If a point is randomly located in an interval ( a , b ) and if Y denotes the location of the point, then Y is assumed to have a uniform distribution over ( a , b ). A plant efficiency expert randomly selects a location along a 500-foot assembly line from which to observe the work habits of the workers on the line. What is the probability that the point she selects is a within 25 feet of the end of the line? b within 25 feet of the beginning of the line? c closer to the beginning of the line than to the end of the line?

Question

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Answer 1

Textbook Question

Chapter 4.4, Problem 48E

If a point is randomly located in an interval (a, b) and if Y denotes the location of the point, then Y is assumed to have a uniform distribution over (a, b). A plant efficiency expert randomly selects a location along a 500-foot assembly line from which to observe the work habits of the workers on the line. What is the probability that the point she selects is

a within 25 feet of the end of the line?
b within 25 feet of the beginning of the line?
c closer to the beginning of the line than to the end of the line?

a.

Expert Solution

To determine

Compute the probability that the point she selects is within 25 feet of the end of the line.

Answer to Problem 48E

The probability that the point she selects is within 25 feet of the end of the line is 0.05.

Explanation of Solution

The probability that the point she selects is within 25 feet of the end of the line is obtained below:

Let random variable Y denotes the location of point and Y is assumed to have uniform distribution over interval (a, b). The length of interval (a, b) = 500 foot.

The probability density function for the uniform distribution on interval (θ1,θ2) is as follows:

f(y)=1θ2−θ1 θ1≤y≤θ2

Here, the interval will be (0, 500) and the length of the interval is 500.

f(y)={1500−0 0≤y≤5000 elsewhere

The probability that the point she selects is within 25 feet of the end of the line is over the interval is as follows:

P(500−25<Y<500)=P(475<Y<500)

P(a<Y<b)=∫θ1θ2f(y)dyP(475<Y<500)=∫4755001500−0dy=∫4755001500dy=1500[y]475500

=1500[500−475]=25500=0.05

Thus, the probability that the point she selects is within 25 feet of the end of the line is 0.05.

b.

Expert Solution

To determine

Compute the probability that the point she selects is within 25 feet of the beginning of the line.

Answer to Problem 48E

The probability that the point she selects is within 25 feet of the beginning of the line is 0.05.

Explanation of Solution

The probability that the point she selects is within 25 feet of the beginning of the line is obtained below:

Let random variable Y denotes the location of point and Y is assumed to have uniform distribution over interval (a, b). The length of interval (a, b) = 500 foot.

The probability density function for the uniform distribution on interval (θ1,θ2) is as follows:

f(y)=1θ2−θ1 θ1≤y≤θ2

Here, the interval will be (0, 500) and the length of the interval is 500.

f(y)={1500−0 0≤y≤5000 elsewhere

The probability that the point she selects is within 25 feet of the beginning of the line is over the interval (0, 25).

P(a<Y<b)=∫θ1θ2f(y)dyP(0<Y<25)=∫0251500−0dy=∫0251500dy=1500[y]025

=1500[25−0]=25500=0.05

Thus, the probability that the point she selects is within 25 feet of the beginning of the line is 0.05.

c.

Expert Solution

To determine

Compute the probability that the point selected by the expert is closer to the beginning of the line than the end of the line.

Answer to Problem 48E

The probability that the point selected by the expert is closer to the beginning of the line than the end of the line is 0.50.

Explanation of Solution

The probability that the point selected by the expert is closer to the beginning of the line than the end of the line is obtained below:

Let random variable Y denotes the location of point and Y is assumed to have uniform distribution over interval (a, b). The length of interval (a, b) = 500 foot.

The probability density function for the uniform distribution on interval (θ1,θ2) is as follows:

f(y)=1θ2−θ1 θ1≤y≤θ2

Here, the interval will be (0, 500) and the length of the interval is 500.

f(y)={1500−0 0≤y≤5000 elsewhere

The probability that the point selected by the expert is closer to the beginning of the line than the end of the line that is the point lies within the first half of the line, which is consider as over the interval (0, 250).

P(a<Y<b)=∫θ1θ2f(y)dyP(0<Y<250)=∫02501500−0dy=∫02501500dy=1500[y]0250

=1500[250−0]=250500=0.50

Thus, the probability that the point selected by the expert is closer to the beginning of the line than the end of the line is 0.50.

