- #1

- 199

- 0

P(|X-80| <= 10)

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

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- Thread starter needhelp83
- Start date

- #1

- 199

- 0

P(|X-80| <= 10)

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

- #2

- 954

- 117

We also know that P(Z<-1) = P(Z>1) due to symmetry. Hope that helps.

- #3

- 199

- 0

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?

- #4

- 954

- 117

Yup.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?

- #5

- 199

- 0

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.

- #6

- 954

- 117

[tex]Z = \frac{X-\mu}{\sigma}[/tex]

Check your standardisation.

Check your standardisation.

- #7

- 199

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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!

- #8

statdad

Homework Helper

- 1,495

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Check your final calculation - you have a glaring error.

- #9

- 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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