BA 275
Winter 2007
Quiz 2
Answer Key
Quiz #2
Name
(please print)
Section
Question 1. Given X ~ N (   3,  2  2 2 ) , Find P(5.48  X  2.56) .
 = -3
=2
Prob = ???
F
-5.48
-2.56
Answer: 0.4796. The z score of –5.48 = –1.24 (that is z = (5.48   ) /   1.24 ). The size
of the area under the curve below z = –1.24 is 0.1075 (from the normal curve table). The z score
of –2.56 is 0.22 and the size of the area to the left of 0.22 is 0.5871. Finally, the shaded area in
the figure above is 0.4796 = 0.5871 – 0.1075.
An automatic machine in a manufacturing process is operating properly if the lengths of an
important subcomponent are normally distributed with mean of 117 cm and standard deviation of
5.2 cm.
Question 2. Find the probability that one selected subcomponent is longer than 120 cm.
Answer: 0.2810. The z score of 120 = (120  117) / 5.2  0.58 . The area to the left of 0.58 is
0.7190. To find the area (i.e., the probability) to the right of 0.58, simply take 1 – 0.7190 =
0.2810.
Question 3. Find the probability that if four subcomponents are randomly selected, all four
have lengths that exceed 120 cm.
Answer: (0.2810)4. From Question 2 above, we know that the chance of getting one
subcomponent with a length longer than 120 cm is 0.2810. To have all 4 subcomponents to
exceed 120 cm, the probability is (0.2810)  (0.2810)  (0.2810)  (0.2810).
Hsieh, P-H
1
BA 275
Winter 2007
Quiz 2
Answer Key
Question 4. Find the probability that if four subcomponents are randomly selected, their mean
length exceeds 120 cm.
Answer: 0.1251. The sampling distribution of the sample mean is a normal distribution.
What is the mean (location) of this normal distribution? It is the same as the population mean
(117 cm). What is the standard deviation (  X , spread) of this normal distribution? It is the
population standard deviation () divided by the squared root of the sample size (n), that is
 X   / n  5.2 / 4  2.6 . So, how likely do you get a sample mean ( X ) that exceeds 120
cm? Find the z score of 120 cm with respect to the normal distribution of the sample
mean, z  (120  117) / 2.6  1.15 . Look up the normal curve table and you will get 0.8749 (this
is the size of the area to the left of z = 1.15). To answer the question, take 1 – 0.8749 = 0.1251.
Question 5. You, as an inspector, randomly selected 25 subcomponents and measured their
mean length to be 120 cm. Based on your sample, do you think the machine is still in control
(i.e., the machine is still producing subcomponents with a mean length of 117 cm), or is already
out of control (the mean length is no longer 117 cm as required)? Support your argument with
numerical evidence.
Again, the sampling distribution of the sample mean is normal with a mean 117 cm and a
standard deviation  X   / n  5.2 / 25  1.04 . What is the chance of getting a sample mean
exceeding 120 cm? The z score of 120 cm, with respect to this normal distribution, is
z  (120  117) / 1.04  2.88 . Use the normal curve table and find the probability to be 0.9980.
In other words, given the mean 117 cm, the chance of getting a sample mean greater than 120 cm
is 1 – 0.9980 = 0.0020, a small chance. Therefore, the process is more likely to be out of control.
Hsieh, P-H
2