   Chapter 6, Problem 19P Essentials of Statistics for the B...

8th Edition
Frederick J Gravetter + 1 other
ISBN: 9781133956570

Solutions

Chapter
Section Essentials of Statistics for the B...

8th Edition
Frederick J Gravetter + 1 other
ISBN: 9781133956570
Textbook Problem

A consumer survey indicates that the average household spends μ = $185 on groceries each week. The distribution of spending amounts is approximately normal with a standard deviation of σ =$25. Based on this distribution. a. What proportion of the population spends more than $200 per week on groceries? b. What is the probability of randomly selecting a family that spends less than$150 per week on groceries? c. How much money do you need to spend on groceries each week to be in the top 20% of the distribution

(a)

To determine

To find: The proportion of the population spends more than $200 per week on groceries. Explanation Given info: The random variable X denote the average household spends and it follows the normal distribution with mean of μ=$185 on groceries each week and standard deviation of σ=$25. Calculation: The formula for z score is, z=Xμσ Substitute the value$200 for X, $185 for μ and$25 for σ.

z=20018525=0

(b)

To determine

To find: The probability of randomly selecting family that spends less than \$150 per week on groceries.

(c)

To determine

To find: The money need to spend on groceries each week to be in the top 20% of the distribution.

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