Download Exam1 - Academic Information System (KFUPM AISYS)

1
KING FAHD UNIVERSITY OF PETROLEUM & MINERALS
DEPARTMENT OF MATHEMATICAL SCIENCES
DHAHRAN, SAUDI ARABIA
STAT 319: PROBABILITY & STATISTICS FOR ENGINEERS & SCIENTISTS
Mid Term Exam, Semester 061
Time: 6.30- 8.30 p.m., Tuesday November 21, 2006
Please CIRCLE the name of your instructor; Write CLEARLY your name,
ID, and Section Number.
Instructors:
Hassen Muttlak, Mohammad Omar, Musawar Malik, Anwar Joarder.
Student Surname:
ID#
Section #
You are allowed to use electronic calculators and other reasonable writing
accessories that help write the exam. Try to define events, formulate
problem and solve.
1. Do not keep your mobile with you during the exam, turn off your mobile
and leave it aside.
2. Check that the exam paper has 9 questions and 6 printed pages (not
including cover page).
Question No
1
Full Marks
15
2
9
3
7
4
5
5
10
6
14
7
6
8
4
9
10
Total
80
Marks Obtained
2
1. Temperature transducers of a certain type are shipped in batches of 50. A
sample of 26 batches was selected, and the number of transducers in
each batch not conforming to design specifications was determined,
resulting in the following data:
0 4 2 1 3
3 1 5 0 2
4 1 2 3 2
3 2 1 0 4
4 5 1
2 1 3
a. (2 Marks) Determine frequencies and relative frequencies for the
observed values of x = number of nonconforming transducers in a batch.
Solution:
x
f
f /n
0
3
0.12
1
6
0.23
2
6
0.23
3
5
0.19
4
4
0.15
5
2
0.08
b. (2 Marks) What proportion of batches in the sample has at most four
nonconforming transducers?
Solution: 1 − 2 /(26) = 12 /13 ≈ 0.92
c. (6 Marks) Calculate mean, median and standard deviation.
Solution: With Sx = 59 and Sx2=189, the TSS = 55.11538. Thus, the mean
and standard deviation are given by y ≈ 2.269230769 and s ≈ 1.484794728
(s 2 ≈ 2.20462) respectively. The ordered sample is given by
000 111111 222222 33333 4444 55 so that the median is given by
Q 2 = y 13.5 = y 13 (0.50) + y 14 (0.50) = 2(0.50) + 2(0.50) = 2
d. (5 Marks) Construct a boxplot and comment on the skewness.
Solution: The outer quartiles are given by
Q1 = y 6.75 = y 6 (0.25) + y 7 (0.75) = 1(0.25) + 1(0.75) = 1,
Q 3 = y 20.25 = y 20 (0.75) + y 21 (0.25) = 3(0.75) + 4(0.25) = 3.25
Min
Max
Q1
Q2
Q3
Since, Q1 = 1,Q 2 = 2, Q 2 = 2,Q 3 = 3.25 it is skewed to the right. One may
14
4244
3 144
42444
3
1
1.25
check further that CS = 3(x − x% ) / s ≈ 0.5439 .
3
2. A system contains two components A and B. The system will function only
if both components function. The probability that A function is 98%, the
probability that B functions is 0.95, and the probability that either A or B
functions is 0.99.
a. (2 Marks) Write out the sample space to solve the following questions:
Solution: The sample space is given by
S = {A ′B ′, AB ′, A ′B , AB }
b. (2 Marks) What is the probability that none of the two components
function?
Solution: P (A ′B ′) = 1 − P (A ∪ B ) = 1 − 0.99 = 0.01
c. (2 Marks) What is the probability that the system functions.
Solution: Given that P (A ) = 0.98, P (B ) = 0.95 and P (AUB ) = 0.99. Then from
P (A ∪ B ) = P (A ) + P (B ) − P (AB ) we have
0.99 = 0.98 + 0.95 − P (AB ) so that P (AB ) = 0.94
d. (3 Marks) Are the components A and B independent?
Solution: Since P (A )P (B ) = 0.931 is different from P (AB ) = 0.94 , the
components are not independent. Alternatively, since
P (AB ) 0.94 94
P (A | B ) =
=
=
≈ 0.989 ≠ P (A ) = 0.98,
P (B )
0.95 95
, we say that the events are dependent.
The above solution can be visualized better in the following table:
A
A′
B
0.94
0.01
0.95
B′
0.04
0.01
0.05
0.98
0.02
3. An automobile service facility specializing in engine tune-ups knows that
50% of all tune-ups are done on four-cylinder automobiles, 40% on sixcylinder automobiles, and 10% on eight-cylinder automobiles. Let X = the
number of cylinders on the next car to be tuned.
a. (2 Marks) What is the pmf (probability mass function) of X?
Solution:
4
6
8
x
f (x )
0.50
0.40
0.10
b. (2 Marks) Find the mean number of cylinders on the car to be tuned up.
