The probability that a piece of tubing selected at random measures greater than 40.05 in. using standard normal distribution if the lengths of a machine precision- cuts tubing are distributed normally with the mean μ = 40.00 in . and standard deviation σ = 0.03 in .

Question

Show that the Laplace equation in Cartesian coordinates: J²u J²u + = 0 მx2 Jy2 can be reduced to the following form in cylindrical polar coordinates: 湯( ди 1 8²u + Or 7,2 მ)2 = 0.

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 8.4, Problem 38E

(a)

To determine

To calculate: The probability that a piece of tubing selected at random measures greater than 40.05 in. using standard normal distribution if the lengths of a machine precision- cuts tubing are distributed normally with the mean μ=40.00 in. and standard deviation σ=0.03 in.

(b)

To determine

To calculate: The probability that a piece of tubing selected at random measures less than 40.045 in. using standard normal distribution if the lengths of a machine precision- cuts tubing are distributed normally with the mean μ=40.00 in. and standard deviation σ=0.03 in.

(c)

To determine

To calculate: The probability that a piece of tubing selected at random measures lies between 39.95 in. and 40.05 in. using standard normal distribution if the lengths of a machine precision- cuts tubing are distributed normally with the mean μ=40.00 in. and standard deviation σ=0.03 in.

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Answer 2

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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 8.4, Problem 38E

(a)

To determine

To calculate: The probability that a piece of tubing selected at random measures greater than 40.05 in. using standard normal distribution if the lengths of a machine precision- cuts tubing are distributed normally with the mean μ=40.00 in. and standard deviation σ=0.03 in.

(b)

To determine

To calculate: The probability that a piece of tubing selected at random measures less than 40.045 in. using standard normal distribution if the lengths of a machine precision- cuts tubing are distributed normally with the mean μ=40.00 in. and standard deviation σ=0.03 in.

(c)

To determine

To calculate: The probability that a piece of tubing selected at random measures lies between 39.95 in. and 40.05 in. using standard normal distribution if the lengths of a machine precision- cuts tubing are distributed normally with the mean μ=40.00 in. and standard deviation σ=0.03 in.

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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 8.4, Problem 38E

(a)

To determine

To calculate: The probability that a piece of tubing selected at random measures greater than 40.05 in. using standard normal distribution if the lengths of a machine precision- cuts tubing are distributed normally with the mean μ=40.00 in. and standard deviation σ=0.03 in.

(b)

To determine

To calculate: The probability that a piece of tubing selected at random measures less than 40.045 in. using standard normal distribution if the lengths of a machine precision- cuts tubing are distributed normally with the mean μ=40.00 in. and standard deviation σ=0.03 in.

(c)

To determine

To calculate: The probability that a piece of tubing selected at random measures lies between 39.95 in. and 40.05 in. using standard normal distribution if the lengths of a machine precision- cuts tubing are distributed normally with the mean μ=40.00 in. and standard deviation σ=0.03 in.

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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 8.4, Problem 38E

(a)

To determine

To calculate: The probability that a piece of tubing selected at random measures greater than 40.05 in. using standard normal distribution if the lengths of a machine precision- cuts tubing are distributed normally with the mean μ=40.00 in. and standard deviation σ=0.03 in.

(b)

To determine

To calculate: The probability that a piece of tubing selected at random measures less than 40.045 in. using standard normal distribution if the lengths of a machine precision- cuts tubing are distributed normally with the mean μ=40.00 in. and standard deviation σ=0.03 in.

(c)

To determine

To calculate: The probability that a piece of tubing selected at random measures lies between 39.95 in. and 40.05 in. using standard normal distribution if the lengths of a machine precision- cuts tubing are distributed normally with the mean μ=40.00 in. and standard deviation σ=0.03 in.

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Answer 5

Chapter 8.4, Problem 38E

(a)

To determine

To calculate: The probability that a piece of tubing selected at random measures greater than 40.05 in. using standard normal distribution if the lengths of a machine precision- cuts tubing are distributed normally with the mean μ=40.00 in. and standard deviation σ=0.03 in.

