Name:_________________________________________
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Instructor:_________________________________________
ISE 205: Engineering Probability & Statistics
EXAM # 2
April 28th, 2013
Duration: 2 hours
1. Please read all the questions before you start. Clarification questions will be answered only
during the first 30 minutes of the exam.
2. Provide your answers in two decimal places.
3. To get full credit for a question you need to JUSTIFY your answer and SHOW ALL YOUR
WORK.
Max
1
2
3
4
5
6
7
8
9
Total
50
1
Mark
1. The probability distribution of the discrete random variable X is shown in the table below.
x
-3
-1
0
4
P(X=x)
a
b
0.15
0.40
Given that E(X) = 0.75,
(a)
Solution:
Using
Find the values of a and b.
 f ( x)  1
 a  b  0.15  0.40  1
 a  b  0.55
x
Using
 x f ( x)  1
 -3a  b  0  0.15  4  0.40  0.75
x
Solving for a and b yields
(b)
a  0.15
and
b  0.40
Find P( X > 2.5)
P( X  2.5) 

f ( x) P( X  4)  0.40
x  2.5
2
 3a  b  0.85
2. Let F(x) be the cumulative discrete distribution function for the random variable X
x  10
 0
0.15 10  x  0

0.35 0  x  5
F (x )  
0.45 5  x  8
0.75 8  x  10

10  x
 1
(a) Find P[ 0 ≤ X < 7]
P (0  X  2.5)  P (X  7)  P (X  0)  F (7)  F (10)  0.45  0.15  0.30
(b) Find P[ X ≥ 3]
P( X  3)  1  P( X  0)  1  F (0)  1  0.35  0.65
(c) Compute the E (5X)
x
-10
0
5
8
10
P(X=x) 0.15 0.20 0.10 0.30 0.25 1.00
xf(x)
-1.5
0
0.5 2.4 2.5 3.9
2
x
100
0
25
64 100
2
x f(x)
15
0 2.5 19.2
25 61.7
E[ X ]   xf ( x)  3.9

E[5 X ]  5E[ X ]  19.5
x
(d) Compute the standard deviation of X.

x
2
f ( x)   2  61.7  3.92  6.8
x
3
3. A class has 30 students and the probability a student fails a course is 0.1. Assume each student is
independent from the other. Compute the following:
a) The probability that more than two students pass the course.
X: number of students who pass the course  X follows binomial with n = 30 and p =0.90
 30 

 30 
 30 
P (X  2)  1  P (X  2)  1    (0.9)0 (0.1)30    (0.9)1 (0.1) 29    (0.9) 2 (0.1) 28 
1
2
 0 

 1  [0]  1
b) The expected number of students who pass the course.
E [X ]  np  30  0.9  27
c) The probability that a student passes the course in at least three trials.
Y: number of trials to pass the course  Y is geometric with p=0.9
P(Y≥3)=1-P(Y≤2)=1-[P(X=1)+P(X=2)]=1-[0.9+0.1x0.9] = 1-0.99=0.01
d) The expected number of trials for a student to pass the course.
E[Y] = 1/p = 1/.9 = 1.11
4
4. A shipment of 15 brake pads of which 3 are defective arrived at a company. A sample of size 4 is
drawn at random from the batch of 15. Compute the following:
(a) The probability the sample does not contain any defective item?
 3 12 
  
0 4
1 495
P (X  0)     
 0.3626
1365
15 
 
4
(b) The probability of having at least one defective.
P (X  1)  1  P (X  0)  1  0.3626  0.6374
5
5. The number of repair requests arriving at a maintenance department follows a Poisson distribution.
The average number of requests is 2 requests per hour. Compute the following:
(a) The probability of at least one request arriving per hour.
P (X  1)  1  P (X  0)  1  e 2  0.8647
(b) The expected number of requests per shift of 10 hours.
10x2=20
(c) The probability of more than one request arriving per shift (A shift is 10 hours).

e 20 20 
P (X  1)  1  [P (X  0)  P (X  1)]  1  e 20 
1
1 

(d) The probability no request for repair arrives in 2 hours?
P (X  0)  e 4  0.01832
(e) The expected time between requests for repair.
1/2=0.5 hours
(f) The probability that the time between requests exceeds the time found in (e)

P (X  0.5) 
 2e
2 x
dx  e 1  0.3679
0.5
6
6. The probability density function of the random variable X is given as:
0.125 1  x  5 and 10  x  14
f (x )  
otherwise
 0
(a) Determine and graph the cumulative distribution function.
0
x 1

 0.125(x  1)
1 x  5

F (x )  
0.5
5  x  10
0.5  0.125(x  10) 10  x  14

1
x  14

1.0
0.5
1
4
10
14
(b) Find P[ 3 ≤ X ≤ 12].
P (3  X  12)  P (X  12)  P (X  3)  0.5  0.125(12  10)  0.125(3  1) =0.5
(c) Compute the expected value of E(X).
0.125(42  12 ) 0.125(142  102 )
E [X ]   0.125xdx   0.125xdx 

 0.9375  6  6.9375
2
2
1
1
4
4
(d) Compute the variance of X.
4
4
1
1
E [X ]   0.125x 2dx   0.125x 2dx   2 
0.125(43  13 ) 0.125(143  103 )

 6.93752
3
3
 2.67  72.67  6.93752  27.2
7
7. Tire pressures on a certain type of car independently follow a normal distribution with mean 1.9 bars
and standard deviation 0.16 bars.
(a) Find the probability that all four tires on a car of this type have pressures between 1.82 bars and
1.92 bars.
 1.92  1.9 
 1.82  1.9 
P (1.82  X  1.92)   

    0.125     0.5   0.54974  0.30854  0.2412
 0.16 
 0.16 
 4
P (all )    0.24124 (1  0.2412) 0  0.00338
 4
(b) 80% of the tires have pressure between 1.9 − b bars and 1.9 + b bars. Find b?
b
 1.9  b  1.9 

 b  0.21
  0.9  z  1.28 
0.16
0.16


8
8. A survey of adults in a certain large town found that 76% of people wore a watch on their left wrist,
15% wore a watch on their right wrist and 9% did not wear a watch.
(a) A random sample of 14 adults was taken. Find the probability that more than 2 adults did not
wear a watch.
Binomial (n=14, p=0.09)
14 

14 
14 
P (X  2)  1  P (X  2)  1    0.0900.9114    0.0910.9113   0.09 20.9112 
1
2
 0 

 1  (0.2670  0.3698  0.23)  0.126
(b) A random sample of 200 adults was taken. Using a suitable approximation, find the probability
that more than 155 wore a watch on their left wrist.
  np  200*.76  152
and
  np (1  p )  36.48
 155.5  152 
P (X  155)  1  P (X  155)  1   
  1  (1.5795)  0.281
36.48 

9
9. The volume filled by an automated machine used for filling cans is normally distributed with mean
3.6 cubic centimeters (cc) and variance 0.04.
(a) What is the probability that a fill volume is less than 4.0 cc.
 4  3.6 
P (X  4)   
  (2)  0.97725
 0.04 
(b) If all cans filled less than 3.2 or more than 4.3cc are rejected. What percentage of cans is
rejected?
(c) P (X  3.2)  P (X  4.3)    3.2  3.6   1    4.3  3.6   (2)  1  (2)  2(1  (2))  0.0455





0.04 

0.04 
(d) At what value should the mean be set so that 99.5 % of all cans exceed 3.85 cc, keeping the
variance the same.
P (X  3.85)  .995  P (X  3.85)  0.005
3.85   

P Z 
  0.005  z  2.57
0.2 

3.85  
 2.57    4.36
0.2
10