The percentage of the U.S. households that had an income of $ 50 , 000 or more when the incomes of each country households in 1995 are normally distributed and the mean household income and standard deviation for each country is as follows, Country United States Canada Switzerland Germany Sweden Mean Household Income ( $ ) 38 , 000 35 , 000 39 , 000 34 , 000 32 , 000 Standard deviation 21 , 000 17 , 000 16 , 000 14 , 000 11 , 000

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Answer 1

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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 9.5, Problem 41E

To determine

To calculate: The percentage of the U.S. households that had an income of $$50,000$ or more when the incomes of each country households in $1995$ are normally distributed and the mean household income and standard deviation for each country is as follows,

Country United States Canada Switzerland Germany Sweden

Mean Household Income $($)$ $38,000$ $35,000$ $39,000$ $34,000$ $32,000$

Standard deviation $21,000$ $17,000$ $16,000$ $14,000$ $11,000$

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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 9.5, Problem 41E

To determine

To calculate: The percentage of the U.S. households that had an income of $$50,000$ or more when the incomes of each country households in $1995$ are normally distributed and the mean household income and standard deviation for each country is as follows,

Country United States Canada Switzerland Germany Sweden

Mean Household Income $($)$ $38,000$ $35,000$ $39,000$ $34,000$ $32,000$

Standard deviation $21,000$ $17,000$ $16,000$ $14,000$ $11,000$

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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 9.5, Problem 41E

To determine

To calculate: The percentage of the U.S. households that had an income of $$50,000$ or more when the incomes of each country households in $1995$ are normally distributed and the mean household income and standard deviation for each country is as follows,

Country United States Canada Switzerland Germany Sweden

Mean Household Income $($)$ $38,000$ $35,000$ $39,000$ $34,000$ $32,000$

Standard deviation $21,000$ $17,000$ $16,000$ $14,000$ $11,000$

Expert Solution & Answer

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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 9.5, Problem 41E

To determine

To calculate: The percentage of the U.S. households that had an income of $$50,000$ or more when the incomes of each country households in $1995$ are normally distributed and the mean household income and standard deviation for each country is as follows,

Country United States Canada Switzerland Germany Sweden

Mean Household Income $($)$ $38,000$ $35,000$ $39,000$ $34,000$ $32,000$

Standard deviation $21,000$ $17,000$ $16,000$ $14,000$ $11,000$

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Answer 5

Chapter 9.5, Problem 41E

To determine

To calculate: The percentage of the U.S. households that had an income of $$50,000$ or more when the incomes of each country households in $1995$ are normally distributed and the mean household income and standard deviation for each country is as follows,

Country United States Canada Switzerland Germany Sweden

Mean Household Income $($)$ $38,000$ $35,000$ $39,000$ $34,000$ $32,000$

Standard deviation $21,000$ $17,000$ $16,000$ $14,000$ $11,000$

Answer 6

Chapter 9.5, Problem 41E

To determine

To calculate: The percentage of the U.S. households that had an income of $$50,000$ or more when the incomes of each country households in $1995$ are normally distributed and the mean household income and standard deviation for each country is as follows,

Country United States Canada Switzerland Germany Sweden

Mean Household Income $($)$ $38,000$ $35,000$ $39,000$ $34,000$ $32,000$

Standard deviation $21,000$ $17,000$ $16,000$ $14,000$ $11,000$

Answer 7

Chapter 9.5, Problem 41E

To determine

To calculate: The percentage of the U.S. households that had an income of $$50,000$ or more when the incomes of each country households in $1995$ are normally distributed and the mean household income and standard deviation for each country is as follows,

Country United States Canada Switzerland Germany Sweden

Mean Household Income $($)$ $38,000$ $35,000$ $39,000$ $34,000$ $32,000$

Standard deviation $21,000$ $17,000$ $16,000$ $14,000$ $11,000$

Answer 8

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Answer 9

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Answer 10

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Answer 11

Ch. 9.1 - In Exercises 110, classify the random variable X...Ch. 9.1 - Prob. 2E Ch. 9.1 - In Exercises 110, classify the random variable X...Ch. 9.1 - In Exercises 110, classify the random variable X...Ch. 9.1 - Prob. 5E Ch. 9.1 - In Exercises 110, classify the random variable X...Ch. 9.1 - Prob. 7E Ch. 9.1 - In Exercises 110, classify the random variable X...Ch. 9.1 - In Exercises 110, classify the random variable X...Ch. 9.1 - In Exercises 110, classify the random variable X...

Solution Summary: The author calculates the percentage of the U.S. households that had an income of 50,000 or more when the incomes of each country households in 1995 are normally distributed.

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Country	United States	Canada	Switzerland	Germany	Sweden
Mean Household Income $($)$	$38,000$	$35,000$	$39,000$	$34,000$	$32,000$
Standard deviation	$21,000$	$17,000$	$16,000$	$14,000$	$11,000$