An industrial oven used to cure sand cores for a factory manufacturing engine bl
ID: 3126076 • Letter: A
Question
An industrial oven used to cure sand cores for a factory manufacturing engine blocks for small cars is able to maintain fairly constant temperatures. The temperature range of the over follows a normal distribution with a mean of 450Fo and a standard deviation of 20Fo. Leslie Larsen, president of the factory, is concerned about the large number of defective cores that have been produced in the past several months. If the over gets hotter than 475Fo, the core is defective.
a. for the probability that the temperature of the oven will range from 430F to 460F , the Z value below the mean is not 0.93
b. for the probability that the temperature of the oven will range from 430F to 460F, the probability is not .62892
c. for the probability that the temperature of the oven will range from 430F to 460F, the Z value above the mean is greater than 0.84
d. for the probability that the temperature of the oven will range from 430F to 460F, the probability is not .5328
e. the probability that the oven will cause a core to be defective, the probability is not .89345
f. for the probability that the temperature of the oven will range from 460F to 470F , the Z value below the mean is not 0.93
g. for the probability that the temperature of the oven will range from 460F to 470F, the probability is not .62892
h. for the probability that the temperature of the oven will range from 460F to 470F, the Z value above the mean is greater than 0.84
i. for the probability that the temperature of the oven will range from 460F to 470F, the probability is not .5328
j. If the oven manufacturer wants to guarantee 2% defect rate, the highest temperature to be tested for this quality control setting is 470F.
Explanation / Answer
mean = 450
standard dev = 20
a)
For x = 430 , z = (430 - 450) /20 = -1 and for x = 460, z = (460 - 450) / 20 = 0.5
Hence P(430 < x < 460) = P(-1 < z < 0.5) = [area to the left of z = 0.5] - [area to the left of -1]
= 0.5 - 0.1587 = 0.3413
e) its defective when it exceed 475
p(x>475) =
For x = 475, z = (475 - 450) / 20 = 1
Hence P(x > 475) = P(z > 1) = [total area] - [area to the left of 1]
= 1 - 0.8413 = 0.1587
f)
For x = 460 , z = (460 - 450) / 20 = 0.5 and for x = 470, z = (470 - 450) / 20 = 1
Hence P(460 < x < 470) = P(0.5< z < 1) = [area to the left of z = 1] - [area to the left of 0.5]
= 0.8413 - 0.6915 = 0.1498
j)
for 2% defect rate means 98accurate
the z score value for 0.98 probability = 2.1
x = mean +z*standard deviation
x = 450 + 2.1*20 = 492