# Thread: Standard Normal Distribution Problem...

1. ## Standard Normal Distribution Problem...

Hi all,

I have a problem that I can't get my head around - maybe one of you bright sparks can help me out!? The question goes:
"A hard drive has a lifetime which is Normally distributed with a mean of 30,000 hours and a standard deviation of 8,000 hours. The warranty is for 2 years and is included in the sale at $100. An extended warranty to cover an extra year is available at a cost of$39.

Q1: Using variable 't' (in hours) to represent the lifetime of the HD, sketch a graph of the Normal distribution and then calculate to probability that any HD dies during its 3rd year of operation.

Q2: Using the answer to Q1 calculate the expected (or average) cost of replacing a HD when it dies during its 3rd year of operation."
I have attempted it and calculated an answer to Q1 (have yet to draw the bell graph), however I am not sure whether this is correct or not :S ...

P(t<=17,520) = P[(t-30,000)/8,000 <= (17520-30,000)/8,000]

= P(Z < -2.60)
= P(Z > 2.60)

= 0.5 - P(0 <= Z <= 2.60)
= 0.5 - 0.495
= 0.005

2. Originally Posted by MrRedNapoleon
Hi all,

I have a problem that I can't get my head around - maybe one of you bright sparks can help me out!? The question goes:
I have attempted it and calculated an answer to Q1 (have yet to draw the bell graph), however I am not sure whether this is correct or not :S ...

P(t<=17,520) = P[(t-30,000)/8,000 <= (17520-30,000)/8,000]

= P(Z < -2.60)
= P(Z > 2.60)

= 0.5 - P(0 <= Z <= 2.60)
= 0.5 - 0.495
= 0.005
What you have calculated is failure within 3 years rather than failure during the third year ....