Download MS125: Probability Assignment 2

MS125: Probability
Assignment 2
To be submitted by 25th of April 2008
Question 1
The random variable x has the following discrete probability distribution:
x
P (X = x)
1
0.1
3
0.2
5
0.4
7
0.2
9
0.1
1. Write down cumulative probability function.
2. Calculate E[X].
3. Calculate V ar[X].
Solution
Cumulative probability function:








F (x) =







0,
0.1,
0.3,
0.7,
0.9,
1,
x < 1;
1 ≤ x < 3;
3 ≤ x < 5;
5 ≤ x < 7;
7 ≤ x < 9;
x ≥ 9.
E[X] = 1 ∗ 0.1 + 3 ∗ 0.2 + 5 ∗ 0.4 + 7 ∗ 0.2 + 9 ∗ 0.1 = 5
E[X 2 ] = 12 ∗ 0.1 + 32 ∗ 0.2 + 52 ∗ 0.4 + 72 ∗ 0.2 + 92 ∗ 0.1 = 29.8
V ar[X] = E[X 2 ] − E[X]2 = 29.8 − 25 = 4.8
Question 2
A machine that produces stampings for automobile engines is malfunctioning
and producing 10% defectives. The defective and nondefective stampings proceed from the machine in a random manner. If the next five stampings are
tested, find the probability that three of them are defective.
1
Solution
Let X be number of defective machines.
X ∼ B(5, 0.1) P (X = 3) = 53 0.13 0.92 = 10 ∗ 0.13 ∗ 0.92 = 0.0081
Question 3
Suppose the number, x, of a company’s employees who are absent on Mondays
has a Poisson probability distribution. Furthermore, assume that the average
number of Monday absentees is 2.6.
1. Find the mean and standard deviation of x, the number of employees
absent on Monday.
2. Find the probability that fewer than two employees are absent on a given
Monday.
3. Find the probability that more than three employees are absent on a given
Monday.
4. Find the probability that exactly four employees are absent on a given
Monday.
Solution
x ∼ P (2.6)
1. E[x] = 2.6
V ar[X] = 2.6,
p
√
Standard deviation: V ar[X] = 2.6 = 1.6125
2. P (x < 2) = P (x ≤ 1) = P (x = 0) + P (x = 1) =
0.0743 + 0.1931 = 0.2674
2.60 e−2.6
0!
+
2.60 e−2.6
0!
=
3. P (x > 3) = 1 − P (x ≤ 3) = 1 − (P (x = 0) + P (x = 1) + P (x = 2) + P (x =
1 −2.6
2 −2.6
3 −2.6
0 −2.6
e
e
e
e
+ 2.6 1!
+ 2.6 2!
+ 2.6 3!
) = 1−(0.0743+0.1931+
3)) = 1−( 2.6 0!
0.2510 + 0.2176) = 0.264
4. P (x = 4) =
2.64 e−2.6
4!
= 0.1414.
Question 4
Suppose the research department of a steel manufacturer believes that one of
the company’s rolling machines is producing sheets of steel of varying thickness.
The thickness is a discrete uniform random variable with values between 150
and 200 millimeters. Any sheets less than 160 millimeters must be scrapped
because they are unacceptable to buyers. Calculate the fraction of steel sheets
produced by this machine that have to be scrapped.
2
Solution
Let X be the thickness
of sheet. P
P159
159
P (X < 160) = i=150 P (X = i) = i=150
1
200−150
=
1
50
P159
i=150
1=
10
50
= 0.2
Question 5
Assume the length of time, X, between charges of a mobile phone is normally
distributed with a mean of 10 hours and a standard deviation of 1.5 hours. Find
the probability that the cell phone will last between 8 and 12 hours between
charges.
Solution
X ∼ N (10, 1.52 )
Z = X−10
1.5 ∼ N (0, 1)
12−10
P (8 < X < 12) = P ( 8−10
1.5 < Z < 1/5 ) = P (−1.33 < Z < 1.33) = P (Z <
1.33) − P (Z < −1.33) = 2 ∗ P (Z < 1.33) − 1 = 0.8164
Question 6
A fair dice is thrown three times. The result of first throw is scored as X1 = 1
if the dice shows 6 and X1 = 0 otherwise; X2 and X3 are scored likewise for the
second and third throws.
