Unifrom Distribution: Measurement Errors Measurement errors from instruments are often modeled using the uniform distribution (see Problem 12). To determine the range of a large public address system, acoustical engineers use a method of triangulation to measure the shock waves sent out by the speakers. The time at which the waves arrive at the sensors must be measured accurately. In this context, a negative error means the signal arrived too early. A positive error means the signal arrived too late. Measurement errors in reading these times have a uniform distribution from -0.05 to +0.05 microseconds (Reference J. Perruzzi and E. Hilliard, "Modeling Time Delay Measurement Errors." Journal of the Acoustical Societyof America, Vol. 75, No. 1, pp. 197-201). What is the probability that such measurements will be in error by (a) less than +0.03 microsecond ( i .e . , − 0.05 ≤ x < 0.03 ) ? (b) more than -0.02 microsecond? (c) between -0.04 and +0.01 microsecond? (d) Find the mean and standard deviation of measurement errors. Measurements from an instrument are called unbiased if the mean of the measurement errors is zero. Would you say the measurements for these acoustical sensors are unbiased? Explain.

Question

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Answer 1

Textbook Question

Chapter 7.1, Problem 13P

Unifrom Distribution: Measurement Errors Measurement errors from instruments are often modeled using the uniform distribution (see Problem 12). To determine the range of a large public address system, acoustical engineers use a method of triangulation to measure the shock waves sent out by the speakers. The time at which the waves arrive at the sensors must be measured accurately. In this context, a negative error means the signal arrived too early. A positive error means the signal arrived too late. Measurement errors in reading these times have a uniform distribution from -0.05 to +0.05 microseconds (Reference J. Perruzzi and E. Hilliard, "Modeling Time Delay Measurement Errors." Journal of the Acoustical Societyof America, Vol. 75, No. 1, pp. 197-201). What is the probability that such measurements will be in error by

(a) less than +0.03 microsecond ( i .e . , − 0.05 ≤ x < 0.03 ) ? (b) more than -0.02 microsecond?

(c) between -0.04 and +0.01 microsecond?

(d) Find the mean and standard deviation of measurement errors. Measurements from an instrument are called unbiased if the mean of the measurement errors is zero. Would you say the measurements for these acoustical sensors are unbiased? Explain.

(a)

Expert Solution

To determine

To find: Theprobability thatsuch measurements will be in error byless than 0.03 microseconds.

Answer to Problem 13P

Solution: Theprobability that such measurements will be in error byless than 0.03 microsecondsis0.8.

Explanation of Solution

Calculation:

Let x be themeasurement errors in reading these times have a uniform distribution from α=−0.05 to β=0.05 microseconds.

Let x is chosen at random from [α, β], the area of the rectangle that lies above [a, b] is the probability that x lies in [a, b]. This area is P(a≤x≤b)=b−aβ−α

We have to find the probability of less than 0.03 microseconds,

P(−0.05≤x≤0.03)=b−aβ−αP(−0.05≤x≤0.03)=0.03−(−0.05)0.05−(−0.05)P(−0.05≤x≤0.03)=0.8

Hence, P(−0.05≤x≤0.03)=0.8.

(b)

Expert Solution

To determine

To find: Theprobability thatsuch measurements will be in error bymore than -0.02 microseconds.

Answer to Problem 13P

Solution: Theprobability that such measurements will be in error by more than -0.02 microsecondsis0.7.

Explanation of Solution

Calculation:

Let x be the measurement errors in reading these times have a uniform distribution from α=−0.05 to β=0.05 microseconds.

Let x is chosen at random from [α, β], the area of the rectangle that lies above [a, b] is the probability that x lies in [a, b]. This area is P(a≤x≤b)=b−aβ−α

We have to find the probability of more than -0.02 microseconds,

P(−0.02≤x≤0.05)=b−aβ−αP(−0.02≤x≤0.05)=0.05−(−0.02)0.05−(−0.05)P(−0.02≤x≤0.05)=0.7

Hence, P(−0.02≤x≤0.05)=0.7.

(c)

Expert Solution

To determine

To find: Theprobability thatsuch measurements will be in error bybetween -0.04 and 0.01 microseconds.

Answer to Problem 13P

Solution: Theprobability that such measurements will be in error bybetween -0.04 and 0.01 microsecondsis0.5.

Explanation of Solution

Calculation:

Let x be the measurement errors in reading these times have a uniform distribution from α=−0.05 to β=0.05 microseconds.

