Question

A) The mean lifetime of a smartphone model is 3 years with a standard deviation of 5 months. Assume normal distribution. What percentage of all smartphones should the manufacturer expect to replace if they offer a warranty period of 30 months?

B) The mean lifetime of a smartphone model is 3 years with a standard deviation of 5 months. Assume that lifetimes are normally distributed. The CEO wants to budget for replacing not more than 5% of the smartphones during the warranty period. What is the maximum length of the warranty period that the company can offer? Explain. (Round down to the nearest month.)

Answer 1

P(X < A) = P(Z < (A - mean)/standard deviation)

Mean = 3 years = 36 months

Standard deviation = 5 months

A) P(lifetime is less than 30 months) = P(X < 30)

= P(Z < (30 - 36)/5)

= P(Z < -1.2)

= 0.1151

Percentage of smartphones the manufacturer expect to replace if they offer a warranty period of 30 months = 11.51%

B) Let the length of warranty be W

P(X < W) = 0.05

P(Z < (W - 36)/5) = 0.05

Take value of Z corresponding to 0.05 from standard normal distribution table

(W - 36)/5 = -1.645

W = 27.775

Maximum length of the warranty period that the company can offer = 27 months