Question

Triple A Battery Company is a major battery manufacturer of batteries and it produces three types of batteries (Type A, B, and C). The batteries are similar in construction but carry a different warranty period. Type A has a 36 month warranty, Type B has a 48 month warranty, and Type C has a 60 month warranty. Regardless of the warranty period, the standard deviation of a battery’s life is 2.5 months. Let’s consider the 36 month battery (Type A) for the following questions.

1. If the average life of a Type A battery is 38 months, then what proportion of the batteries will fail before the warranty period?
2. If a Type A battery is chosen at random, what is the probability that battery will last at least 42 months? Assume the mean life time of the battery is 38 months.
3. Each year, the company establishes a budget for warranty costs. If a battery does not last until the warranty period expires, then they have to replace the battery at no charge to the customer. This year they expect to replace 2% of the Type A batteries. What is the cutoff point, in months, for this 2%?
4. Given the scenario in the previous problem, what would the mean lifetime have to be for the company to stay under budget for the Type A batteries? In other words, what would the mean have to be if only 2% of the population is less than 36 months?
5. The company has found that if their Type A batteries last at least 6 months beyond the warranty period (44 months or more), then the customer is sure to buy their battery as a replacement when it fails. What portion of the customers can they count on as being repeat customers? Assume the mean life is 38 months.

Answer 1

a)batteries will fail before the warranty period if they fail fail before 36 months

so,

µ =    38              
σ =    2.5000              
left tailed                  
X ≤    36              
                  
Z =   (X - µ ) / σ =   -0.80          
                  
P(X ≤   36   ) = P(Z ≤   -0.80   ) =   0.2119

0.2119 proportion of the batteries will fail before the warranty period

b)

µ =    38                          
σ =    2.50                          
right tailed                              
X ≥   42                          
                              
Z =   (X - µ ) / σ =   1.60                      
                              
P(X ≥   42   ) = P(Z ≥   1.60   ) =   P ( Z <   -1.60   ) =    0.0548

5.48%  is the probability that battery will last at least 42 months

c)

µ =    38                  
σ =    2.5                  
proportion=   0.02                  
                      
Z value at    0.02   =   -2.054   (excel formula =NORMSINV(α))      
z=(x-µ)/σ                      
so, X=zσ+µ=   -2.054   *   2.5   +   38  
X   =   32.8656              
so, cutoff point, in months, for this 2% is 32.86 months

d)

mean,µ=x-Z_0.02σ = 36-(-2.054)*2.5 = 41.13

so, mean would be 41.13 months

e)

µ =    38                          
σ =    2.50                          
right tailed                              
X ≥   44                          
                              
Z =   (X - µ ) / σ =   2.40                      
                              
P(X ≥   44   ) = P(Z ≥   2.40   ) =   P ( Z <   -2.40   ) =    0.0082