# Normal Random Variable Probability

1. Aug 25, 2009

### needhelp83

If X is a normal rv with mean 80 and standard deviation 10, compute the following probabilities by standardizing:

P(|X-80| <= 10)

I know how to determine the probability without absolute value, but this confuses me. Any help?

2. Aug 25, 2009

### Fightfish

From |X-80| <= 10, you can get -10 <= (X-80) <= 10, -1 <= Z <= 1
We also know that P(Z<-1) = P(Z>1) due to symmetry. Hope that helps.

3. Aug 26, 2009

### needhelp83

I do understand partially. I remember this from just basic algebra. Afterwards, would I solve for P(70 <= X <= 90) where I added 80 to both sides?

4. Aug 26, 2009

### Fightfish

Yup.

5. Aug 27, 2009

### needhelp83

Ok I am not getting this to work out correctly:

P(|X-80|) $$\leq$$ 10)

P(-10 $$\leq$$ X-80 $$\leq$$ 10)

P( $$\frac{-10-80}{10}$$ $$\leq$$ Z $$\leq$$ $$\frac{10-80}{10}$$)

-9 $$\leq$$ Z $$\leq$$ -7

Ok, so these aren't on the Z table, so I am definitely not doing something right. What am I now missing? The 80 is my mean and 10 is my s.d.

6. Aug 27, 2009

### Fightfish

$$Z = \frac{X-\mu}{\sigma}$$

7. Aug 27, 2009

### needhelp83

P(-10 $$\leq$$ X-80 $$\leq$$ 10)

P (70 $$\leq$$ X $$\leq$$ 90)

P ($$\frac{70-80}{10}$$ $$\leq$$ Z $$\leq$$ $$\frac{90-80}{10}$$)

.1587-.8413 = -0.6826

I am funny! What a silly mistake. Should be right now!

8. Aug 27, 2009

Check your final calculation - you have a glaring error.

9. Aug 27, 2009

### needhelp83

Wow, this problem kicked my butt!