Statistics-Percentiles of the Distribution Engineering Design - Spl 8 . Hyundai
ID: 3050599 • Letter: S
Question
Statistics-Percentiles of the Distribution Engineering Design - Spl 8 . Hyundai advertises a 10 year-100,000 mile power-train warranty for its cars. You work for a competitor and want to see if you can meet this for your vehicles. You have collected engine performance data and determined that the life is approximately normally distributed with a mean of 112,300 miles and a standard deviation of 4800 miles. (a.) What percentage of the engines will fail by 100,000 miles? (b.) At what mileage should you set the warranty for your vehicles if you are willing to repair 190 at no cost to the consumer? 2. You have tested bearings for a particular machine. This resulted in a mean of 40,570,000 cycles with a standard deviation of 887,500 cycles (assume approximately normally distributed). (a.) What percentage of the bearings will last for 38,000,000 cycles? (b.) If the manufacturer is willing to accept a 5% failure rate, what warranty time should he put on the bearings? Measurements are made on a particular part. The results are approximately normally distributed with a mean of 1.250 inches and a standard deviation of 0.006 inches. (a.) What is the probability that the part will be less than 1.240 inches? (b.) What is the probability that the part will be at least 1.258 inches? 3.Explanation / Answer
1)
mean = 112300 , s = 4800
a)
P(X < 100000)
z = ( x - mean) / s
= ( 100000 - 112300) / 4800
= -2.5625
P(X < 100000) = P(z <-2.5626) = 0.00519
b) z value at 1% for left tail = -2.3263
z = (x - mean) / s
-2.3263 = ( x - 112300)/ 4800
x = 101133.76
2)
mean = 40570000 , s = 887500
a)
P(X < 38000000)
z = ( x - mean) / s
= ( 38000000 - 40570000 ) /887500
= -2.89577
P(X < 38000000) = P(z <-2.89577) = 0.0019
b) z value at 1% for left tail = -2.3263
z = (x - mean) / s
-1.64485 = ( x - 40570000 )/ 887500
x = 39110192.41
3)
mean = 1.250 , s = 0.006
a)
P(X < 1.240)
z = ( x - mean) / s
= ( 1.240 - 1.250) / 0.006
= -1.667
P(X < 1.240) = P(z < -1.667) = 0.0478
b)
P(X > 1.258)
z = ( x - mean) / s
= ( 1.258 - 1.250) / 0.006
= 1.333
P(X > 1.258) = P(z > 1.333) = 0.0912