Let x is chosen at random from [α, β], the area of the rectangle that lies above [a, b] is the probability that x lies in [a, b]. This area is P(a≤x≤b)=b−aβ−α

We have to find the probability between -0.04 and 0.01 microseconds,

P(−0.04≤x≤0.01)=b−aβ−αP(−0.04≤x≤0.01)=0.01−(−0.04)0.05−(−0.05)P(−0.04≤x≤0.01)=0.5

Hence, P(−0.04≤x≤0.01)=0.5.

(d)

Expert Solution

To determine

To find: Themean and standard deviation of measurement errors and whether the measurements for these acoustical sensors are unbiased.

Answer to Problem 13P

Solution:

The mean of measurement errors is 0. The standard deviation of measurement errors is 0.029. Yes, the measurements for these acoustical sensors are unbiased.

Explanation of Solution

Calculation:

Let x be the measurement errors in reading these times have a uniform distribution from α=−0.05 to β=0.05 microseconds.

We have to find the mean,

μ=α+β2μ=0.05+(−0.05)2μ=0

Therefore, the mean is 0 microsecond.

We have to find the standard deviation,

σ=β−α12σ=0.05−(−0.05)12σ≈0.029

Therefore, thestandard deviationis 0.029microsecond.

Yes, since the obtained mean of the measurement errors is zero so the measurements for these acoustical sensors are unbiased.

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Answer 2

Textbook Question

Chapter 7.1, Problem 13P

Unifrom Distribution: Measurement Errors Measurement errors from instruments are often modeled using the uniform distribution (see Problem 12). To determine the range of a large public address system, acoustical engineers use a method of triangulation to measure the shock waves sent out by the speakers. The time at which the waves arrive at the sensors must be measured accurately. In this context, a negative error means the signal arrived too early. A positive error means the signal arrived too late. Measurement errors in reading these times have a uniform distribution from -0.05 to +0.05 microseconds (Reference J. Perruzzi and E. Hilliard, "Modeling Time Delay Measurement Errors." Journal of the Acoustical Societyof America, Vol. 75, No. 1, pp. 197-201). What is the probability that such measurements will be in error by

(a) less than +0.03 microsecond ( i .e . , − 0.05 ≤ x < 0.03 ) ? (b) more than -0.02 microsecond?

(c) between -0.04 and +0.01 microsecond?

(d) Find the mean and standard deviation of measurement errors. Measurements from an instrument are called unbiased if the mean of the measurement errors is zero. Would you say the measurements for these acoustical sensors are unbiased? Explain.

(a)

Expert Solution

To determine

To find: Theprobability thatsuch measurements will be in error byless than 0.03 microseconds.

Answer to Problem 13P

Solution: Theprobability that such measurements will be in error byless than 0.03 microsecondsis0.8.

Explanation of Solution

Calculation:

Let x be themeasurement errors in reading these times have a uniform distribution from α=−0.05 to β=0.05 microseconds.

Let x is chosen at random from [α, β], the area of the rectangle that lies above [a, b] is the probability that x lies in [a, b]. This area is P(a≤x≤b)=b−aβ−α

We have to find the probability of less than 0.03 microseconds,

P(−0.05≤x≤0.03)=b−aβ−αP(−0.05≤x≤0.03)=0.03−(−0.05)0.05−(−0.05)P(−0.05≤x≤0.03)=0.8

Hence, P(−0.05≤x≤0.03)=0.8.

(b)

Expert Solution

To determine

To find: Theprobability thatsuch measurements will be in error bymore than -0.02 microseconds.

Answer to Problem 13P

Solution: Theprobability that such measurements will be in error by more than -0.02 microsecondsis0.7.

Explanation of Solution

Calculation:

Let x be the measurement errors in reading these times have a uniform distribution from α=−0.05 to β=0.05 microseconds.

Let x is chosen at random from [α, β], the area of the rectangle that lies above [a, b] is the probability that x lies in [a, b]. This area is P(a≤x≤b)=b−aβ−α

We have to find the probability of more than -0.02 microseconds,

P(−0.02≤x≤0.05)=b−aβ−αP(−0.02≤x≤0.05)=0.05−(−0.02)0.05−(−0.05)P(−0.02≤x≤0.05)=0.7

Hence, P(−0.02≤x≤0.05)=0.7.

