Exam
Probability Final and Statistics, GENG200
Spring 2010
SAMPLE SOLUTION
Duration
Start time
Date
th
Sun. 13 June 2010
08:00 am
120 min
INSTRUCTIONS
1. All writings must be on this booklet only.
2. Read and observe the following rules:
a. No candidate shall be permitted to enter the examination
room after the expiration of onehalf hour, or to leave
during the first halfhour of the examination.
b. Candidates are not permitted to ask questions of the
invigilators, except in cases of supposed errors or
ambiguities in examination questions.
Caution Candidates guilty of any of the following, or similar,
dishonest practices shall be immediately dismissed from the
examination and shall be liable to disciplinary action:
Making use of any books, papers or memoranda, cell
phones, calculators, audio or visual cassette players or
other memory aid devices, other than as authorized by the
examiners.
Speaking or communicating with other candidates.
Purposely exposing written papers to the view of other
candidates.
The plea of accident or forgetfulness shall not be received.
3. Show all your work. Justify your answers. Partial credit is
possible for an answer, but only if you show the intermediate
steps in obtaining the answer.
No. of pages
5
Question
(1)
(2)
(3)
(4)
(5)
(6)
(7)
(8)
Total
STUDENT NAME :_________________________________
STUDENT ID
:_________________________________
Answer All the following questions
Mark
/3
/4
/4
Bonus
/4
/4
/5
/6
/
/ 30
Student Name:
Student ID:
Question 1:
[3 Marks]
A lot of 100 electronic components is inspected by choosing a sample of five components. Assume that 10 of the
components do not conform to customer requirements.
a) How many different samples are possible?
b) How many samples of five contains exactly one nonconforming component?
c) How many samples of five contain at least one nonconforming component?
Solution:
2-41.
a) From equation 2-4, the number of samples of size five is
100
( ) = 5!95!! = 7.5288e + 007
100
5
b) There are 10 ways of selecting one nonconforming chip and there are
( ) = 490!! = 2.5552e + 006
!86
90
4
ways of selecting four conforming chips. Therefore, the number of samples that contain exactly one
( )= 2.5552e+007
90
nonconforming chip is 10 4
c) The number of samples that contain at least one nonconforming chip is the total number of samples
( ) minus the number of samples that contain no nonconforming chips ( ) .
140
5
That is
130
5
100
( ) ( ) = 5!95!! 590!! = 3.1338e + 007
!85
100
5
90
5
[4 Marks]
Question 2:
The probability distribution for the random variable x is given by
2
0.1
x
P (x )
3.5
0.2
4
0.3
4.5
5
0.15
a) What is the value of ?
b) The variable x is sampled 10,000 times. How many times would you expect x to have a value between 3.5 and
4.6, inclusive?
c) Calculate the expected value and the standard deviation of x.
d) Calculate the expected value and the standard deviation of the random variable x+4.
Answer:
a)
b)
c)
= 1 - [ 0.1 + 0.2 + 0.3 + 0.15] = 0.25
P(3.5 x 4.6) = 0.2+0.3+0.25=0.75. Expected value = 10000x(0.75) = 7500
E [ x ] = xi P ( xi )
i
E[ x ] = = 2 0.1 + 3.5 0.2 + 4 0.3 + 4.5 0.25 + 5 0.15 = 3.975
[ x] =
(x
i
) 2 P ( xi )
i
d)
E[ x + 4] = ( xi + 4) P( xi ) = xi P( xi ) + 4 = + 4 ,
i
[ x + 4] =
(x + 4 )
i
i
2
P ( xi ) = [ x]
i
Page 2 of 6
Student Name:
Student ID:
Question 3:
[4 Marks]
Let X denote the number of bits received in error in a digital communication channel, and assume that X is a binomial
random variable with p = 0.001. If 1000 bits are transmitted, determine the following:
a) P(X = 1)
b)
1
c)
2
d) Mean and variance of X.
