Question 983115: The life of a smartphone is normally distributed with a mean 3 years and standard deviation 4 months. The manufacturer will replace it if it breaks during the guarantee period.
a. If the manufacturer guarantees the smartphone for 28 months what fraction of the phones will he probably have to replace?
b. How long should the guarantee period be if the manufacturer does not want to replace more than 5% of the smartphones? (Round to the nearest month).
I'm running into some issues with this one.. I believe the mean should be calculated as 36 months, and that z(28) = (28-32)/4 = -1?
Please help. Thanks so much!
Answer by jim_thompson5910(35256)   (Show Source):
You can put this solution on YOUR website! a)
X = length of lifespan of the phone (in months) from the time it's made to the time it breaks.
3 years = 3*12 = 36 months
mu = 36 months
sigma = 4 months
If the phone breaks before the guarantee period, then the phone is replaced for free.
Basically, the phone is replaced for free if X < 28 because the lifespan is less than the guaranteed amount. We need to find P(X < 28)
x = 28
z = (x-mu)/sigma
z = (28-36)/4
z = -8/4
z = -2
P(X < 28) is the same as P(Z < -2)
Use a table to find that P(Z < -2) = 0.02275
So, P(X < 28) = 0.02275
The proportion of phones that will be replaced for free is 0.02275, which is 2.275% of the entire population of phones sold.
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b)
Use invNorm on a TI calculator to find that invNorm(0.05) = -1.645 approximately
This means that P(Z < -1.645) = 0.05
Use this z-score to find the raw score x.
z = (x-mu)/sigma
-1.645 = (x-36)/4
-1.645*4 = x-36
-6.58 = x-36
-6.58+36 = x
29.42 = x
x = 29.42
If the cutoff period is 29.42 months, which rounds to 29 months, then 5% or less of the phones will be replaced for free.
Note: we round down to ensure that the percentage is less than 5%. If we rounded up, then the percentage would most likely be larger than 5%
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