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Normal Random Variable Probability

  • Thread starter needhelp83
  • Start date
199
0
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?
 
954
117
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.
 
199
0
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.
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?
 
954
117
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?
Yup.
 
199
0
Ok I am not getting this to work out correctly:

P(|X-80|) [tex]\leq[/tex] 10)

P(-10 [tex]\leq[/tex] X-80 [tex]\leq[/tex] 10)

P( [tex]\frac{-10-80}{10}[/tex] [tex]\leq[/tex] Z [tex]\leq[/tex] [tex]\frac{10-80}{10}[/tex])

-9 [tex]\leq[/tex] Z [tex]\leq[/tex] -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.
 
954
117
[tex]Z = \frac{X-\mu}{\sigma}[/tex]
Check your standardisation.
 
199
0
P(-10 [tex]\leq[/tex] X-80 [tex]\leq[/tex] 10)

P (70 [tex]\leq[/tex] X [tex]\leq[/tex] 90)

P ([tex]\frac{70-80}{10}[/tex] [tex]\leq[/tex] Z [tex]\leq[/tex] [tex]\frac{90-80}{10}[/tex])

.1587-.8413 = -0.6826

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

statdad

Homework Helper
1,492
32
Check your final calculation - you have a glaring error.
 
199
0
Check your final calculation - you have a glaring error.
Wow, this problem kicked my butt!

ANSWER IS:
0.8413 - 0.1587 =0.6826

Thanks for the help guys!
 

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