easy normal approx question

Xer0 · #1 04-08-2011, 12:42 PM

I feel like this is an easy problem and I'm just overlooking something. The answer provided is 0.58 and I can't seem to get it.

$X \sim N(5, 49)$ find $P(|X-7|)\ge4$

Adding 7 to 4 and subtracting 7 to 4 then standardizing gives me

$P(-1.14 \le Z \le .86)$ but I'm thinking this is wrong because it doesn't give me 0.58. Thanks for the help!

carrytheCrøss · #2 04-08-2011, 01:06 PM

Quote:

Originally Posted by Xer0

I feel like this is an easy problem and I'm just overlooking something. The answer provided is 0.58 and I can't seem to get it.

$X \sim N(5, 49)$ find $P(|X-7|)\ge4$

Adding 7 to 4 and subtracting 7 to 4 then standardizing gives me

$P(-1.14 \le Z \le .86)$ but I'm thinking this is wrong because it doesn't give me 0.58. Thanks for the help!

You should re-write the initial inequality as P(X-7>4) and P(X-7<-4), or, P(X>11) and P(X<3). From there you should get the answer of .58 (I did).

Xer0 · #3 04-08-2011, 01:45 PM

Thanks!

Thread Tools
Show Printable Version Email this Page
Display Modes
Switch to Linear Mode Hybrid Mode Switch to Threaded Mode