Question

A company manufactures a wash-down motor that is used in the food processing industry. The motor is marketed with a warranty that guarantees it will be replaced free of charge if it fails within the first 13,000 hours of operation. On average, the wash-down motors operate for 16,000 hours with a standard deviation of 2,450 hours before failing. The number of operating hours before failure is approximately normally distributed. a. What is the probability that a wash-down motor will have to be replaced free of charge? b. What percentage of wash-down motors can be expected to operate for more than 20,900 hours? If the company wants to design a wash-down motor so that no more than 1% are replaced free of charge, what would the average hours of operation before failure have to be if the standard deviation remains at 2,450 hours? C. a. The probability that a wash-down motor will have to be replaced free of charge is (Round to four decimal places as needed.) b. It can be expected that % of wash-down motors will operate for more than 20,900 hours. (Round to two decimal places as needed.) hours. c. The average hours of operation before failure would have to be (Round to the nearest hour as needed.)

Answer 1

Mean, p= 16,000 Sol let x be random variable denotes the number of operating hours before failure. Sländard deviatio, o = 2,450 XNN(M= 16000, 5= 2450) Then Z= X-M X- 16000 follows standard Normal. 2450 a) Here, we need to find P(X4 13,000) 13000 - 16000 P(x +13,000) - PC 24 50 Plz C-1-2245) 0.1103821 to from Standard Normal 0.110 4 Here, to find P(x > 20,900) need ع لا 20900 - 16000 P(x> 20900) = 24 50 PLZ72) PIEC2) = - 0.97225 = 0.02275 2.2&%. expected It Can be will operate that 2.28% of wash - down motors than 20, 900 hours for more

Here we need to find xo such 15 at 0.01 P(x < xo) = 1%= p(x-4. 20 - 16000 0:0) 2450 xor 16000 Plzc ) - 2450 20- 16000 = -2.326 (from Z-lable] 24 50 = 16000 - 2450 *2.326 70 = 10301.3 10301 The average hours of operation before failure would have to be 10301 hourd.