(c)

Expert Solution

To determine

To find: Theprobability thatsuch measurements will be in error bybetween -0.04 and 0.01 microseconds.

Answer to Problem 13P

Solution: Theprobability that such measurements will be in error bybetween -0.04 and 0.01 microsecondsis0.5.

Explanation of Solution

Calculation:

Let x be the measurement errors in reading these times have a uniform distribution from α=−0.05 to β=0.05 microseconds.

Let x is chosen at random from [α, β], the area of the rectangle that lies above [a, b] is the probability that x lies in [a, b]. This area is P(a≤x≤b)=b−aβ−α

We have to find the probability between -0.04 and 0.01 microseconds,

P(−0.04≤x≤0.01)=b−aβ−αP(−0.04≤x≤0.01)=0.01−(−0.04)0.05−(−0.05)P(−0.04≤x≤0.01)=0.5

Hence, P(−0.04≤x≤0.01)=0.5.

(d)

Expert Solution

To determine

To find: Themean and standard deviation of measurement errors and whether the measurements for these acoustical sensors are unbiased.

Answer to Problem 13P

Solution:

The mean of measurement errors is 0. The standard deviation of measurement errors is 0.029. Yes, the measurements for these acoustical sensors are unbiased.

Explanation of Solution

Calculation:

Let x be the measurement errors in reading these times have a uniform distribution from α=−0.05 to β=0.05 microseconds.

We have to find the mean,

μ=α+β2μ=0.05+(−0.05)2μ=0

Therefore, the mean is 0 microsecond.

We have to find the standard deviation,

σ=β−α12σ=0.05−(−0.05)12σ≈0.029

Therefore, thestandard deviationis 0.029microsecond.

Yes, since the obtained mean of the measurement errors is zero so the measurements for these acoustical sensors are unbiased.

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Answer 3

Textbook Question

Chapter 7.1, Problem 13P

Unifrom Distribution: Measurement Errors Measurement errors from instruments are often modeled using the uniform distribution (see Problem 12). To determine the range of a large public address system, acoustical engineers use a method of triangulation to measure the shock waves sent out by the speakers. The time at which the waves arrive at the sensors must be measured accurately. In this context, a negative error means the signal arrived too early. A positive error means the signal arrived too late. Measurement errors in reading these times have a uniform distribution from -0.05 to +0.05 microseconds (Reference J. Perruzzi and E. Hilliard, "Modeling Time Delay Measurement Errors." Journal of the Acoustical Societyof America, Vol. 75, No. 1, pp. 197-201). What is the probability that such measurements will be in error by

(a) less than +0.03 microsecond ( i .e . , − 0.05 ≤ x < 0.03 ) ? (b) more than -0.02 microsecond?

(c) between -0.04 and +0.01 microsecond?

(d) Find the mean and standard deviation of measurement errors. Measurements from an instrument are called unbiased if the mean of the measurement errors is zero. Would you say the measurements for these acoustical sensors are unbiased? Explain.

(a)

Expert Solution

To determine

To find: Theprobability thatsuch measurements will be in error byless than 0.03 microseconds.

Answer to Problem 13P

Solution: Theprobability that such measurements will be in error byless than 0.03 microsecondsis0.8.

Explanation of Solution

Calculation:

Let x be themeasurement errors in reading these times have a uniform distribution from α=−0.05 to β=0.05 microseconds.

Let x is chosen at random from [α, β], the area of the rectangle that lies above [a, b] is the probability that x lies in [a, b]. This area is P(a≤x≤b)=b−aβ−α

We have to find the probability of less than 0.03 microseconds,

P(−0.05≤x≤0.03)=b−aβ−αP(−0.05≤x≤0.03)=0.03−(−0.05)0.05−(−0.05)P(−0.05≤x≤0.03)=0.8

Hence, P(−0.05≤x≤0.03)=0.8.

(b)

Expert Solution

To determine

To find: Theprobability thatsuch measurements will be in error bymore than -0.02 microseconds.

Answer to Problem 13P

Solution: Theprobability that such measurements will be in error by more than -0.02 microsecondsis0.7.

Explanation of Solution

Calculation:

Let x be the measurement errors in reading these times have a uniform distribution from α=−0.05 to β=0.05 microseconds.