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Answer 2

Textbook Question

Chapter 4.4, Problem 48E

If a point is randomly located in an interval (a, b) and if Y denotes the location of the point, then Y is assumed to have a uniform distribution over (a, b). A plant efficiency expert randomly selects a location along a 500-foot assembly line from which to observe the work habits of the workers on the line. What is the probability that the point she selects is

a within 25 feet of the end of the line?
b within 25 feet of the beginning of the line?
c closer to the beginning of the line than to the end of the line?

a.

Expert Solution

To determine

Compute the probability that the point she selects is within 25 feet of the end of the line.

Answer to Problem 48E

The probability that the point she selects is within 25 feet of the end of the line is 0.05.

Explanation of Solution

The probability that the point she selects is within 25 feet of the end of the line is obtained below:

Let random variable Y denotes the location of point and Y is assumed to have uniform distribution over interval (a, b). The length of interval (a, b) = 500 foot.

The probability density function for the uniform distribution on interval (θ1,θ2) is as follows:

f(y)=1θ2−θ1 θ1≤y≤θ2

Here, the interval will be (0, 500) and the length of the interval is 500.

f(y)={1500−0 0≤y≤5000 elsewhere

The probability that the point she selects is within 25 feet of the end of the line is over the interval is as follows:

P(500−25<Y<500)=P(475<Y<500)

P(a<Y<b)=∫θ1θ2f(y)dyP(475<Y<500)=∫4755001500−0dy=∫4755001500dy=1500[y]475500

=1500[500−475]=25500=0.05

Thus, the probability that the point she selects is within 25 feet of the end of the line is 0.05.

b.

Expert Solution

To determine

Compute the probability that the point she selects is within 25 feet of the beginning of the line.

Answer to Problem 48E

The probability that the point she selects is within 25 feet of the beginning of the line is 0.05.

Explanation of Solution

The probability that the point she selects is within 25 feet of the beginning of the line is obtained below:

Let random variable Y denotes the location of point and Y is assumed to have uniform distribution over interval (a, b). The length of interval (a, b) = 500 foot.

The probability density function for the uniform distribution on interval (θ1,θ2) is as follows:

f(y)=1θ2−θ1 θ1≤y≤θ2

Here, the interval will be (0, 500) and the length of the interval is 500.

f(y)={1500−0 0≤y≤5000 elsewhere

The probability that the point she selects is within 25 feet of the beginning of the line is over the interval (0, 25).

P(a<Y<b)=∫θ1θ2f(y)dyP(0<Y<25)=∫0251500−0dy=∫0251500dy=1500[y]025

=1500[25−0]=25500=0.05

Thus, the probability that the point she selects is within 25 feet of the beginning of the line is 0.05.

c.

Expert Solution

To determine

Compute the probability that the point selected by the expert is closer to the beginning of the line than the end of the line.

Answer to Problem 48E

The probability that the point selected by the expert is closer to the beginning of the line than the end of the line is 0.50.

Explanation of Solution

The probability that the point selected by the expert is closer to the beginning of the line than the end of the line is obtained below:

Let random variable Y denotes the location of point and Y is assumed to have uniform distribution over interval (a, b). The length of interval (a, b) = 500 foot.

The probability density function for the uniform distribution on interval (θ1,θ2) is as follows:

f(y)=1θ2−θ1 θ1≤y≤θ2

Here, the interval will be (0, 500) and the length of the interval is 500.

f(y)={1500−0 0≤y≤5000 elsewhere

The probability that the point selected by the expert is closer to the beginning of the line than the end of the line that is the point lies within the first half of the line, which is consider as over the interval (0, 250).

P(a<Y<b)=∫θ1θ2f(y)dyP(0<Y<250)=∫02501500−0dy=∫02501500dy=1500[y]0250

=1500[250−0]=250500=0.50

Thus, the probability that the point selected by the expert is closer to the beginning of the line than the end of the line is 0.50.

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Answer 3

Textbook Question

Chapter 4.4, Problem 48E

If a point is randomly located in an interval (a, b) and if Y denotes the location of the point, then Y is assumed to have a uniform distribution over (a, b). A plant efficiency expert randomly selects a location along a 500-foot assembly line from which to observe the work habits of the workers on the line. What is the probability that the point she selects is

a within 25 feet of the end of the line?
b within 25 feet of the beginning of the line?
c closer to the beginning of the line than to the end of the line?

a.

Expert Solution

To determine

Compute the probability that the point she selects is within 25 feet of the end of the line.

Answer to Problem 48E

The probability that the point she selects is within 25 feet of the end of the line is 0.05.

