Question
Asked Oct 2, 2019

An expensive watch is powered by a 3-volt lithium battery expected to last 5 years. Suppose the life of the battery has a standard deviation of .6 year and is normally distributed

Compute the length-of-life value for which 10% of the watch's batteries last longer .

check_circleExpert Solution
Step 1

Given Data
Mean = 5
Standard devi...

P
0.10
=
X - 5
NORM.S.INV (0.1))
Ø1 (0.10) 1.28(From Excel
=
0.6
1.28 x 0.6 + 5 = 4.232
X
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P 0.10 = X - 5 NORM.S.INV (0.1)) Ø1 (0.10) 1.28(From Excel = 0.6 1.28 x 0.6 + 5 = 4.232 X

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