Question

For the year 2010, 33% of taxpayers with adjusted gross incomes between $30,000 and $60,000 itemized deductions on their federal income tax return (The Wall Street Journal, October 25, 2012). The mean amount of deductions for this population of taxpayers was $16,642. Assume the standard deviation is $2400. a.) What is the probability that a sample of taxpayers from this income group who have itemized deductions will show a sample mean within $200 of the population mean for each of the following sample sizes: 30, 50, 100, and 400? b.) What is the advantage of a larger sample size when attempting to estimate the population mean? When I work out n=30 for a.) I get .46, but according to the book the answer is P(-.46<= z <= .46) = .6772 - .3228 = .3544 How are they getting the numbers .6772 and .3228??

          For the year 2010, 33% of taxpayers with adjusted gross incomes
between $30,000 and $60,000 itemized deductions on their federal
income tax return (The Wall Street Journal, October 25,
2012). The mean amount of deductions for this population of
taxpayers was $16,642. Assume the standard deviation is $2400.
a.) What is the probability that a sample of taxpayers from this
income group who have itemized deductions will show a sample mean
within $200 of the population mean for each of the following sample
sizes: 30, 50, 100, and 400?
b.) What is the advantage of a larger sample size when
attempting to estimate the population mean?
When I work out n=30 for a.) I get .46, but according to the
book the answer is P(-.46<= z <= .46) = .6772 - .3228 =
.3544
How are they getting the numbers .6772 and .3228??

Added by Gregory W.

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