Posted by Chrystal on Thursday, March 19, 2009 at 3:03pm.
The monthly income of residents of Daisy City is normally distributed with a mean of $3000 and a standard deviation of $500.
a. What is the random variable?
b. The mayor of Daisy City makes $2250 a month. What percentage of Daisy City's residents has incomes that are more than the mayor?
c. Individuals with incomes of less than $1985 per monthe are exempt from city taxes. What percentage of residents is exempt from city taxes.
d. What are the minimum and the maximum incomes of the middle 95% of the residents
e. Two hundred residents have incomes of at least $4440 per month. What is the population of Daisy City.
Quantitative Methods II - Damon, Thursday, March 19, 2009 at 4:41pm
You have a normal distribution table handy. I do not so all I can do is outline the methods.
mean = 3000
sigma = 500
a. the monthly income
b. how far below mean is the mayor's income?
3000 - 2250 = 750
then how many sigmas below mean is this?
750/500 = 1.5 sigma (called "z") below mean
so now the question becomes how what percent of a normal distribution is between 1.5 sigma (called "z") below mean and + infinity
this is of course 100% - the percent between -oo and 1.5 sigma below mean
My guess not having the table handy is about 7% for between -oo and mean -1.5 sigma
so about 93 % have more income
c. 3000 - 1985 = 1015 below mean
1015/500 = 2.03 sigma below mean
look for between -oo and -2.03
This is tiny, around 2 1/2 percent if I had a table
d. middle 95% is from about F(z) = 2.5% to F(z)=97.5% or F(z) = .025 to .975
That would be from about mean -2 sigma to mean + 2 sigma (use table of z versus F(z) of course. Do not trust my guess)
so from mean -2*500 to mean + 2*500
or from $2000 to $4000
4440 - 3000 = 1440
1440/500 = 2.88 sigma above mean
so what percent below mean + 2.88 sigma?
F(z)about .995 (remember guessing
.005 or about .5% have incomes above 4440
.005 n = 200
n = 200/.005 = 40,000
remember I am beign very approximate guessing normal distribution table.
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