1. The lifetimes of TVs produced by Company B are normally distributed with a me

Question

1. The lifetimes of TVs produced by Company B are normally distributed with a mean of 70 months and a standard deviation of 10 months.

a) If company B wants to replace only 1% of its TVs, what should the length of the warranty be?

b) Now a new manufacturing technology will reduce the standard deviation by a factor of 2. Assuming company B does not change its warranty policy from (a), what is the % of tvs needs to replace?

c) If replacing a tv costs $200, and this new technology costs $1M to implement, how many tvs company B should sell so that the cost savings (replacing fewer tvs) will equal the cost of implementation of the new technology?

Please let me know the type of distribution and explain why you take that approach to solve the problem so I can understand this further

Explanation / Answer

The lifetime of TVs are normaly distributed.

Mean = 70 months

sd = 10 months

a)

company wants to replace 1% of tv.

Finding the z value from the above percentiloe from the z table = z = -2.3263

z = (x - mean) / sd

-2.3263 = x - 70 / 10

x = 46.737

Therefore the length of the warrenty should be 46.737 months

b)

standard deviation reduced by a factor of 2.

New sd = 5

now the new z = (46.737 - 70 )/ 5

z = -4.6526

finding the percentile from z tabel from the above z value = 0.0002

Hence only 0.0002% of the TVs should be replaced

c)

replacing tv cost = $200

new technology cost = $1M

Let the number of tv sold = x

Before 1% of tv would be replaced.

Replaced tv = x/100

With new tech 0.0002% tv would be replaced

number of replaced tv = 0.000002x

Difference in tv replaced = (0.001-0.000002)x

Total cost saved = 200 * (0.000998) x = 1000000

therefore x = 5,010,020 number of TVs to be sold to recover cost of investment

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