Midterm and solution MAE 3020, Information Processing for Automation
Midterm Exam, Fall 1998
(1) In a certain assembly plant, three machines, B1, B2 and B3 makes 30%, 45% and 25%,
respectively, of the products. It is known from past experience that 2%, 3% and 2% of the
products made by each machine, respectively, are defective. Suppose that a finished
product is randomly selected, what is the probability that it is defective? If the product is
defective, what is the probability that it was made by machine B3?
S: Let P(Bi) be the probability of a product being made by machine Bi, and P(F, Bi) be
the probability of probability of a defective product made by machine Bi, then
P(B1) = 0.3, P(B2) = 0.45, P(B3) = 0.25
P(F, B1) = 0.02, P(F, B2) = 0.03, P(F, B3) = 0.02
Consequently, the probability that a randomly selected product is defective is:
P(F) = (0.3)(0.02) + (0.45)(0.03) + (0.25)(0.02) = 0.0245
If the product is defective, then using the Baysian formula, the probability that the
product is made by machine B3 is:
P(B3 / F) = P(B3, F) / P(F) = (0.25)(0.02) / 0.0245 = 0.204
(2) A lot contains seven components including 4 good components and 3 bad
components. A sample of 3 components is taken by a quality control inspector. What is
the probability that one of the components is a bad component? What is probability that
at least one component is bad component? What is the expected number of bad
components?
S: Note that this is not a binomial distribution problem because the sample space is
small (n = 7). The probability that one of three selected components is a bad
component can be determined by counting:
 4  3 
  
2 1
18
p     
35
7
 
 3
The probability that at least one component is bad component is equal to:
 4  3 
  
3 0
4 31
p  1      1 

35 35
7
 
 3
Finally, it may be noted that the probability distribution is hypergeometry. Hence, the
mean of the number of bad components is:
kn 3  3


 1.286
N
7
(3) An appliance dealer sells three different models of upright freezers that having 13.5,
15.9 and 19.1 cubic feet of storage space. Let X be the amount of storage space purchased
by the next customer to buy a freezer. Suppose X has the probability distribution as
follows:
x
p(x)
13.5
0.2
15.9
0.5
19.1
0.3
Furthermore, suppose the price of a freezer having capacity X cubic feet is 25X - 8.5.
(a) What is the expected price paid by the next customer?
(b) What is the variance of the price paid by the next customer?
(c) Suppose that while the rated capacity of a freezer is X, the actual capacity is h(X) = X
- 0.01X2. What is the expected actual capacity of the freezer purchased by the next
customer?
S: (a) The expected price paid by the next customer is:
E(25X – 8.5) = 25E(X) – 8.5 = (25)(16.38) – 8.5 = 401
(b) The variance of the price paid by the next customer is:
V(25X – 8.5) = (25)2V(X) = (625)(3.9936) = 2495.75
(c) The expected actual capacity of the freezer is:
E(h(X)) =  h(X) f(X)
= [(13.5)-(0.01)(13.5)2](0.2) + [(15.9)-(0.01)(15.9)2](0.5) +
[(19.1)-(0.01)(19.1)2](0.3)
= 13.66
(4) The Rockwell hardness of a metal is determined by impressing a hardened point into
the surface of the metal and then measuring the depth of penetration of the point. Suppose
the Rockwell hardness of a particular alloy is normally distributed with mean 70 and
standard deviation 3.
(a) If a specimen is acceptable only if its hardness is between 67 and 75, what is the
probability that a randomly chosen specimen has an acceptable hardness?
(b) If the acceptable range of hardness was (70 - c, 70 + c), for what value of c would
95% of all specimen have acceptable hardness?
(c) If the acceptable range is as in part (a) and the hardness of each ten randomly selected
specimens is independently determined, what is the expected number of acceptable
specimen among ten?
(d) What is the probability that at most eight of ten independently selected specimens
have a hardness of less than 73.84?
S: (a) Let X be the hardness of the metal, then X ~ N(70, 3), and therefore:
75  70 
 67  70
Z
P(67  X  75) = P
 = 0.95221 – 0.15866 = 0.79355
3 
 3
c
 z 0.05 2 , since z0.025 = 1.96, c = (1.96)(3) = 5.88
(b)

(c) Because the sampling is independent, its distribution is Binomial with p = 0.79,
therefore the mean is  = (10)(0.79)  8
(d) First P(X  73.84) = 0.89973, then, use the binomial distribution:
P(p = 0.9, n = 10, x = 9) = 0.38742
P(p = 0.9, n = 10, x = 10) = 0.34868
Thus, the probability that at most eight of ten specimens have a hardness of less than
73.84 is:
1 – 0.38742 – 0.34868 = 0.264
(5) Consider the following measurements of the heat producing capacity of the coal
produced by two mines (in million of calories per ton);
Mine 1:
Mine 2:
8260
7950
8130
7890
8350
7900
8070
8140
8340
7920
7840
Assuming that the populations are normally distributed, can it be concluded that the two
population variances are equal?
S: From the samples, we have:
n1 = 5, s12 = 15750
n2 = 6, s22 = 10920
Note that the ratio of the variances conforms the F-distribution, the F-value is:
s2
F  12  1.44
s2
On the other hand, from the F-distribution table, it can be seen that:
F0.1(n1, n2) = F0.1(5, 6) = 3.1075.
In other words, if the variance is really different (with a probability of 1 – a = 90%),
the F-value will be great than 3.01075. Since F = 1.44 &lt; F0.05(n1, n2) = 3.1075, we can
say the variances are the same with a 90% probability. In fact, F = 1.44 correspond to
a probability of 66.8%. We really cannot say the variances are different.
(6) An electrical firm manufactures light bulbs that have a length of life that is
approximately normally distributed with a standard deviation of 40 hours. If a sample of
30 bulbs has an average life of 780 hours, find a 95% confidence interval for the
population mean of all bulbs produced by the firm.
S: Since n = 30, x =780,  = 40, z0.025 = 1.96, and the confidence interval is determined
by:
x  za 2

n
   x  za 2
Thus:
7.65.686    794.314

n