Explanation of Solution

The probability that the point she selects is within 25 feet of the end of the line is obtained below:

Let random variable Y denotes the location of point and Y is assumed to have uniform distribution over interval (a, b). The length of interval (a, b) = 500 foot.

The probability density function for the uniform distribution on interval (θ1,θ2) is as follows:

f(y)=1θ2−θ1 θ1≤y≤θ2

Here, the interval will be (0, 500) and the length of the interval is 500.

f(y)={1500−0 0≤y≤5000 elsewhere

The probability that the point she selects is within 25 feet of the end of the line is over the interval is as follows:

P(500−25<Y<500)=P(475<Y<500)

P(a<Y<b)=∫θ1θ2f(y)dyP(475<Y<500)=∫4755001500−0dy=∫4755001500dy=1500[y]475500

=1500[500−475]=25500=0.05

Thus, the probability that the point she selects is within 25 feet of the end of the line is 0.05.

b.

Expert Solution

To determine

Compute the probability that the point she selects is within 25 feet of the beginning of the line.

Answer to Problem 48E

The probability that the point she selects is within 25 feet of the beginning of the line is 0.05.

Explanation of Solution

The probability that the point she selects is within 25 feet of the beginning of the line is obtained below:

Let random variable Y denotes the location of point and Y is assumed to have uniform distribution over interval (a, b). The length of interval (a, b) = 500 foot.

The probability density function for the uniform distribution on interval (θ1,θ2) is as follows:

f(y)=1θ2−θ1 θ1≤y≤θ2

Here, the interval will be (0, 500) and the length of the interval is 500.

f(y)={1500−0 0≤y≤5000 elsewhere

The probability that the point she selects is within 25 feet of the beginning of the line is over the interval (0, 25).

P(a<Y<b)=∫θ1θ2f(y)dyP(0<Y<25)=∫0251500−0dy=∫0251500dy=1500[y]025

=1500[25−0]=25500=0.05

Thus, the probability that the point she selects is within 25 feet of the beginning of the line is 0.05.

c.

Expert Solution

To determine

Compute the probability that the point selected by the expert is closer to the beginning of the line than the end of the line.

Answer to Problem 48E

The probability that the point selected by the expert is closer to the beginning of the line than the end of the line is 0.50.

Explanation of Solution

The probability that the point selected by the expert is closer to the beginning of the line than the end of the line is obtained below:

Let random variable Y denotes the location of point and Y is assumed to have uniform distribution over interval (a, b). The length of interval (a, b) = 500 foot.

The probability density function for the uniform distribution on interval (θ1,θ2) is as follows:

f(y)=1θ2−θ1 θ1≤y≤θ2

Here, the interval will be (0, 500) and the length of the interval is 500.

f(y)={1500−0 0≤y≤5000 elsewhere

The probability that the point selected by the expert is closer to the beginning of the line than the end of the line that is the point lies within the first half of the line, which is consider as over the interval (0, 250).

P(a<Y<b)=∫θ1θ2f(y)dyP(0<Y<250)=∫02501500−0dy=∫02501500dy=1500[y]0250

=1500[250−0]=250500=0.50

Thus, the probability that the point selected by the expert is closer to the beginning of the line than the end of the line is 0.50.

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Answer 4

Textbook Question

Chapter 4.4, Problem 48E

If a point is randomly located in an interval (a, b) and if Y denotes the location of the point, then Y is assumed to have a uniform distribution over (a, b). A plant efficiency expert randomly selects a location along a 500-foot assembly line from which to observe the work habits of the workers on the line. What is the probability that the point she selects is

a within 25 feet of the end of the line?
b within 25 feet of the beginning of the line?
c closer to the beginning of the line than to the end of the line?

a.

Expert Solution

To determine

Compute the probability that the point she selects is within 25 feet of the end of the line.

Answer to Problem 48E

The probability that the point she selects is within 25 feet of the end of the line is 0.05.

Explanation of Solution

The probability that the point she selects is within 25 feet of the end of the line is obtained below:

Let random variable Y denotes the location of point and Y is assumed to have uniform distribution over interval (a, b). The length of interval (a, b) = 500 foot.

