Company XYZ know that replacement times for the portable MP3 players it produces

Question

Company XYZ know that replacement times for the portable MP3 players it produces are normally distributed with a mean of 4 years and a standard deviation of 0.6 years.

Find the probability that a randomly selected portable MP3 player will have a replacement time less than 2.6 years?
P(X < 2.6 years) =

Enter your answer accurate to 4 decimal places. Answers obtained using exact z-scores or z-scores rounded to 3 decimal places are accepted.

If the company wants to provide a warranty so that only 2.2% of the portable MP3 players will be replaced before the warranty expires, what is the time length of the warranty?
warranty = years

Enter your answer as a number accurate to 1 decimal place. Answers obtained using exact z-scores or z-scores rounded to 3 decimal places are accepted.

Explanation / Answer

a)

We first get the z score for the critical value. As z = (x - u) / s, then as          
          
x = critical value =    2.6      
u = mean =    4      
          
s = standard deviation =    0.6      
          
Thus,          
          
z = (x - u) / s =    -2.333333333      
          
Thus, using a table/technology, the left tailed area of this is          
          
P(z <   -2.333333333   ) =    0.009815329 = 0.0098 [ANSWER]

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b)

First, we get the z score from the given left tailed area. As          
          
Left tailed area =    0.022      
          
Then, using table or technology,          
          
z =    -2.014090812      
          
As x = u + z * s,          
          
where          
          
u = mean =    4      
z = the critical z score =    -2.014090812      
s = standard deviation =    0.6      
          
Then          
          
x = critical value =    2.791545513 YEARS = 2.8 years [ANSWER, WARRANTY]  

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