Let x is chosen at random from [α, β], the area of the rectangle that lies above [a, b] is the probability that x lies in [a, b]. This area is P(a≤x≤b)=b−aβ−α

We have to find the probability of more than -0.02 microseconds,

P(−0.02≤x≤0.05)=b−aβ−αP(−0.02≤x≤0.05)=0.05−(−0.02)0.05−(−0.05)P(−0.02≤x≤0.05)=0.7

Hence, P(−0.02≤x≤0.05)=0.7.

(c)

Expert Solution

To determine

To find: Theprobability thatsuch measurements will be in error bybetween -0.04 and 0.01 microseconds.

Answer to Problem 13P

Solution: Theprobability that such measurements will be in error bybetween -0.04 and 0.01 microsecondsis0.5.

Explanation of Solution

Calculation:

Let x be the measurement errors in reading these times have a uniform distribution from α=−0.05 to β=0.05 microseconds.

Let x is chosen at random from [α, β], the area of the rectangle that lies above [a, b] is the probability that x lies in [a, b]. This area is P(a≤x≤b)=b−aβ−α

We have to find the probability between -0.04 and 0.01 microseconds,

P(−0.04≤x≤0.01)=b−aβ−αP(−0.04≤x≤0.01)=0.01−(−0.04)0.05−(−0.05)P(−0.04≤x≤0.01)=0.5

Hence, P(−0.04≤x≤0.01)=0.5.

(d)

Expert Solution

To determine

To find: Themean and standard deviation of measurement errors and whether the measurements for these acoustical sensors are unbiased.

Answer to Problem 13P

Solution:

The mean of measurement errors is 0. The standard deviation of measurement errors is 0.029. Yes, the measurements for these acoustical sensors are unbiased.

Explanation of Solution

Calculation:

Let x be the measurement errors in reading these times have a uniform distribution from α=−0.05 to β=0.05 microseconds.

We have to find the mean,

μ=α+β2μ=0.05+(−0.05)2μ=0

Therefore, the mean is 0 microsecond.

We have to find the standard deviation,

σ=β−α12σ=0.05−(−0.05)12σ≈0.029

Therefore, thestandard deviationis 0.029microsecond.

Yes, since the obtained mean of the measurement errors is zero so the measurements for these acoustical sensors are unbiased.

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Answer 4

Textbook Question

Chapter 7.1, Problem 13P

Unifrom Distribution: Measurement Errors Measurement errors from instruments are often modeled using the uniform distribution (see Problem 12). To determine the range of a large public address system, acoustical engineers use a method of triangulation to measure the shock waves sent out by the speakers. The time at which the waves arrive at the sensors must be measured accurately. In this context, a negative error means the signal arrived too early. A positive error means the signal arrived too late. Measurement errors in reading these times have a uniform distribution from -0.05 to +0.05 microseconds (Reference J. Perruzzi and E. Hilliard, "Modeling Time Delay Measurement Errors." Journal of the Acoustical Societyof America, Vol. 75, No. 1, pp. 197-201). What is the probability that such measurements will be in error by

(a) less than +0.03 microsecond ( i .e . , − 0.05 ≤ x < 0.03 ) ? (b) more than -0.02 microsecond?

(c) between -0.04 and +0.01 microsecond?

(d) Find the mean and standard deviation of measurement errors. Measurements from an instrument are called unbiased if the mean of the measurement errors is zero. Would you say the measurements for these acoustical sensors are unbiased? Explain.

(a)

Expert Solution

To determine

To find: Theprobability thatsuch measurements will be in error byless than 0.03 microseconds.

Answer to Problem 13P

Solution: Theprobability that such measurements will be in error byless than 0.03 microsecondsis0.8.

Explanation of Solution

Calculation:

Let x be themeasurement errors in reading these times have a uniform distribution from α=−0.05 to β=0.05 microseconds.

Let x is chosen at random from [α, β], the area of the rectangle that lies above [a, b] is the probability that x lies in [a, b]. This area is P(a≤x≤b)=b−aβ−α

We have to find the probability of less than 0.03 microseconds,

P(−0.05≤x≤0.03)=b−aβ−αP(−0.05≤x≤0.03)=0.03−(−0.05)0.05−(−0.05)P(−0.05≤x≤0.03)=0.8

Hence, P(−0.05≤x≤0.03)=0.8.

(b)

Expert Solution

To determine

To find: Theprobability thatsuch measurements will be in error bymore than -0.02 microseconds.

Answer to Problem 13P

Solution: Theprobability that such measurements will be in error by more than -0.02 microsecondsis0.7.

