Answer:
a)  P [ X ≤ 12000 ]  = 0.0262    or    2.62 %
b) P [ X > 16100]  = 0,0228   or  2,28 %
X = 13479,2 hours   or   13500 hours Â
Step-by-step explanation:
Normal Distribution
μ₀  = 14000   hours  and     σ  =  1050   hours
a) Probability of X Â (replacement of motor free of charge)
Z = ( X - μ₀ ) /  σ
Z Â = Â ( 12000- 14000) / 1050
Z Â = Â - 2000/1050
Z = - 1,904 Â Â
From z Table we get
for  z = -1.904    P [ X ≤ 12000 ]  = 0.0262
P [ X ≤ 12000 ]  = 0.0262    or    2.62 %
b) What % of motors can be expected to operate more than 16100 hours
z = ( 16100 - 14000) / 1050
z = 2100/1050   ⇒  2
From z table we find with z = 2 total number f motor operating up to 16100 hours
From z table   we find 0,9772
Then probability of motors operating for more than 16100 hours is:
P [ X > 16100] Â = Â 1 Â - Â 0.9772 Â Â = Â 0,0228
Then
P [ X > 16100]  = 0,0228   or  2,28 %
c) The average hours of operation before failures if only 1% of motors would be replaced free of charge is:
1%  = 0,01   Probability then z score is  z = 0,4960
0,4960 Â = Â ( 14000 - X ) / 1050
0,4960 * 1050 = ( 14000 - X )
520,8 - 14000 = Â - X
X = 13479,2 hours  as is not common such offer company could set up guarantee for  13500 h