculator, x=68.26

For right 25^t^h percentile,

∫x∞1 2πσe −12 ( x−μ )2=0.2

Using calculator, x=81.73

Now, between 68.26 and 81.73 middle 60% of the scores of statistics exam will lie

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Answer 4

Question

Features Features Normal distribution is characterized by two parameters, mean (µ) and standard deviation (σ). When graphed, the mean represents the center of the bell curve and the graph is perfectly symmetric about the center. The mean, median, and mode are all equal for a normal distribution. The standard deviation measures the data's spread from the center. The higher the standard deviation, the more the data is spread out and the flatter the bell curve looks. Variance is another commonly used measure of the spread of the distribution and is equal to the square of the standard deviation.

Chapter 7.2, Problem 50E

(a)

To determine

The 40^t^h percentile of the diameter.

(a)

Expert Solution

Answer to Problem 50E

The 40^t^h percentile of the diameter is 72.97

Explanation of Solution

It is given in the question that,

μ=75σ=8

Probability density function of normal distribution is f(x,σ,μ)=12πσe−12( x−μ)2:−∞<x<∞

Now, for determining 60^t^h percentile of the diameter,

∫−∞x1 2πσe −12 ( x−μ )2=0.4

Using calculator, x=72.97

(b)

To determine

The 65^t^h percentile of the diameter.

(b)

Expert Solution

Answer to Problem 50E

The 65^t^h percentile of the diameter is 78.08

Explanation of Solution

It is given in the question that,

μ=75σ=8

Probability density function of normal distribution is f(x,σ,μ)=12πσe−12( x−μ)2:−∞<x<∞

Now, for determining 32^t^h percentile of the diameter,

∫−∞x1 2πσe −12 ( x−μ )2=0.65

Using calculator, x=78.08

(c)

To determine

The minimum score above which top 10% of students will lie

(c)

Expert Solution

Answer to Problem 50E

The minimum score above which top 10% of students will lie is 85.25

Explanation of Solution

It is given in the question that,

μ=75σ=8

Probability density function of normal distribution is f(x,σ,μ)=12πσe−12( x−μ)2:−∞<x<∞

For determining the minimum score above which top 10% of students will lie, one has to find 90^t^h percentile of the score of statistics exam.

Now,

∫−∞x1 2πσe −12 ( x−μ )2=0.90

Using calculator, x=85.25

The minimum score above which top 10% of students will lie is 85.25

(d)

To determine

The range in which middle 60% of the diameters will lie.

(d)

Expert Solution

Answer to Problem 50E

Between 68.26 25.04 and 81.73 middle 60% of the scores of statistics exam will lie.

Explanation of Solution

It is given in the question that,

μ=25.1σ=0.08

Probability density function of normal distribution is f(x,σ,μ)=12πσe−12( x−μ)2:−∞<x<∞

For determining the range in which middle 60% of the scores will lie, one has to find 20^t^hpercentile from left side and 20^t^h percentile from right side.

For left 25^t^h percentile,

∫−∞x1 2πσe −12 ( x−μ )2=0.2

Using calculator, x=68.26

For right 25^t^h percentile,

∫x∞1 2πσe −12 ( x−μ )2=0.2

Using calculator, x=81.73

Now, between 68.26 and 81.73 middle 60% of the scores of statistics exam will lie

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

Answer 5

Chapter 7.2, Problem 50E

(a)

To determine

The 40^t^h percentile of the diameter.

(a)

Expert Solution

Answer to Problem 50E

The 40^t^h percentile of the diameter is 72.97

Explanation of Solution

It is given in the question that,

μ=75σ=8

Probability density function of normal distribution is f(x,σ,μ)=12πσe−12( x−μ)2:−∞<x<∞

Now, for determining 60^t^h percentile of the diameter,

∫−∞x1 2πσe −12 ( x−μ )2=0.4

Using calculator, x=72.97

(b)

To determine

The 65^t^h percentile of the diameter.

