The replacement times for an iPod are normally distributed with a mean of 9.1 years and a standard deviation of 1.4 years.
(a) Find the probability that a randomly selected player will have a replacement time less than 10.0 years.
(b) If you want to provide a warranty so that only 2% of the iPod will be replaced before the warranty expires, what is he time length of the warranty?
(a) Find the probability that a randomly selected player will have a replacement time less than 10.0 years.
(b) If you want to provide a warranty so that only 2% of the iPod will be replaced before the warranty expires, what is he time length of the warranty?
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z = (X-Mean)/S.D
a) z value corresponding to X=10.0 is
z = (10.0-9.1)/1.4 = + 0.64
The area under the standard normal curve left to z = + 0.64 indicates the required probability.
Therefore required probability = 0.5000+0.2389 = 0.7389 or 73.89%
b) 2% = 0.0200 is the area under the standard normal curve in the extreme left tail.
z value corresponding to 0.5000-0.0200 = 0.4800 area is to be located from the table.
The corresponding z value is - 2.05
z value is -ve because it lies on the left side of the normal curve.
- 2.05 = (X-9.1)/1.4
- 2.87 = X-9.1
X = 9.1-2.87 = 6.23
Therefore the length of the warrant = 6.23 years.
a) z value corresponding to X=10.0 is
z = (10.0-9.1)/1.4 = + 0.64
The area under the standard normal curve left to z = + 0.64 indicates the required probability.
Therefore required probability = 0.5000+0.2389 = 0.7389 or 73.89%
b) 2% = 0.0200 is the area under the standard normal curve in the extreme left tail.
z value corresponding to 0.5000-0.0200 = 0.4800 area is to be located from the table.
The corresponding z value is - 2.05
z value is -ve because it lies on the left side of the normal curve.
- 2.05 = (X-9.1)/1.4
- 2.87 = X-9.1
X = 9.1-2.87 = 6.23
Therefore the length of the warrant = 6.23 years.