# Standard deviation word problem

• Oct 17th 2010, 12:12 PM
caligirl350
Standard deviation word problem
Household income does not tend to follow a normal distribution in a particular state, yet average income is approximately $45,000/year in this state, with a standard deviation of about$9000. At least what percentage of household incomes in this state is likely to be between $9,000 to$81,000/year

Not sure what formula to use?

My husband tried to help me and he came up with 99%. Anyone?
• Oct 17th 2010, 12:28 PM
dwsmith
(1) draw a graph with 45,000 at the 0

(2) put 81,000 and 9,000 on the graph with corresponding z scores
$z=\frac{x-\bar x}{\sigma}$

(3) integrate $\frac{1}{\sqrt{2\pi}}*\int_{low \ z \ score}^{high \ z \ score}e^{\frac{-x^2}{2}}dx$
• Oct 17th 2010, 01:03 PM
caligirl350
I dont understand? My book is not showing any way to work this out
• Oct 17th 2010, 01:24 PM
dwsmith
In order to find the area under the normal probability density curve, you need the lower and upper bounds traditionally called z-scores.
A z-score is obtained by subtracting the given x value minus the mean and then dividing by the standard deviation.
Once you have obtained you lower and upper bounds, you can consult the the t-distribution for samples over 120 but your z-scores are so far out they won't be in the table; therefore, you need to integrate.
• Oct 17th 2010, 01:56 PM
dwsmith
$\bar x=45,000 \ \mbox{given}$

$\sigma=9,000 \ \mbox{given}$

$z_1=\frac{x-\bar x}{\sigma}=\frac{9,000-45,000}{9,000}=-4$

You can find $z_2$ now.

$\frac{1}{\sqrt{2\pi}}*\int_{-4}^{z_2}e^{\frac{-x^2}{2}} \mbox{dx}$
• Oct 17th 2010, 02:00 PM
caligirl350
I am trying to figure out the percentage?
• Oct 17th 2010, 02:27 PM
dwsmith
The percentage = the area under the curve
• Oct 17th 2010, 02:28 PM
dwsmith
When you eventually solve this integral, you would receive a decimal.

$\frac{1}{\sqrt{2\pi}}*\int_{-4}^{z_2}e^{\frac{-x^2}{2}} \mbox{dx}$

$=.999936657516*100%$ $=99.9936657516%$
• Oct 17th 2010, 02:36 PM
caligirl350
Thanks I came up with 99% for my answer.