(b)

Expert Solution

Answer to Problem 50E

The 65^t^h percentile of the diameter is 78.08

Explanation of Solution

It is given in the question that,

μ=75σ=8

Probability density function of normal distribution is f(x,σ,μ)=12πσe−12( x−μ)2:−∞<x<∞

Now, for determining 32^t^h percentile of the diameter,

∫−∞x1 2πσe −12 ( x−μ )2=0.65

Using calculator, x=78.08

(c)

To determine

The minimum score above which top 10% of students will lie

(c)

Expert Solution

Answer to Problem 50E

The minimum score above which top 10% of students will lie is 85.25

Explanation of Solution

It is given in the question that,

μ=75σ=8

Probability density function of normal distribution is f(x,σ,μ)=12πσe−12( x−μ)2:−∞<x<∞

For determining the minimum score above which top 10% of students will lie, one has to find 90^t^h percentile of the score of statistics exam.

Now,

∫−∞x1 2πσe −12 ( x−μ )2=0.90

Using calculator, x=85.25

The minimum score above which top 10% of students will lie is 85.25

(d)

To determine

The range in which middle 60% of the diameters will lie.

(d)

Expert Solution

Answer to Problem 50E

Between 68.26 25.04 and 81.73 middle 60% of the scores of statistics exam will lie.

Explanation of Solution

It is given in the question that,

μ=25.1σ=0.08

Probability density function of normal distribution is f(x,σ,μ)=12πσe−12( x−μ)2:−∞<x<∞

For determining the range in which middle 60% of the scores will lie, one has to find 20^t^hpercentile from left side and 20^t^h percentile from right side.

For left 25^t^h percentile,

∫−∞x1 2πσe −12 ( x−μ )2=0.2

Using calculator, x=68.26

For right 25^t^h percentile,

∫x∞1 2πσe −12 ( x−μ )2=0.2

Using calculator, x=81.73

Now, between 68.26 and 81.73 middle 60% of the scores of statistics exam will lie

Answer 6

Chapter 7.2, Problem 50E

(a)

To determine

The 40^t^h percentile of the diameter.

(a)

Expert Solution

Answer to Problem 50E

The 40^t^h percentile of the diameter is 72.97

Explanation of Solution

It is given in the question that,

μ=75σ=8

Probability density function of normal distribution is f(x,σ,μ)=12πσe−12( x−μ)2:−∞<x<∞

Now, for determining 60^t^h percentile of the diameter,

∫−∞x1 2πσe −12 ( x−μ )2=0.4

Using calculator, x=72.97

Answer 7

Chapter 7.2, Problem 50E

(a)

To determine

The 40^t^h percentile of the diameter.

Answer 8

(a)

Expert Solution

Answer to Problem 50E

The 40^t^h percentile of the diameter is 72.97

Answer 9

(a)

Expert Solution

Answer 10

(a)

Expert Solution

Answer 11

Expert Solution

Answer 12

Explanation of Solution

It is given in the question that,

μ=75σ=8

Probability density function of normal distribution is f(x,σ,μ)=12πσe−12( x−μ)2:−∞<x<∞

Now, for determining 60^t^h percentile of the diameter,

∫−∞x1 2πσe −12 ( x−μ )2=0.4

Using calculator, x=72.97

Answer 13

(b)

To determine

The 65^t^h percentile of the diameter.

(b)

Expert Solution

Answer to Problem 50E

The 65^t^h percentile of the diameter is 78.08

Explanation of Solution

It is given in the question that,

μ=75σ=8

Probability density function of normal distribution is f(x,σ,μ)=12πσe−12( x−μ)2:−∞<x<∞

Now, for determining 32^t^h percentile of the diameter,

∫−∞x1 2πσe −12 ( x−μ )2=0.65

Using calculator, x=78.08

Answer 14

(b)

To determine

The 65^t^h percentile of the diameter.

Answer 15

(b)

Expert Solution

Answer to Problem 50E

The 65^t^h percentile of the diameter is 78.08

Answer 16

(b)

Expert Solution

Answer 17

(b)

Expert Solution

Answer 18

Expert Solution

Answer 19

Explanation of Solution

It is given in the question that,

μ=75σ=8

Probability density function of normal distribution is f(x,σ,μ)=12πσe−12( x−μ)2:−∞<x<∞

Now, for determining 32^t^h percentile of the diameter,

∫−∞x1 2πσe −12 ( x−μ )2=0.65

Using calculator, x=78.08

Answer 20

(c)

To determine

The minimum score above which top 10% of students will lie

(c)

Expert Solution

Answer to Problem 50E

The minimum score above which top 10% of students will lie is 85.25

Explanation of Solution

It is given in the question that,

μ=75σ=8

Probability density function of normal distribution is f(x,σ,μ)=12πσe−12( x−μ)2:−∞<x<∞

For determining the minimum score above which top 10% of students will lie, one has to find 90^t^h percentile of the score of statistics exam.