Answer:
3-120.
1000
1
999
= 1) =
1 0.001 (0.999) = 0.3681
1000
999
0
b) P( X 1) = 1 P( X = 0) = 1
0 0.001 (0.999 ) = 0.6319
a) P ( X
1000
1000
1000
1000
999
2
998
0.0010 (0.999 ) +
0.0011 (0.999 ) +
c ) P ( X 2) =
0
1
2 0.001 0.999
= 0.9198
d ) E ( X ) = 1000(0.001) = 1
V ( X ) = 1000(0.001)(0.999) = 0.999
[Bonus]
Question 4:
The function
,
a joint probability density function over the range 0 < x < 3 and x <y < x+2.
Determine the following:
a) The value of c.
b)
1,
2.
c) Expected value of X (
).
d) Variance of X (Var[X]).
e) Conditional probability distribution of Y given that X = 1.
Answer:
5-18.
3 x+2
3
a) c ( x + y )dydx = xy + y
2
2
0x
0
x+2
x
3
[
dx = x( x + 2) + ( x +22) x 2
2
0
]dx = c (4x + 2)dx = [2x + 2x]
3
x2
2
2
0
3
0
= 24c
Therefore, c = 1/24.
b) P(X < 1, Y < 2) equals the integral of f XY ( x, y ) over the following region.
y
2
0
12
x
0
Then,
12
P ( X < 1, Y < 2) =
1
1
1
( x + y)dydx = 24 xy +
24 0 x
0
y2
2
2
3
dx =
x
12
1
2
2 x + 2 32x dx = 24 x + 2 x
24 0
x3
2
= 0.10417
0
1
c)
3 x+2
E( X ) =
1
24
0
1
3
x( x + y)dydx = 24 x
x
0
2
y+
xy 2
2
x+2
x
dx =
3
3
1
1 4x 3
15
2
2
x (4 + 2 x)dx = 24 3 + x 0 = 8
24 0
Page 3 of 6
Student Name:
Student ID:
d)
3 x+2
V (X ) =
=
1
24
0
x
2
3
1
15
x 2 ( x + y )dydx =
x3 y +
24
8
0
x2 y2
2
x+2
x
2
3
x4
1
15
15
3
2
dx =
(3x + 4 x + 4 x 4 )dx 8
24 0
8
2
2
x 15
1 3x
4x
31707
+
+ 2x2
=
24 4
3
20 0 8
320
4
e) f Y 1 ( y ) =
3
f XY (1, y )
f X (1)
53
=
1
(1+ y )
24
11
+
6 12
=
1+ y
6
for 1 < y < 3.
See the following graph,
y
f
2
Y|1
(y) defined over this line segment
1
0
12
0
x
[4 Marks]
Question 5:
The amount of time that a customer spends waiting at an airport check-in counter is a random variable with mean 8.2
minutes and a standard deviation 1.5 minutes. Suppose that a random sample of size n = 49 customers is observed.
Find the probability that the average time waiting in line for these customers is:
a) Less than 10 minutes.
b) Between 5 and 10 minutes.
Answer:
7-10
X = 8.2
n = 49
minutes
X = 15
.
minutes
X =
X
n
=
15
.
= 0.2143
49
X = X = 8.2 mins
Using the central limit theorem, X is approximately normally distributed.
a) P ( X < 10) = P ( Z < 10 8.2 ) = P ( Z < 8.4) = 1
0.2143
b)
P (5 < X < 10) =
5 8 .2
P ( 0.2143 <
Z<
10 8.2
)
0.2143
= P ( Z < 8.4) P ( Z < 14.932) = 1 0 = 1
Page 4 of 6
Student Name:
Student ID:
Question 6:
[4 Marks]
/ is a biased estimator of .
a) Show that
b) Find the amount of bias in the estimator
c) What happens to the bias as the sample size n increases?