4
Solution: E (X ) = 4(0.50) + 6(0.40) + 8(0.10) = 5.2 . To view this one may
consider that the population is
4
4
4
4
4
6
6
6
6
8
c. (3 Marks) Find E (X 2 − 10.4X + 27.04)
Solution:
E (X 2 ) = 42 (0.50) + 62 (0.40) + 82 (0.10) = 28.8,
E (X 2 − 10.4X + 27.04) = 28.8 − 10.4(5.2) + 27.04 = 1.76
4. Automobiles arrive at a vehicle equipment inspection station according to
a Poisson process with an average of 10 cars per hour.
Let X be the autos that arrive in t hours, and λt be the expected no. of
autos arriving in t hours.
a. (3 Marks) What is the probability that exactly seven cars will arrive in the
next hour?
(10 × 1)7 −10×1
e
≈ 0.090079
Solution: P (X = 7) =
7!
b. (2 Marks) what is the probability that at least two cars will arrive in the
next 30 minutes.
Solution:
P (X = 2,3, L)
= 1 − P (X = 0,1) = 1 − P (X = 0) − P (X = 1)
(10 × 0.5)0 −10×0.5 (10 × 0.5)1 −10×0.5
e
e
= 1−
−
0!
1!
= 1 − 6e −5 ≈ 0.959572
5. A college professor always finishes his lectures within 2 minutes after the
bell rings to end the period and to end the lecture. Let X = the time that
elapses between the bell and the end of the lecture and suppose the pdf of
X is
f (x ) = k x 2 , 0 ≤ x ≤ 2
a. (3 Marks) Find the value of the normalizing constant k.
2
Solution: Since 1 = ∫ kx 2 dx = k x 3 / 3 = k (8 / 3 − 0) . So, k = 3 / 8 = 0.375
2
0
0
5
b. (3 Marks) What is the probability that the lecture continues beyond the bell
rings for between 60 and 90 seconds?
1.5
Solution: P (1 < X < 1.5) =
∫ kx
dx = kx 3 / 3
2
1.5
1
1
=
19k
≈ 0.296875
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c. (4 Marks) Find the average time that the lecturer continues beyond the
bell rings.
2
2
0
0
Solution: µ = E (X ) = ∫ xf (x )dx = ∫ x (kx 2 )dx = kx 4 / 4 = 4k = 3 / 2 .
2
0
6. When a new machine is functioning properly, only 3% of the items
produced are defective. Assume that we will randomly select n parts
produced on the machine and that we are interested in the number of
defective parts found.
Solution: Let Y be the number of defective parts in the sample. Then
B (n , p ) = B (n , 0.03) .
a. (4 Marks) What is the probability that at least two parts in a sample of
n = 5 are defective?
Solution:
P (Y = 2,3, L) = 1 − P (Y = 0,1)
= 1 − 0.975 − 5(0.03)(0.97) 4
≈ 0.008472
b. (5 Marks) What is the expected number of defective parts in a sample of
100(= n ) parts? What is the standard deviation of the number of defective
parts in the sample?
Solution:
µ = np = 100(0.03) = 3,
σ = npq = 100(0.03)(0.97) = 2.91 ≈ 1.70587
c. (5 Marks) What is the probability that at most 9 parts in a sample of
200(= n ) parts are defective?
Solution:
Y ~ B (n , p ) ≈ N (np , npq ) = N (6,58.2)
6
P (Y ≤ 9 | binomial, n = 200, p = 0.03)
≈ P (Y ≤ 9.5 | normal, µ =np ,σ 2 =npq )
⎛ Y − 6 9.5 − 6 ⎞
=P⎜
≤
⎟
5.82 ⎠
⎝ 5.82
= P ( Z ≤ 1.4508) ≈ P ( Z ≤ 1.45)
= 0.9265
Note that the probability by binomial mass function is 0.919221 approximately.
7. (6 Marks) A manufacturer manufactures personal computers at two plants,
one in Riyadh and the other in Jeddah. The Riyadh plant has 40
employees; the Jedda plant has 20. A random sample of 10 employees is
to be asked to fill out a benefits questionnaire. What is the probability that
4 employees in the sample work at the Jedda Plant?
Solution: Let X be the number of employees in the sample who worrk in
Jedda. Then
⎛ 20 ⎞ ⎛ 40 ⎞
⎜ ⎟⎜ ⎟
4
6
246050
P (X = 4) = ⎝ ⎠ ⎝ ⎠ =
≈ 0.246663
997513
⎛ 60 ⎞
⎜ ⎟
⎝ 10 ⎠
If somebody by mistake use binomial probability mass function then
n = 10, p = 20 / 60 = 1/ 3, P (X = 4) ≈ 0.2276 .
8. (4 Marks) The lifetime of a particular circuit has an exponential distribution
with a mean of 3 years. What is the probability that the circuit lasts longer
than 4 years?