The probability density function for the uniform distribution on interval (θ1,θ2) is as follows:

f(y)=1θ2−θ1 θ1≤y≤θ2

Here, the interval will be (0, 500) and the length of the interval is 500.

f(y)={1500−0 0≤y≤5000 elsewhere

The probability that the point she selects is within 25 feet of the end of the line is over the interval is as follows:

P(500−25<Y<500)=P(475<Y<500)

P(a<Y<b)=∫θ1θ2f(y)dyP(475<Y<500)=∫4755001500−0dy=∫4755001500dy=1500[y]475500

=1500[500−475]=25500=0.05

Thus, the probability that the point she selects is within 25 feet of the end of the line is 0.05.

b.

Expert Solution

To determine

Compute the probability that the point she selects is within 25 feet of the beginning of the line.

Answer to Problem 48E

The probability that the point she selects is within 25 feet of the beginning of the line is 0.05.

Explanation of Solution

The probability that the point she selects is within 25 feet of the beginning of the line is obtained below:

Let random variable Y denotes the location of point and Y is assumed to have uniform distribution over interval (a, b). The length of interval (a, b) = 500 foot.

The probability density function for the uniform distribution on interval (θ1,θ2) is as follows:

f(y)=1θ2−θ1 θ1≤y≤θ2

Here, the interval will be (0, 500) and the length of the interval is 500.

f(y)={1500−0 0≤y≤5000 elsewhere

The probability that the point she selects is within 25 feet of the beginning of the line is over the interval (0, 25).

P(a<Y<b)=∫θ1θ2f(y)dyP(0<Y<25)=∫0251500−0dy=∫0251500dy=1500[y]025

=1500[25−0]=25500=0.05

Thus, the probability that the point she selects is within 25 feet of the beginning of the line is 0.05.

c.

Expert Solution

To determine

Compute the probability that the point selected by the expert is closer to the beginning of the line than the end of the line.

Answer to Problem 48E

The probability that the point selected by the expert is closer to the beginning of the line than the end of the line is 0.50.

Explanation of Solution

The probability that the point selected by the expert is closer to the beginning of the line than the end of the line is obtained below:

Let random variable Y denotes the location of point and Y is assumed to have uniform distribution over interval (a, b). The length of interval (a, b) = 500 foot.

The probability density function for the uniform distribution on interval (θ1,θ2) is as follows:

f(y)=1θ2−θ1 θ1≤y≤θ2

Here, the interval will be (0, 500) and the length of the interval is 500.

f(y)={1500−0 0≤y≤5000 elsewhere

The probability that the point selected by the expert is closer to the beginning of the line than the end of the line that is the point lies within the first half of the line, which is consider as over the interval (0, 250).

P(a<Y<b)=∫θ1θ2f(y)dyP(0<Y<250)=∫02501500−0dy=∫02501500dy=1500[y]0250

=1500[250−0]=250500=0.50

Thus, the probability that the point selected by the expert is closer to the beginning of the line than the end of the line is 0.50.

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Answer 5

Textbook Question

Answer 6

Textbook Question

Answer 7

a.

Expert Solution

To determine

Compute the probability that the point she selects is within 25 feet of the end of the line.

Answer to Problem 48E

The probability that the point she selects is within 25 feet of the end of the line is 0.05.

Explanation of Solution

The probability that the point she selects is within 25 feet of the end of the line is obtained below:

Let random variable Y denotes the location of point and Y is assumed to have uniform distribution over interval (a, b). The length of interval (a, b) = 500 foot.

The probability density function for the uniform distribution on interval (θ1,θ2) is as follows:

f(y)=1θ2−θ1 θ1≤y≤θ2

Here, the interval will be (0, 500) and the length of the interval is 500.

f(y)={1500−0 0≤y≤5000 elsewhere

The probability that the point she selects is within 25 feet of the end of the line is over the interval is as follows:

P(500−25<Y<500)=P(475<Y<500)

P(a<Y<b)=∫θ1θ2f(y)dyP(475<Y<500)=∫4755001500−0dy=∫4755001500dy=1500[y]475500

=1500[500−475]=25500=0.05

Thus, the probability that the point she selects is within 25 feet of the end of the line is 0.05.

b.

Expert Solution

To determine

Compute the probability that the point she selects is within 25 feet of the beginning of the line.

Answer to Problem 48E

The probability that the point she selects is within 25 feet of the beginning of the line is 0.05.

Explanation of Solution

The probability that the point she selects is within 25 feet of the beginning of the line is obtained below:

Let random variable Y denotes the location of point and Y is assumed to have uniform distribution over interval (a, b). The length of interval (a, b) = 500 foot.

