SOLUTION: 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 d

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Question 363541: 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)
I would like to see the process.
Found 2 solutions by ewatrrr, stanbon:
Answer by ewatrrr(24785) About Me  (Show Source):
You can put this solution on YOUR website!


Hi, 

*Note: z+=+%28x+-+mu%29%2Fsigma

mu=3 and sigma+=+.16

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) About Me  (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)
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z(3.3) = (3.3-3)/0.16 = 1.875
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P(x > 3.3) = P(z > 1.875) = 0.0304
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You would not be the first person to
find that the "book" answer was wrong.
Cheers,
Stan H.