# Normal distribution

• Sep 18th 2013, 10:52 PM
downthesun01
Normal distribution
In an analysis of healthcare data, ages have been rounded to the nearest multiple of 5 years. The difference between the true age and the rounded age is assumed to be uniformly distributed on the interval from −2.5 years to 2.5 years. The healthcare data are based on a random sample of 48 people.

What is the approximate probability that the mean of the rounded ages is within

0.25 years of the mean of the true ages?

Here's my solution:

$\displaystyle \text{Looking for: }Pr[-2.5 \leq\bar{x}\leq 2.5]$

$\displaystyle E[X] = \frac{2.5+(-2.5)}{2}=0$

$\displaystyle var[X]=\frac{(2.5-(-2.5))^2}{12}=\frac{25}{12}$

$\displaystyle E[\bar{x}]=0*48$

$\displaystyle var[\bar{x}]=\frac{48*25}{2}=600$

$\displaystyle \sigma_{\bar{x}}=\frac{600}{\sqrt{48}}$

The solution in the book as $\displaystyle \sigma_{\bar{x}}$ as $\displaystyle \frac{\frac{25}{12}}{\sqrt{48}}$

I don't understand why
• Sep 19th 2013, 05:54 AM
HallsofIvy
Re: Normal distribution
Quote:

Originally Posted by downthesun01
In an analysis of healthcare data, ages have been rounded to the nearest multiple of 5 years. The difference between the true age and the rounded age is assumed to be uniformly distributed on the interval from −2.5 years to 2.5 years. The healthcare data are based on a random sample of 48 people.

What is the approximate probability that the mean of the rounded ages is within

0.25 years of the mean of the true ages?

Here's my solution:

$\displaystyle \text{Looking for: }Pr[-2.5 \leq\bar{x}\leq 2.5]$

$\displaystyle E[X] = \frac{2.5+(-2.5)}{2}=0$

$\displaystyle var[X]=\frac{(2.5-(-2.5))^2}{12}=\frac{25}{12}$

$\displaystyle E[\bar{x}]=0*48$

$\displaystyle var[\bar{x}]=\frac{48*25}{2}=600$

How did you get this? What happened to the "12" in the denominator before?

Quote:

$\displaystyle \sigma_{\bar{x}}=\frac{600}{\sqrt{48}}$

Quote:

The solution in the book as $\displaystyle \sigma_{\bar{x}}$ as $\displaystyle \frac{\frac{25}{12}}{\sqrt{48}}$

I don't understand why
• Sep 19th 2013, 04:22 PM
downthesun01
Re: Normal distribution
Sorry, it should be

$\displaystyle \frac{48*25}{12}=100$

To find the the standard deviation of $\displaystyle \bar{x}$ don't I have to multiply Var[X] by n? Then, take the square root of that, and divide it by the square root of n?
• Sep 20th 2013, 02:42 AM
downthesun01
Re: Normal distribution
I figured out my mistake. Thanks