I. The Readylite Company produces a flashlight in which product managers are try

Question

I. The Readylite Company produces a flashlight in which product managers are trying to decide how long a warranty to issue. If the managers believe the life of the flashlight followsa normal distribution with a mean of 3 years and a standard deviation of 1.5 years): a. What percentage of flashlights sold can they anticipate b. What percentage of flashlights sold can they anticipate c. What percentage of flashlights sold can they anticipate will be returned within the first year? will be returned within two years? will be returned between the first year and the third year?

Explanation / Answer

Anticipated flashlight life mean: 3 years

Therefore, the Z for flashlights returned with one year:

Z= x-u/s

Where, x= number of flashlights returned within one year

u = mean value, and

s = standard deviation

Hence,

Z= 1-3/1.5 = -2/1.5 = -1.33

Now,

P(x<1) = P(Z<-1.33)

Therefore probability is = 0.9176 (from normal distribution table)

We calculate the percentage of flashlights that can be returned in the first year.

0.9176* 100 = 91.76 % of flashlights can be returned in the first year

Z= 2-3/1.5= -0.67

Now,

P(x<2) = P (Z< -0.67)

P = 0.25143

Hence the percentage anticipated to be returned will be = 25%

Z= 3-3/1.5= 0

Now,

P(x<3) = P (Z< 0)

P = 0.50000

Percentage = 50%

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