The probability density function for the uniform distribution on interval (θ1,θ2) is as follows:

f(y)=1θ2−θ1 θ1≤y≤θ2

Here, the interval will be (0, 500) and the length of the interval is 500.

f(y)={1500−0 0≤y≤5000 elsewhere

The probability that the point she selects is within 25 feet of the beginning of the line is over the interval (0, 25).

P(a<Y<b)=∫θ1θ2f(y)dyP(0<Y<25)=∫0251500−0dy=∫0251500dy=1500[y]025

=1500[25−0]=25500=0.05

Thus, the probability that the point she selects is within 25 feet of the beginning of the line is 0.05.

c.

Expert Solution

To determine

Compute the probability that the point selected by the expert is closer to the beginning of the line than the end of the line.

Answer to Problem 48E

The probability that the point selected by the expert is closer to the beginning of the line than the end of the line is 0.50.

Explanation of Solution

The probability that the point selected by the expert is closer to the beginning of the line than the end of the line is obtained below:

Let random variable Y denotes the location of point and Y is assumed to have uniform distribution over interval (a, b). The length of interval (a, b) = 500 foot.

The probability density function for the uniform distribution on interval (θ1,θ2) is as follows:

f(y)=1θ2−θ1 θ1≤y≤θ2

Here, the interval will be (0, 500) and the length of the interval is 500.

f(y)={1500−0 0≤y≤5000 elsewhere

The probability that the point selected by the expert is closer to the beginning of the line than the end of the line that is the point lies within the first half of the line, which is consider as over the interval (0, 250).

P(a<Y<b)=∫θ1θ2f(y)dyP(0<Y<250)=∫02501500−0dy=∫02501500dy=1500[y]0250

=1500[250−0]=250500=0.50

Thus, the probability that the point selected by the expert is closer to the beginning of the line than the end of the line is 0.50.

Answer 8

a.

Expert Solution

To determine

Compute the probability that the point she selects is within 25 feet of the end of the line.

Answer to Problem 48E

The probability that the point she selects is within 25 feet of the end of the line is 0.05.

Explanation of Solution

The probability that the point she selects is within 25 feet of the end of the line is obtained below:

Let random variable Y denotes the location of point and Y is assumed to have uniform distribution over interval (a, b). The length of interval (a, b) = 500 foot.

The probability density function for the uniform distribution on interval (θ1,θ2) is as follows:

f(y)=1θ2−θ1 θ1≤y≤θ2

Here, the interval will be (0, 500) and the length of the interval is 500.

f(y)={1500−0 0≤y≤5000 elsewhere

The probability that the point she selects is within 25 feet of the end of the line is over the interval is as follows:

P(500−25<Y<500)=P(475<Y<500)

P(a<Y<b)=∫θ1θ2f(y)dyP(475<Y<500)=∫4755001500−0dy=∫4755001500dy=1500[y]475500

=1500[500−475]=25500=0.05

Thus, the probability that the point she selects is within 25 feet of the end of the line is 0.05.

Answer 9

a.

Expert Solution

Answer 10

a.

Expert Solution

Answer 11

Expert Solution

Answer 12

To determine

Compute the probability that the point she selects is within 25 feet of the end of the line.

Answer 13

Answer to Problem 48E

The probability that the point she selects is within 25 feet of the end of the line is 0.05.

Answer 14

Explanation of Solution

The probability that the point she selects is within 25 feet of the end of the line is obtained below:

Let random variable Y denotes the location of point and Y is assumed to have uniform distribution over interval (a, b). The length of interval (a, b) = 500 foot.

The probability density function for the uniform distribution on interval (θ1,θ2) is as follows:

f(y)=1θ2−θ1 θ1≤y≤θ2

Here, the interval will be (0, 500) and the length of the interval is 500.

f(y)={1500−0 0≤y≤5000 elsewhere

The probability that the point she selects is within 25 feet of the end of the line is over the interval is as follows:

P(500−25<Y<500)=P(475<Y<500)

P(a<Y<b)=∫θ1θ2f(y)dyP(475<Y<500)=∫4755001500−0dy=∫4755001500dy=1500[y]475500

=1500[500−475]=25500=0.05

Thus, the probability that the point she selects is within 25 feet of the end of the line is 0.05.

Answer 15

b.

Expert Solution

To determine

Compute the probability that the point she selects is within 25 feet of the beginning of the line.

Answer to Problem 48E

The probability that the point she selects is within 25 feet of the beginning of the line is 0.05.

