Answer: 0.0227
Step-by-step explanation:
Mean expectancy hours M = 2400 hours
Standard deviation S = 300 hours
Number of hours used per year = 1000 hours
Number of hours in 3 years N = 3*1000 = 3000 hours
Probability of lasting more than 3000 hours = 1 - Probability of lasting up to 3000 hours
P(z>3000) = 1 - P(z< 3000)
P(z<3000) = P[(N-M)/S < (3000-2400)/300 ]
P(z<3000) = P(z< 600/300)
P(z<3000) = P(z< 2) = ¢(2)
P(z<3000) = 0.9773
Therefore,
P(z>3000) = 1 - P(z<3000)
P(z>3000) = 1 - 0.9773
P(z>3000) = 0.0227
The probability of the component lasting more than 3 years (3000 hours) is 0.0227