Let Y1 = X1 + X2 and Y2 = X1 − X3 .
1. Calculate the joint probabilities in the bivariate distribution of the pair
(Y1 , Y2 ) and display in an appropriate table.
2. Find the marginal probability distributions of Y1 and Y2 .
3. Calculate the means and variances of Y1 and Y2 .
4. Calculate the covariance of Y1 and Y2 .
5. Find the conditional distribution of Y1 given Y2 = 0.
6. Find the conditional mean of Y1 given Y2 = 0.
7. State with justification if Y1 and Y2 are independent.
Solution
1. P (X1 = 1) = P (X2 = 1) = P (X3 = 1) =
P (X1 = 0) = P (X2 = 0) = P (X3 = 0) =
3
1
6
5
6
X1
0
0
0
0
1
1
1
1
Y2
X2
0
0
1
1
0
0
1
1
X3
0
1
0
1
0
1
0
1
Y1 = X1 + X2
0
0
1
1
1
1
2
2
0
Y1
1
2
P
-1
25
216
5
216
0
30
216
0
125
216
30
216
1
216
156
216
1
0
25
216
5
216
30
216
P
150
216
60
216
6
216
1
Y2 = X1 − X3
0
-1
0
-1
1
0
1
0
5
6
5
6
5
6
5
6
1
6
1
6
1
6
1
6
×
×
×
×
×
×
×
×
5
6
5
6
1
6
1
6
5
6
5
6
1
6
1
6
P
× 56
× 16
× 56
× 16
× 56
× 16
× 56
× 16
=
=
=
=
=
=
=
=
125
216
25
216
25
216
5
216
25
216
5
216
5
216
1
216
2. Marginal probability distribution of Y1 :
y1
0
1
2
60
6
P (Y1 = y1 ) 150
216
216
216
Marginal probability distribution of Y2 :
y2
-1
0
1
P (Y2 = y2 )
30
216
156
216
30
216
3.
E[Y1 ] =
E[Y12 ] =
V ar[Y1 ] =
E[Y2 ] =
E[Y22 ] =
V ar[Y2 ] =
150
60
6
1
+1×
+2×
=
216
216
216
3
60
6
7
150
+ 12 ×
+ 22 ×
=
02 ×
216
216
216
18
1
5
7
− =
E[Y12 ] − (E[Y1 ])2 =
18 9
18
30
156
30
(−1) ×
+0×
+1×
=0
216
216
216
156
30
5
30
+0×
+ 12 ×
=
(−1)2 ×
216
216
216
18
5
E[Y22 ] − (E[Y2 ])2 =
18
0×
25
5
+ 1 × 216
+2×
4. Cov[Y1 , Y2 ] = E[Y1 Y2 ] − E[Y1 ]E[Y2 ] = E[Y1 Y2 ] = 1(−1) 216
5
5
1 × 216 = 36
4
5.
P (Y1 = 0|Y2 = 0)
=
P (Y1 = 1|Y2 = 0)
=
P (Y1 = 2|Y2 = 0)
=
125/216
125
P (Y1 = 0, Y2 = 0)
=
=
P (Y2 = 0)
156/216
156
P (Y1 = 1, Y2 = 0)
30/216
30
5
=
=
=
P (Y2 = 0)
156/216
156
26
P (Y1 = 2, Y2 = 0)
1/216
1
=
=
P (Y2 = 0)
156/216
156
6.
E[Y1 |Y2 = 0] = 0 + 1 ×
1
8
5
+2×
=
26
156
39
7. Y1 and Y2 are not independent because, for example, P (Y1 = 0, Y2 = 1) =
0 6= P (Y1 = 0)P (Y2 = 1)
Question 7
x1 , . . . , xn is a random sample of size n from random variable with probability
density function:
f (x) = (1 + θ)xθ ,
0<x<1
Find θ̂, the maximum likelihood estimator of θ.
Solution
Qn
Qn
Qn
Qn
θ
θ
M L = L(θ) = i=1 (1 + θ)xθi = ( Qi=1 (1 + θ))
(1 + θ)n ( i=1 xi )
i=1 xi = P
n
n
l(θ) = ln L(θ) = n ln(1 + θ) + θ ln i=1 xi = n ln(1 + θ) + θ i=1 ln xi
Pn
dl(θ)
n
+ i=1 ln xi = 0
dθ = 1+θ
P
n
n
i=1 ln xi
1+θ = −
θ̂ = − Pn n ln x − 1 = −1/ln¯x − 1
i=1
i
Question 8
A manufacturer of cereal wants to test the performance of one of its filling
machines. The machine is designed to discharge a mean amount of 12 ounces
per box, and the manufacturer wants to detect any departure from this setting.