Explanation of Solution

The probability that the point she selects is within 25 feet of the beginning of the line is obtained below:

Let random variable Y denotes the location of point and Y is assumed to have uniform distribution over interval (a, b). The length of interval (a, b) = 500 foot.

The probability density function for the uniform distribution on interval (θ1,θ2) is as follows:

f(y)=1θ2−θ1 θ1≤y≤θ2

Here, the interval will be (0, 500) and the length of the interval is 500.

f(y)={1500−0 0≤y≤5000 elsewhere

The probability that the point she selects is within 25 feet of the beginning of the line is over the interval (0, 25).

P(a<Y<b)=∫θ1θ2f(y)dyP(0<Y<25)=∫0251500−0dy=∫0251500dy=1500[y]025

=1500[25−0]=25500=0.05

Thus, the probability that the point she selects is within 25 feet of the beginning of the line is 0.05.

Answer 16

b.

Expert Solution

Answer 17

b.

Expert Solution

Answer 18

Expert Solution

Answer 19

To determine

Compute the probability that the point she selects is within 25 feet of the beginning of the line.

Answer 20

Answer to Problem 48E

The probability that the point she selects is within 25 feet of the beginning of the line is 0.05.

Answer 21

Explanation of Solution

The probability that the point she selects is within 25 feet of the beginning of the line is obtained below:

Let random variable Y denotes the location of point and Y is assumed to have uniform distribution over interval (a, b). The length of interval (a, b) = 500 foot.

The probability density function for the uniform distribution on interval (θ1,θ2) is as follows:

f(y)=1θ2−θ1 θ1≤y≤θ2

Here, the interval will be (0, 500) and the length of the interval is 500.

f(y)={1500−0 0≤y≤5000 elsewhere

The probability that the point she selects is within 25 feet of the beginning of the line is over the interval (0, 25).

P(a<Y<b)=∫θ1θ2f(y)dyP(0<Y<25)=∫0251500−0dy=∫0251500dy=1500[y]025

=1500[25−0]=25500=0.05

Thus, the probability that the point she selects is within 25 feet of the beginning of the line is 0.05.

Answer 22

c.

Expert Solution

To determine

Compute the probability that the point selected by the expert is closer to the beginning of the line than the end of the line.

Answer to Problem 48E

The probability that the point selected by the expert is closer to the beginning of the line than the end of the line is 0.50.

Explanation of Solution

The probability that the point selected by the expert is closer to the beginning of the line than the end of the line is obtained below:

Let random variable Y denotes the location of point and Y is assumed to have uniform distribution over interval (a, b). The length of interval (a, b) = 500 foot.

The probability density function for the uniform distribution on interval (θ1,θ2) is as follows:

f(y)=1θ2−θ1 θ1≤y≤θ2

Here, the interval will be (0, 500) and the length of the interval is 500.

f(y)={1500−0 0≤y≤5000 elsewhere

The probability that the point selected by the expert is closer to the beginning of the line than the end of the line that is the point lies within the first half of the line, which is consider as over the interval (0, 250).

P(a<Y<b)=∫θ1θ2f(y)dyP(0<Y<250)=∫02501500−0dy=∫02501500dy=1500[y]0250

=1500[250−0]=250500=0.50

Thus, the probability that the point selected by the expert is closer to the beginning of the line than the end of the line is 0.50.

Answer 23

c.

Expert Solution

Answer 24

c.

Expert Solution

Answer 25

Expert Solution

Answer 26

To determine

Compute the probability that the point selected by the expert is closer to the beginning of the line than the end of the line.

Answer 27

Answer to Problem 48E

The probability that the point selected by the expert is closer to the beginning of the line than the end of the line is 0.50.

Answer 28

Explanation of Solution

The probability that the point selected by the expert is closer to the beginning of the line than the end of the line is obtained below:

Let random variable Y denotes the location of point and Y is assumed to have uniform distribution over interval (a, b). The length of interval (a, b) = 500 foot.

The probability density function for the uniform distribution on interval (θ1,θ2) is as follows:

f(y)=1θ2−θ1 θ1≤y≤θ2

Here, the interval will be (0, 500) and the length of the interval is 500.

f(y)={1500−0 0≤y≤5000 elsewhere

The probability that the point selected by the expert is closer to the beginning of the line than the end of the line that is the point lies within the first half of the line, which is consider as over the interval (0, 250).

P(a<Y<b)=∫θ1θ2f(y)dyP(0<Y<250)=∫02501500−0dy=∫02501500dy=1500[y]0250