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
Step-by-step explanation:
Normal Distribution
μ₀ = 14000 hours and σ = 1050 hours
a) Probability of X (replacement of motor free of charge)
Z = ( X - μ₀ ) / σ
Z = ( 12000- 14000) / 1050
Z = - 2000/1050
Z = - 1,904
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] = 1 - 0.9772 = 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 = ( 14000 - X ) / 1050
0,4960 * 1050 = ( 14000 - X )
520,8 - 14000 = - X
X = 13479,2 hours as is not common such offer company could set up guarantee for 13500 h
