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 12,000 hours of operation. On average, the wash-down motors operate for 14,000 hours with a standard deviation of 1, 050 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 16, 100 hours?c. 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 1, 050 hours? 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 16, 100 hours (Round to two decimal places as needed.)c. The average hours of operation before failure would have to be hours.(Round to the nearest hour as needed.)

Answer 1

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