Explanation of Solution

Calculation:

Let x be the measurement errors in reading these times have a uniform distribution from α=−0.05 to β=0.05 microseconds.

Let x is chosen at random from [α, β], the area of the rectangle that lies above [a, b] is the probability that x lies in [a, b]. This area is P(a≤x≤b)=b−aβ−α

We have to find the probability of more than -0.02 microseconds,

P(−0.02≤x≤0.05)=b−aβ−αP(−0.02≤x≤0.05)=0.05−(−0.02)0.05−(−0.05)P(−0.02≤x≤0.05)=0.7

Hence, P(−0.02≤x≤0.05)=0.7.

(c)

Expert Solution

To determine

To find: Theprobability thatsuch measurements will be in error bybetween -0.04 and 0.01 microseconds.

Answer to Problem 13P

Solution: Theprobability that such measurements will be in error bybetween -0.04 and 0.01 microsecondsis0.5.

Explanation of Solution

Calculation:

Let x be the measurement errors in reading these times have a uniform distribution from α=−0.05 to β=0.05 microseconds.

Let x is chosen at random from [α, β], the area of the rectangle that lies above [a, b] is the probability that x lies in [a, b]. This area is P(a≤x≤b)=b−aβ−α

We have to find the probability between -0.04 and 0.01 microseconds,

P(−0.04≤x≤0.01)=b−aβ−αP(−0.04≤x≤0.01)=0.01−(−0.04)0.05−(−0.05)P(−0.04≤x≤0.01)=0.5

Hence, P(−0.04≤x≤0.01)=0.5.

(d)

Expert Solution

To determine

To find: Themean and standard deviation of measurement errors and whether the measurements for these acoustical sensors are unbiased.

Answer to Problem 13P

Solution:

The mean of measurement errors is 0. The standard deviation of measurement errors is 0.029. Yes, the measurements for these acoustical sensors are unbiased.

Explanation of Solution

Calculation:

Let x be the measurement errors in reading these times have a uniform distribution from α=−0.05 to β=0.05 microseconds.

We have to find the mean,

μ=α+β2μ=0.05+(−0.05)2μ=0

Therefore, the mean is 0 microsecond.

We have to find the standard deviation,

σ=β−α12σ=0.05−(−0.05)12σ≈0.029

Therefore, thestandard deviationis 0.029microsecond.

Yes, since the obtained mean of the measurement errors is zero so the measurements for these acoustical sensors are unbiased.

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

Answer 5

Textbook Question

Answer 6

Textbook Question

Answer 7

(a)

Expert Solution

To determine

To find: Theprobability thatsuch measurements will be in error byless than 0.03 microseconds.

Answer to Problem 13P

Solution: Theprobability that such measurements will be in error byless than 0.03 microsecondsis0.8.

Explanation of Solution

Calculation:

Let x be themeasurement errors in reading these times have a uniform distribution from α=−0.05 to β=0.05 microseconds.

Let x is chosen at random from [α, β], the area of the rectangle that lies above [a, b] is the probability that x lies in [a, b]. This area is P(a≤x≤b)=b−aβ−α

We have to find the probability of less than 0.03 microseconds,

P(−0.05≤x≤0.03)=b−aβ−αP(−0.05≤x≤0.03)=0.03−(−0.05)0.05−(−0.05)P(−0.05≤x≤0.03)=0.8

Hence, P(−0.05≤x≤0.03)=0.8.

(b)

Expert Solution

To determine

To find: Theprobability thatsuch measurements will be in error bymore than -0.02 microseconds.

Answer to Problem 13P

Solution: Theprobability that such measurements will be in error by more than -0.02 microsecondsis0.7.

Explanation of Solution

Calculation:

Let x be the measurement errors in reading these times have a uniform distribution from α=−0.05 to β=0.05 microseconds.

Let x is chosen at random from [α, β], the area of the rectangle that lies above [a, b] is the probability that x lies in [a, b]. This area is P(a≤x≤b)=b−aβ−α

We have to find the probability of more than -0.02 microseconds,

P(−0.02≤x≤0.05)=b−aβ−αP(−0.02≤x≤0.05)=0.05−(−0.02)0.05−(−0.05)P(−0.02≤x≤0.05)=0.7

Hence, P(−0.02≤x≤0.05)=0.7.

(c)

Expert Solution

To determine

To find: Theprobability thatsuch measurements will be in error bybetween -0.04 and 0.01 microseconds.

