2333MidtermS 2014

Topics: Normal distribution, Probability theory, Failure, Temperature, Conditional probability / Pages: 7 (957 words) / Published: Jul 5th, 2015
Name:
INDE 2333
Engineering Statistics I
Midterm Examination
Fall 2014
Total Points: 100

Professor: Qianmei Feng
Time: 80 Minutes

Please show all of your work, including the methods used and step-by-step calculations. You will not be graded based on only the final answers but based on the overall process to obtain the answer.
Please use these sheets to answer your questions and no attachments are necessary.
Problem 1
(a) The manager of a small plant wishes to determine the number of ways she can assign personnel to the first shift. She has 18 persons who can serve as operators of the production equipment, 10 persons who can serve as maintenance personnel, and 4 persons who can be supervisors. If the shift requires 6 operators, 2 maintenance persons, and 1 supervisor, in how many different ways can it be staffed?
[8 points]

18 10  4 
     18564  45  4  3,341,520
 6  2  1 

(b) Suppose A and B are not mutually exclusive events, and we have P(A)=0.35, P(B)=0.40,
P(AB)=0.18. Compute the following probabilities:
i) P (AB)=?

[4 points]
P (AB)=P[A]+P[B]-P[AB]=0.35+0.40-0.18=0.57

ii) P(AB)=?

P[A  B] 

P[ A  B] 0.18

 0.45
P[ B]
0.40

1 of 6

[4 points]

Name:
Problem 2
A diagnostic test for a certain disease is said to be 90% accurate in that, if a person has the disease, the test will detect it with probability 0.9. Also, if a person does not have the disease, the test will report that he or she does not have it with probability 0.9. Only 1% of the population has the disease in question. If a person is chosen at random from the population and the diagnostic test indicates that she has the disease, what is the conditional probability that she does, in fact, have the disease?
[15 points]
Define these events:
D: person has the disease
H: test says person has the disease
P ( H D )  0.9; P ( H D )  0.9; P ( D )  0.01; P ( D )  0.99
Using Bayes’ Theorem,

P( D H ) 

P( H D) P( D)
P( H D) P( D)  P( H D ) P( D )



0.9  0.01

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