1. ## 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:

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

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

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

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

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

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

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

I don't understand why

2. ## Re: Normal distribution

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:

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

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

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

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

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

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

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

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

I don't understand why

3. ## Re: Normal distribution

Sorry, it should be

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

To find the the standard deviation of $\bar{x}$ 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?

4. ## Re: Normal distribution

I figured out my mistake. Thanks