Answer to Problem 13P

Solution: Theprobability that such measurements will be in error bybetween -0.04 and 0.01 microsecondsis0.5.

Explanation of Solution

Calculation:

Let x be the measurement errors in reading these times have a uniform distribution from α=−0.05 to β=0.05 microseconds.

Let x is chosen at random from [α, β], the area of the rectangle that lies above [a, b] is the probability that x lies in [a, b]. This area is P(a≤x≤b)=b−aβ−α

We have to find the probability between -0.04 and 0.01 microseconds,

P(−0.04≤x≤0.01)=b−aβ−αP(−0.04≤x≤0.01)=0.01−(−0.04)0.05−(−0.05)P(−0.04≤x≤0.01)=0.5

Hence, P(−0.04≤x≤0.01)=0.5.

(d)

Expert Solution

To determine

To find: Themean and standard deviation of measurement errors and whether the measurements for these acoustical sensors are unbiased.

Answer to Problem 13P

Solution:

The mean of measurement errors is 0. The standard deviation of measurement errors is 0.029. Yes, the measurements for these acoustical sensors are unbiased.

Explanation of Solution

Calculation:

Let x be the measurement errors in reading these times have a uniform distribution from α=−0.05 to β=0.05 microseconds.

We have to find the mean,

μ=α+β2μ=0.05+(−0.05)2μ=0

Therefore, the mean is 0 microsecond.

We have to find the standard deviation,

σ=β−α12σ=0.05−(−0.05)12σ≈0.029

Therefore, thestandard deviationis 0.029microsecond.

Yes, since the obtained mean of the measurement errors is zero so the measurements for these acoustical sensors are unbiased.

Answer 8

(a)

Expert Solution

To determine

To find: Theprobability thatsuch measurements will be in error byless than 0.03 microseconds.

Answer to Problem 13P

Solution: Theprobability that such measurements will be in error byless than 0.03 microsecondsis0.8.

Explanation of Solution

Calculation:

Let x be themeasurement errors in reading these times have a uniform distribution from α=−0.05 to β=0.05 microseconds.

Let x is chosen at random from [α, β], the area of the rectangle that lies above [a, b] is the probability that x lies in [a, b]. This area is P(a≤x≤b)=b−aβ−α

We have to find the probability of less than 0.03 microseconds,

P(−0.05≤x≤0.03)=b−aβ−αP(−0.05≤x≤0.03)=0.03−(−0.05)0.05−(−0.05)P(−0.05≤x≤0.03)=0.8

Hence, P(−0.05≤x≤0.03)=0.8.

Answer 9

(a)

Expert Solution

Answer 10

(a)

Expert Solution

Answer 11

Expert Solution

Answer 12

To determine

To find: Theprobability thatsuch measurements will be in error byless than 0.03 microseconds.

Answer 13

Answer to Problem 13P

Solution: Theprobability that such measurements will be in error byless than 0.03 microsecondsis0.8.

Answer 14

Explanation of Solution

Calculation:

Let x be themeasurement errors in reading these times have a uniform distribution from α=−0.05 to β=0.05 microseconds.

Let x is chosen at random from [α, β], the area of the rectangle that lies above [a, b] is the probability that x lies in [a, b]. This area is P(a≤x≤b)=b−aβ−α

We have to find the probability of less than 0.03 microseconds,

P(−0.05≤x≤0.03)=b−aβ−αP(−0.05≤x≤0.03)=0.03−(−0.05)0.05−(−0.05)P(−0.05≤x≤0.03)=0.8

Hence, P(−0.05≤x≤0.03)=0.8.

Answer 15

(b)

Expert Solution

To determine

To find: Theprobability thatsuch measurements will be in error bymore than -0.02 microseconds.

Answer to Problem 13P

Solution: Theprobability that such measurements will be in error by more than -0.02 microsecondsis0.7.

Explanation of Solution

Calculation:

Let x be the measurement errors in reading these times have a uniform distribution from α=−0.05 to β=0.05 microseconds.

Let x is chosen at random from [α, β], the area of the rectangle that lies above [a, b] is the probability that x lies in [a, b]. This area is P(a≤x≤b)=b−aβ−α

We have to find the probability of more than -0.02 microseconds,

P(−0.02≤x≤0.05)=b−aβ−αP(−0.02≤x≤0.05)=0.05−(−0.02)0.05−(−0.05)P(−0.02≤x≤0.05)=0.7

Hence, P(−0.02≤x≤0.05)=0.7.

