Question

Assume that the life of an engine follows a normal distribution with an average life of 12 years and a standard deviation of 2 years. The manufacturer replaces without additional charge all the motors that fail during the warranty.

a) What is the probability that an engine will fail after 14 years?
b) What is the probability that the engine lasts between 4 to 13 years?
c) A contractor buys 400 engines which is the probability that 70 or more of those that I bought have a duration of more than 14 years?
d) If the manufacturer wishes to replace only 1% of the engines it sells, what is the warranty time that you must offer?

Answer 1

a)

for normal distribution z score =(X-μ)/σx
here mean=       μ=	12
std deviation   =σ=	2.000

probability that an engine will fail after 14 years:

probability =

P(X>14)

=

P(Z>1)=

1-P(Z<1)=

1-0.8413=

0.1587

b)

probability that the engine lasts between 4 to 13 years:

probability =

P(4<X<13)

=

P(-4<Z<0.5)=

0.6915-0=

0.6915

c)

for 400 engines ;expected motors that have duration more than 14 years=np=400*0.1587= 63.48

and std deviaiton =sqrT(np(1-p))=7.308

hence from normal approximation and continuity correction:

probability that 70 or more of those that I bought have a duration of more than 14 years:

P(X>=70)=P(Z>(69.5-63.48)/7.308)=P(Z>0.82)=0.2061

d)

for 1 percentile crtiical z =-2.33

hence corresponding value =mean+z*std deviation =12-2.33*2=7.34 Years