The Readylite Company produces a flashlight in which product managers are trying

Question

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 follows a normal distribution with a mean of two and one-half years and a standard deviation of six months:

A. What percentage of flashlights sold can they anticipate will be returned within the first two years?

B. What percentage of flashlights sold can they anticipate will be returned within the first two and one-half years?

Explanation / Answer

Let the Z value of the probability of the flashlight being returned within 2 years = Z1

Z value of the flashlight being returned within 2 and half years = Z2

Given are the following information about life of the flashlight :

Mean = m = 2.5 years

Standard deviation = S = 0.5 years ( i.e 6 months )

Therefore,

Mean life + Zvalue x Standard deviation of life = Years within which flashlight is being returned

Or, 2.5 + 0.5xZ1 = 2

Or, Z1 = - 1

Similarly 2.5 + Z2 x 0,5 = 2.5

Or, Z2 = 0

As derived from Standard normal distribution table :

Probability corresponding to Z = -1 will be 0.15866

Probability corresponding to Z = 0 will be 0.5

Thus percentage of flashlight which can be expected to be returned within first 2 years

= Probability corresponding to Z = - 1 x 100

= 0.1586 x 100

= 15.86%

Percentage of flashlight which can be anticipated to be returned within 2.5 years

= Probability corresponding to Z = 0 x 100

= 0.5 x 100

= 50%

PERCENTAGE OF FLASHILIGHTS SOLD CAN BE ANTICIPATED TOBE RETURNED WITHIN FIRST 2 YEARS = 15.86 %

PERCENTAGE OF FLASHLIGHT WHICH CAN BE ANTICIPATED TO BE RETURNED WITHIN FIRST 2.5 YEARS = 50%

PERCENTAGE OF FLASHILIGHTS SOLD CAN BE ANTICIPATED TOBE RETURNED WITHIN FIRST 2 YEARS = 15.86 %

PERCENTAGE OF FLASHLIGHT WHICH CAN BE ANTICIPATED TO BE RETURNED WITHIN FIRST 2.5 YEARS = 50%

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