Answer:
The amount of time that a customer spends waiting at an airport check-in counter is a random variable with mean 8.2
minutes
2
(X i X )
n
7-20
Show that
i =1
n
is a biased estimator of 2 :
a)
n
2
(X i X )
n
n
n
2
= 1 E ( X nX )2 = 1 E X 2 nE X 2 = 1
2 + 2 n 2 +
E i=1
i
i
i=1
n i=1
n
n
n i =1
n
2
1
1
= n 2 + n 2 n 2 2 = ((n 1) 2 ) = 2
n
n
n
()
(
()
(
)
)
(X X ) is a biased estimator of 2 .
i
2
n
b) Bias
(
X 2 nX
= E i
n
)
2
2
=2
2
n
2 =
2
n
c) Bias decreases as n increases.
Question 7:
[5 Marks]
The manufacturer produces piston rings for automobile engine. It is known that ring diameter is normally distributed
with = 0.001 millimeters. A random sample of 15 rings has a mean diameter of
74.036 millimeters.
a) Construct a 99% two-sided confidence interval on the mean piston ring diameter.
b) Construct a 99% lower-confidence bound on the mean piston ring diameter.
c) Compare the lower bound of this confidence interval in (b) with the one in part (a).
Answer:
8-11
a)
99% Two-sided CI on the true mean piston ring diameter
For = 0.01, z/2 = z0.005 = 2.58 , and x = 74.036, = 0.001, n=15
x z0.005
x + z0.005
n
n
b)
0.001
0.001
74.036 2.58
74.036 + 2.58
15
15
74.0353 74.0367
99% One-sided CI on the true mean piston ring diameter
For = 0.01, z = z0.01 =2.33 and x = 74.036, = 0.001, n=15
x z 0.01
n
0.001
74.036 2.33
15
74.0354
c)
The lower bound of the one sided confidence interval is less than the lower bound of the two-sided
confidence. This is because the Type I probability of 99% one sided confidence interval (or = 0.01)
in the left tail (or in the lower bound) is greater than Type I probability of 99% two-sided confidence
interval (or /2 = 0.005) in the left tail.
Page 5 of 6
Student Name:
Student ID:
Question 8:
[6 Marks]
The life in hours of a battery is known to be approximately normally distributed, with standard deviation = 1.25
hours. A random sample of 10 batteries has a mean life of
40.5 hours.
a) Is there evidence to support the claim that battery life exceeds 40 hours? Use = 0.05.
b) What is the P-value for the test in part (a)?
c) What is the -error for the test in part (a) if the true mean life is 42 hours?
d) What sample size would be required to ensure that does not exceed 0.10 if the true mean life is 44 hours?
e) Explain how you could answer the question in part (a) by calculating an appropriate confidence bound on life.
Solution:
9-39
a.) 1) The parameter of interest is the true mean battery life in hours, .
2) H0 : = 40
3) H1 : > 40
4) = 0.05
x
5) z0 =
/ n
6) Reject H0 if z0 > z where z0.05 = 1.65
7) x = 40.5 , = 1.25
z0 =
40.5 40
1.25 / 10
= 1.26
8) Since 1.26 < 1.65 do not reject H0 and conclude the battery life is not significantly different greater than 40 at = 0.05.
1 (1.26) = 1 0.8962 = 0.1038
40 42
c) = z 0.05 +
1.25 / 10
b) P-value =
= (1.65 + 5.06)
= (-3.41)
0.000325
d)
n=
(z
+ z ) 2
2
2
=
(z 0.05 + z 0.10 )2 2
( 40 44) 2
=
(1.65 + 1.29) 2 (1.25) 2
= 0.844, n 1
( 4) 2
e) 95% Confidence Interval
x + z 0 .05 / n
40 . 5 + 1 . 65 (1 . 25 ) / 10
39 . 85
The lower bound of the 90 % confidence interval must be greater than 40 to verify that the true mean exceeds 40 hours.
Page 6 of 6