A major television manufacturer has determined that its 40-inch LED televisions

Question

A major television manufacturer has determined that its 40-inch LED televisions have a mean service life that can be modeled by a normal distribution with a mean of six years and a standard deviation of one-half year.

a. What probability can you assign to service lives of at least (1) Five years? (2) Six years? (3) Seven and one-half years?

b If the manufactuerer offers service contracts of four years on these televisions what percentage can be expected to fail from wear-out during the service period?

c. What service life the manufacturer must specify, so that at least .9772 of the products would not be in need of warranty service?

Explanation / Answer

Answer to question a :

Given are following data :

Mean service life = m = 6 years

Standard deviation of service life = Sd = 1.5 years

Let z value for probability that service life will be upto 5 years = Z1

Therefore ,

M + Z1xSd = 5

Or, 6 + 1.5.Z1 = 5

Or, Z1 = - 0.666 ( - 0.67 rounded to 2 decimal places )

Corresponding probability from standard normal distribution table = 0.25143

Therefore , probability that service life will be at least 5 years = 1 – 0.25143 = 0.74857

Let z value for probability that service life will be upto 6 years = Z2

Therefore ,

M + Z2xSd = 6

Or, 6 + 1.5.Z2 = 6

Or, Z2 = 0

Corresponding probability from standard normal distribution table = 0.5

Therefore , probability that service life will be at least 6 years = 1 – 0.5 = 0.5

Let z value for probability that service life will be upto 7.5 years = Z3

Therefore ,

M + Z3xSd = 7.5

Or, 6 + 1.5.Z3= 7.5

Or, Z3 = 1

Corresponding probability from standard normal distribution table = 0.84134

Therefore, probability that service life will be at least 7 years = 1 – 0.84134 = 0.15855

PROBABILITY THAT SERVICE LIFE WILL BE AT LEAST 5 YEARS IS 0.74857

PROBABILITY THAT SERVICE LIFE WILL BE AT LEAST 6 YEARS IS 0.5

PROBABILITY THAT SERIVCE LIFE WILL BE AT LEAST 7.5 YEARS 0.15855

Answer to question b :

Let Z value corresponding to the probability that the service life of the televisions will be maximum 4 years = Z4

Therefore ,

M + Z4 x Sd = 4

6 + 1.5.Z4 = 4

Or, 1.5.Z4 = - 2

Or, Z4 = - 2/1.5

Or, Z4 = - 1.33

Corresponding probability for Z4 = - 1.33 as derived from standard normal distribution table will be 0.09176

Therefore , percentage of televisions expected to wear out during service period = 0.09176 x 100 = 9.176 %

Answer to question c :

Given that 0.9772 of the products would not be in need of warranty service, i.e. probability that the product will not fail is 0.9772

Hence probability that the product will fail = 1 – 0.9772 = 0.0228

Corresponding Z value = NORMSINV ( 0.0228) = - 1.999

Therefore , service life manufacturer must specify

= m + Z x Sd

= 6 - 1.999 x 1.5

= 6 - 2.998

= 3.002 ( 3 rounded to nearest whole number )

Thus it is for a service life of 3 years only 2.28 % of products will fail and 97.72 % will not fail and therefore would not be in need of warranty service

MANUFACTURER MUST SPECIFY A SERVICE LIFE OF 3 YEARS

PROBABILITY THAT SERVICE LIFE WILL BE AT LEAST 5 YEARS IS 0.74857

PROBABILITY THAT SERVICE LIFE WILL BE AT LEAST 6 YEARS IS 0.5

PROBABILITY THAT SERIVCE LIFE WILL BE AT LEAST 7.5 YEARS 0.15855

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