Now,

∫−∞x1 2πσe −12 ( x−μ )2=0.90

Using calculator, x=85.25

The minimum score above which top 10% of students will lie is 85.25

Answer 21

(c)

To determine

The minimum score above which top 10% of students will lie

Answer 22

(c)

Expert Solution

Answer to Problem 50E

The minimum score above which top 10% of students will lie is 85.25

Answer 23

(c)

Expert Solution

Answer 24

(c)

Expert Solution

Answer 25

Expert Solution

Answer 26

Explanation of Solution

It is given in the question that,

μ=75σ=8

Probability density function of normal distribution is f(x,σ,μ)=12πσe−12( x−μ)2:−∞<x<∞

For determining the minimum score above which top 10% of students will lie, one has to find 90^t^h percentile of the score of statistics exam.

Now,

∫−∞x1 2πσe −12 ( x−μ )2=0.90

Using calculator, x=85.25

The minimum score above which top 10% of students will lie is 85.25

Answer 27

(d)

To determine

The range in which middle 60% of the diameters will lie.

(d)

Expert Solution

Answer to Problem 50E

Between 68.26 25.04 and 81.73 middle 60% of the scores of statistics exam will lie.

Explanation of Solution

It is given in the question that,

μ=25.1σ=0.08

Probability density function of normal distribution is f(x,σ,μ)=12πσe−12( x−μ)2:−∞<x<∞

For determining the range in which middle 60% of the scores will lie, one has to find 20^t^hpercentile from left side and 20^t^h percentile from right side.

For left 25^t^h percentile,

∫−∞x1 2πσe −12 ( x−μ )2=0.2

Using calculator, x=68.26

For right 25^t^h percentile,

∫x∞1 2πσe −12 ( x−μ )2=0.2

Using calculator, x=81.73

Now, between 68.26 and 81.73 middle 60% of the scores of statistics exam will lie

Answer 28

(d)

To determine

The range in which middle 60% of the diameters will lie.

Answer 29

(d)

Expert Solution

Answer to Problem 50E

Between 68.26 25.04 and 81.73 middle 60% of the scores of statistics exam will lie.

Answer 30

(d)

Expert Solution

Answer 31

(d)

Expert Solution

Answer 32

Expert Solution

Answer 33

Explanation of Solution

It is given in the question that,

μ=25.1σ=0.08

Probability density function of normal distribution is f(x,σ,μ)=12πσe−12( x−μ)2:−∞<x<∞

For determining the range in which middle 60% of the scores will lie, one has to find 20^t^hpercentile from left side and 20^t^h percentile from right side.

For left 25^t^h percentile,

∫−∞x1 2πσe −12 ( x−μ )2=0.2

Using calculator, x=68.26

For right 25^t^h percentile,

∫x∞1 2πσe −12 ( x−μ )2=0.2

Using calculator, x=81.73

Now, between 68.26 and 81.73 middle 60% of the scores of statistics exam will lie

Answer 34

The 40 t h percentile of the diameter.

Concept explainers

Videos

The 40^t^h percentile of the diameter.

Answer to Problem 50E

Explanation of Solution

The 65^t^h percentile of the diameter.

Answer to Problem 50E

Explanation of Solution

The minimum score above which top 10% of students will lie

Answer to Problem 50E

Explanation of Solution

The range in which middle 60% of the diameters will lie.

Answer to Problem 50E

Explanation of Solution

Want to see more full solutions like this?

Chapter 7 Solutions

The 40 t h percentile of the diameter.

Concept explainers

Videos

The 40th percentile of the diameter.

Answer to Problem 50E

Explanation of Solution

The 65th percentile of the diameter.

Answer to Problem 50E

Explanation of Solution

The minimum score above which top 10% of students will lie

Answer to Problem 50E

Explanation of Solution

The range in which middle 60% of the diameters will lie.

Answer to Problem 50E

Explanation of Solution

Want to see more full solutions like this?

Chapter 7 Solutions

The 40^t^h percentile of the diameter.

The 65^t^h percentile of the diameter.