# Thread: Standard deviation word problem

1. ## 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?

2. (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 $\frac{1}{\sqrt{2\pi}}*\int_{low \ z \ score}^{high \ z \ score}e^{\frac{-x^2}{2}}dx$

3. I dont understand? My book is not showing any way to work this out

4. 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.

5. $\bar x=45,000 \ \mbox{given}$

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

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

You can find $z_2$ now.

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

6. I am trying to figure out the percentage?

7. The percentage = the area under the curve

8. When you eventually solve this integral, you would receive a decimal.

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

$=.999936657516*100%$ $=99.9936657516%$

9. Thanks I came up with 99% for my answer.