Solution: Let Y be the lifetime of a circuit. Then
f (y ) =
1
β
e −y /β , 0 < y < ∞
where β = 3 years, and
∞
1
P (Y > 4) = ∫ e − y / 3dy = e −4 / 3 ≈ 0.2636
3
3
9. The fill volume of cans filled by a certain machine is normally distributed
with mean 12.05 oz and standard deviation 0.03 oz.
7
Let X ~ N (12.05, 0.032 ) .
a. (4 Marks) What proportion of the can contains less than 12 oz?
Solution:
⎛ X − 12.05 12 − 12.05 ⎞
P (X < 12) = ⎜
<
⎟ = P ( Z < −1.66667) ≈ 0.0475
0.03 ⎠
⎝ 0.03
b. (3 Marks) If you select 5 cans what is the probability that more than two of
them will have content less than 12 oz?
Solution:
P (Y = 3, 4,5)
= 1 − P (Y = 0,1, 2)
⎡⎛ 5 ⎞
⎤
⎛ 5⎞
⎛ 5⎞
= 1 − ⎢⎜ ⎟ (0.0475)0 0.95255 + ⎜ ⎟ (0.0475)1 0.95254 + ⎜ ⎟ (0.0475) 2 0.95253 ⎥
⎝1⎠
⎝ 2⎠
⎣⎝ 0 ⎠
⎦
= 1 − (0.784016 + 0.19549 + 0.019498) ≈ 0.000997
c. (3 Marks) Suppose that the process mean can be adjusted through
calibration leaving standard deviation the same. That is for the process,
the mean is unknown say µ and standard deviation is σ = 0.03 oz. What
value should the mean be set to so that 99% of the cans will contain 12 oz
or more?
Solution:
P (X ≥ 12) = 0.99,
12 − µ
) = 0.99,
0.03
12 − µ
P (Z <
) = 0.01,
0.03
P (Z ≥
From Table P ( Z < −2.33) = 0.01 so that
12 − µ
= −2.33,
0.03
µ = 12.0699
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3*.A system contains two components A and B. The system will function
only if both components function. The probability that A function is 98%,
the probability that B functions is 0.95, and the probability that either A or
B functions is 0.99. Are the events independent?
Solution: The sample space is given by
S = {A ′B ′, AB ′, A ′B , AB }
P (A ′B ′) = 1 − P (A ∪ B ) = 1 − 0.99 = 0.01 . Given that P (A ) = 0.98, P (B ) = 0.95 and
P (AUB ) = 0.99. Then from
P (A ∪ B ) = P (A ) + P (B ) − P (AB ) we have
0.99 = 0.98 + 0.95 − P (AB ) so that P (AB ) = 0.94
Since P (A )P (B ) = 0.931 is different from P (AB ) = 0.94 , the components are
not independent. Alternatively, since P (A | B ) ≠ P (A ) , we say that the events
are dependent.
The above solution can be visualized better in the following table:
A
A′
B
0.94
0.01
0.95
B′
0.04
0.01
0.05
0.98
0.02
Further Analyses:
Method 1:
P (AB ) 0.94 94
=
=
≈ 0.989,
P (B )
0.95 95
P (A ) = 0.98,
P (AB ′) 0.04
P (A | B ′) =
=
≈ 0.80
P (B ′)
0.05
P (A | B ) =
The Joarder and Al-Sabah inequality is given by
P (A | B ) > P (A ) > P (A | B ′)
If B functions, it is more likely that A functions. The event A depends on B with
dependence coefficient
∆=
P (A | B ) − P (A | B ′) 0.989 − 0.80
=
≈ 0.1929
P (A )
0.98
P (AB ) 0.94 47
=
=
≈ 0.9592,
P (A )
0.98 49
P (B ) = 0.95,
P (BA ′) 0.01
P (B | A ′) =
=
≈ 0.50
P (A ′) 0.02
P (B | A ) =
9
The Joarder and Al-Sabah inequality is given by
P (B | A ) > P (B ) > P (B | A ′)
If A functions, it is more likely that B functions. The event B depends on A with
dependence coefficient
∆=
P (B | A ) − P (B | A ′) 0.9592 − 0.50
=
≈ 0.4834 .
P (B )
0.95
Method 2:
Let us define coprobability of A and B by
Cop (A , B ) = P (AB ) − P (A )P (B ) = 0.94 − 0.98(0.95) = 0.009
which lies between 0 and 1. Then the event A depends on B with dependence
coefficient
δ (A : B ) =
Cop (A , B ) 0.009
=
≈ 0.00947
P (B )
0.950
Similarly, the event B depends on A with dependence coefficient
δ (B : A ) =
Cop (B , A ) 0.009
=
≈ 0.00918
P (A )
0.98
Note that δ (A : B ) = P (A | B ) − P ( A ) =
δ (B : A ) = P (B | A ) − P (B ) =
94
− 0.98 ≈ 0.00947, and
95
47
− 0.95 = 0.00918.
49