(i) He randomly samples 100 boxes from production run and obtains that the
sample mean fill of 100 boxes is x̄ = 11.5 and standard deviation is s2 = 6.25.
Test the null hypothesis H0 : µ = 12 against the alternative hypothesis H0 :
µ 6= 12. (ii) The next day the manufacture samples 10 boxes and gets x̄ = 11.6
and s2 = 1.6. Test the null hypothesis H0 : µ = 12 against the alternative
hypothesis H0 : µ 6= 12.
5
Solution
(i) Single sample with unknown variance and large sample size.
H0 : µ = 12
H0 : µ 6= 12
Two-tailed test.
Test statistics:
s2 =√6.25
s = 6.25 = 2.5
x̄−µ
√ = 11.5−12
√
= −2
Z = s/
n
2.5/ 100
p = P (|Z| > 2) = 2 ∗ P (Z > 2) = 2 ∗ (1 − P (Z < 2)) = 2 ∗ (1 − 0.9772) = 0.0456
Moderate evidence against the null hypothesis that mean is equal to 12.
(ii) Single sample with unknown variance and small sample size.
H0 : µ = 12
H1 : µ 6= 12
s2 =√0.16
s = 0.16 = 0.4
Test statistics:
x̄−12
√ = 11.6−12
√
= −3.16
T = s/
n
0.4/ 10
p = P (|T9 | > 3.16) = 2∗P (T9 > 3.16) = 2∗(1−P (T9 < 3.16)) = 2∗(1−0.9946) =
0.0108
There is a moderate evidence against the null hypothesis.
Question 9
An economist decided to test the hypothesis that higher retail prices are being
charged for Japanese automobiles in Japan than in the United States. She
obtained random samples of 50 retail sales in the United States and 30 retail
sales in Japan over the same time period and for the same model of automobile,
converted the Japanese sales prices from yen to dollars using current conversion
rates, and obtained the summary information shown in Table below.
Sample size
Sample mean
Sample standard deviation
USA Sales
50
$ 16,545
$ 1,989
Japan Sales
30
$ 17,243
$ 1,843
Test the null hypothesis H0 : µU = µJ against the alternative H0 : µU > µJ .
Solution
Two samples with unknown variances and large sample sizes.
H0 : µU = µJ
H1 : µU > µJ
One-tailed test.
Test statistics:
= √ 16545−17243
Z = p 2x̄U −x̄J 2
= −1.59
2
2
Su /nU +Sj /nJ
1989 /50+1843 /30
p = P (|Z| > 1.59) = 1 − P (Z < 1.59) = 1 − 0.9441 = 0.0559
6
There is a weak evidence against the null hypothesis, mean retail prices are
equal.
Question 10
In order to create a behavioral profile of travelers, agency interviewed 5,026
travelers in the Tampa Bay region. Two of the characteristics they investigated
were the travelers’ education level and their use of the Internet to seek travel
information. The table below summarizes the results of the interviews.
College Degree or More
Less than a College Degree
USE INTERNET
Yes
No
1,072
1,287
640
2,027
Test if travelers who use the Internet to search for travel information are
likely to be people who are college educated.
Solution
H0 : The Internet use and the education are independent.
H1 : The Internet use and the education are not independent.
Observed frequencies:
USE INTERNET
Yes
No
1072
1287
640
2027
1712
3314
College Degree or More
Less than a College Degree
Total
Total
2359
2667
5026
Expected frequencies:
College Degree or More
Less than a College Degree
USE INTERNET
Yes
No
803.54
1555.46
908.47
1758.54
Test statistics:
2
2
2
2
2
P4
i)
χ2 = i=1 (Oi −E
= (1072−803.54)
+ (1287−1555.46)
+ (640−908.47)
+ (2027−1758.54)
=
Ei
803.54
1555.46
908.47
1758.54
256.35
ν = (2 − 1)(2 − 1) = 1
p = P (χ21 > 256.35)
P (χ21 > 12.12) = 0.0005
p < 0.0005
There a strong evidence against null hypothesis. The Internet use and education
are not independent.
7