Hi, 

*Note: 

 and 

z =( 3.3 - 3.00)/.16

z = 1.875

P(z = 1.875) = .9697  (use table or Exel function =NORMSDIST(1.875))

P(x <3.3) = .9697

P(x >3.3) = 1 - .9697 = .0303



 

Answer by stanbon(75887)   (Show Source): You can put this solution on YOUR website!
 I have the answer to this problem and I am sure I am using the correct formula, but I am not getting the correct answer of .227 

The lifetime of a machine component is normally distributed with a mean of 3 and a standard deviation of .16. What is the probability that X is greater than 3.3

X ~ N(3, .16)

-------------

z(3.3) = (3.3-3)/0.16 = 1.875

---

P(x > 3.3) = P(z > 1.875) = 0.0304

------

You would not be the first person to 

find that the "book" answer was wrong.

Cheers,

Stan H. 

 

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