Normal distribution with random sampling problem. Screenshot attached.

Question

anonymous · Answer

Since the problem states that this scenario follows a normal distribution, then the first I thing I did was standardize $\bar{X}$

Standardization:
$$\huge Z=\frac{ \bar{X}-\mu }{ \frac{ \sigma }{ \sqrt{n}} }$$
After standardizing, I got:$$P(-0.08 \le Z \le 0.08)$$However when I plugged in $\Phi (0.08)- \Phi (-0.08)$ from the standard normal distribution tables, I got the wrong answer.

ganeshie8 · Answer

could you try these once

a) 0.5763

b) 0.2119

anonymous · Answer

The first one was correct! But b). was not

ganeshie8 · Answer

How about these

a) 0.5763

b) 0.1587

anonymous · Answer

Yup! That one's good now

ganeshie8 · Answer

there is an arithmetic error in your work : 

$P(-0.08 \le Z \le 0.08)$

should be changed to 

$P(-0.8 \le Z \le 0.8)$

anonymous · Answer

Interesting. Oops :) 
At least I had the right process -- I thought I was going crazy for getting this wrong! X) Thanks!

ganeshie8 · Answer

Happens :)
btw, you may use wolfram for these calculations

ganeshie8 · Answer

http://www.wolframalpha.com/input/?i=P%28-0.01%2F%280.05%2F4%29%3CX%3C0.01%2F%280.05%2F4%29%29

anonymous · Answer

Ah, I'll do that! Thanks!