Answer 16

(b)

Expert Solution

Answer 17

(b)

Expert Solution

Answer 18

Expert Solution

Answer 19

To determine

To find: Theprobability thatsuch measurements will be in error bymore than -0.02 microseconds.

Answer 20

Answer to Problem 13P

Solution: Theprobability that such measurements will be in error by more than -0.02 microsecondsis0.7.

Answer 21

Explanation of Solution

Calculation:

Let x be the measurement errors in reading these times have a uniform distribution from α=−0.05 to β=0.05 microseconds.

Let x is chosen at random from [α, β], the area of the rectangle that lies above [a, b] is the probability that x lies in [a, b]. This area is P(a≤x≤b)=b−aβ−α

We have to find the probability of more than -0.02 microseconds,

P(−0.02≤x≤0.05)=b−aβ−αP(−0.02≤x≤0.05)=0.05−(−0.02)0.05−(−0.05)P(−0.02≤x≤0.05)=0.7

Hence, P(−0.02≤x≤0.05)=0.7.

Answer 22

(c)

Expert Solution

To determine

To find: Theprobability thatsuch measurements will be in error bybetween -0.04 and 0.01 microseconds.

Answer to Problem 13P

Solution: Theprobability that such measurements will be in error bybetween -0.04 and 0.01 microsecondsis0.5.

Explanation of Solution

Calculation:

Let x be the measurement errors in reading these times have a uniform distribution from α=−0.05 to β=0.05 microseconds.

Let x is chosen at random from [α, β], the area of the rectangle that lies above [a, b] is the probability that x lies in [a, b]. This area is P(a≤x≤b)=b−aβ−α

We have to find the probability between -0.04 and 0.01 microseconds,

P(−0.04≤x≤0.01)=b−aβ−αP(−0.04≤x≤0.01)=0.01−(−0.04)0.05−(−0.05)P(−0.04≤x≤0.01)=0.5

Hence, P(−0.04≤x≤0.01)=0.5.

Answer 23

(c)

Expert Solution

Answer 24

(c)

Expert Solution

Answer 25

Expert Solution

Answer 26

To determine

To find: Theprobability thatsuch measurements will be in error bybetween -0.04 and 0.01 microseconds.

Answer 27

Answer to Problem 13P

Solution: Theprobability that such measurements will be in error bybetween -0.04 and 0.01 microsecondsis0.5.

Answer 28

Explanation of Solution

Calculation:

Let x be the measurement errors in reading these times have a uniform distribution from α=−0.05 to β=0.05 microseconds.

Let x is chosen at random from [α, β], the area of the rectangle that lies above [a, b] is the probability that x lies in [a, b]. This area is P(a≤x≤b)=b−aβ−α

We have to find the probability between -0.04 and 0.01 microseconds,

P(−0.04≤x≤0.01)=b−aβ−αP(−0.04≤x≤0.01)=0.01−(−0.04)0.05−(−0.05)P(−0.04≤x≤0.01)=0.5

Hence, P(−0.04≤x≤0.01)=0.5.

Answer 29

(d)

Expert Solution

To determine

To find: Themean and standard deviation of measurement errors and whether the measurements for these acoustical sensors are unbiased.

Answer to Problem 13P

Solution:

The mean of measurement errors is 0. The standard deviation of measurement errors is 0.029. Yes, the measurements for these acoustical sensors are unbiased.

Explanation of Solution

Calculation:

Let x be the measurement errors in reading these times have a uniform distribution from α=−0.05 to β=0.05 microseconds.

We have to find the mean,

μ=α+β2μ=0.05+(−0.05)2μ=0

Therefore, the mean is 0 microsecond.

We have to find the standard deviation,

σ=β−α12σ=0.05−(−0.05)12σ≈0.029

Therefore, thestandard deviationis 0.029microsecond.

Yes, since the obtained mean of the measurement errors is zero so the measurements for these acoustical sensors are unbiased.

Answer 30

(d)

Expert Solution

Answer 31

(d)

Expert Solution

Answer 32

Expert Solution

Answer 33

To determine

To find: Themean and standard deviation of measurement errors and whether the measurements for these acoustical sensors are unbiased.

Answer 34

Answer to Problem 13P

Solution:

The mean of measurement errors is 0. The standard deviation of measurement errors is 0.029. Yes, the measurements for these acoustical sensors are unbiased.