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#1
04-08-2011, 12:42 PM
 Xer0 Join Date: Mar 2011 Posts: 9
easy normal approx question

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!
#2
04-08-2011, 01:06 PM
 carrytheCrøss Member CAS Join Date: May 2005 Posts: 2,761

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).
__________________
a wonderful post
#3
04-08-2011, 01:45 PM
 Xer0 Join Date: Mar 2011 Posts: 9

Thanks!

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