(b)

To determine

To calculate: The probability that a piece of tubing selected at random measures less than 40.045 in. using standard normal distribution if the lengths of a machine precision- cuts tubing are distributed normally with the mean μ=40.00 in. and standard deviation σ=0.03 in.

(c)

To determine

To calculate: The probability that a piece of tubing selected at random measures lies between 39.95 in. and 40.05 in. using standard normal distribution if the lengths of a machine precision- cuts tubing are distributed normally with the mean μ=40.00 in. and standard deviation σ=0.03 in.

Answer 6

Chapter 8.4, Problem 38E

(a)

To determine

To calculate: The probability that a piece of tubing selected at random measures greater than 40.05 in. using standard normal distribution if the lengths of a machine precision- cuts tubing are distributed normally with the mean μ=40.00 in. and standard deviation σ=0.03 in.

Answer 7

Chapter 8.4, Problem 38E

(a)

To determine

To calculate: The probability that a piece of tubing selected at random measures greater than 40.05 in. using standard normal distribution if the lengths of a machine precision- cuts tubing are distributed normally with the mean μ=40.00 in. and standard deviation σ=0.03 in.

Answer 8

(b)

To determine

To calculate: The probability that a piece of tubing selected at random measures less than 40.045 in. using standard normal distribution if the lengths of a machine precision- cuts tubing are distributed normally with the mean μ=40.00 in. and standard deviation σ=0.03 in.

Answer 9

(b)

To determine

To calculate: The probability that a piece of tubing selected at random measures less than 40.045 in. using standard normal distribution if the lengths of a machine precision- cuts tubing are distributed normally with the mean μ=40.00 in. and standard deviation σ=0.03 in.

Answer 10

(c)

To determine

To calculate: The probability that a piece of tubing selected at random measures lies between 39.95 in. and 40.05 in. using standard normal distribution if the lengths of a machine precision- cuts tubing are distributed normally with the mean μ=40.00 in. and standard deviation σ=0.03 in.

Answer 11

(c)

To determine

To calculate: The probability that a piece of tubing selected at random measures lies between 39.95 in. and 40.05 in. using standard normal distribution if the lengths of a machine precision- cuts tubing are distributed normally with the mean μ=40.00 in. and standard deviation σ=0.03 in.

Answer 12

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Ch. 8.1 - Determine whether the following experiments...Ch. 8.1 - Prob. 2CP Ch. 8.1 - Determine whether the following experiments...Ch. 8.1 - Prob. 4CP Ch. 8.1 - 1. Suppose a fair die is rolled 18 times. (a) What...Ch. 8.1 - 2. Suppose a fair coin is tossed 12 times. (a)...Ch. 8.1 - Prob. 3E Ch. 8.1 - Prob. 4E Ch. 8.1 - Prob. 5E Ch. 8.1 - Prob. 6E

The probability that a piece of tubing selected at random measures greater than 40.05 in. using standard normal distribution if the lengths of a machine precision- cuts tubing are distributed normally with the mean μ = 40.00 in . and standard deviation σ = 0.03 in .

Solution Summary: The author calculates the probability that a piece of tubing selected at random measures greater than 40.05 in. using standard normal distribution.

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To calculate: The probability that a piece of tubing selected at random measures greater than 40.05 in. using standard normal distribution if the lengths of a machine precision- cuts tubing are distributed normally with the mean μ=40.00 in. and standard deviation σ=0.03 in.

To calculate: The probability that a piece of tubing selected at random measures less than 40.045 in. using standard normal distribution if the lengths of a machine precision- cuts tubing are distributed normally with the mean μ=40.00 in. and standard deviation σ=0.03 in.

To calculate: The probability that a piece of tubing selected at random measures lies between 39.95 in. and 40.05 in. using standard normal distribution if the lengths of a machine precision- cuts tubing are distributed normally with the mean μ=40.00 in. and standard deviation σ=0.03 in.

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