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Department of Mechanical Engineering
05.06.2013
Statistics and Probability Final Exam Solutions
Name:
Question
Point
ID:
1
2
3
4
5
Score
Please read before starting
1-It is strictly forbidden to pass any table, formula sheet or calculator to other students
during the exam.
2-You may use the tables, the formula sheet and a calculator.
3-You have 100 minutes to complete the exam.
QUESTION 1: X and Y are the continuous random variables with the joint density function
as given below:
𝑥(1 + 3𝑦 2 )
,
𝑓𝑋𝑌 (𝑥, 𝑦) = {
4
0,
0 ≤ 𝑥 < 2 𝑎𝑛𝑑 0 ≤ 𝑦 < 1
𝑒𝑙𝑠𝑒𝑤ℎ𝑒𝑟𝑒
a) Find the marginal density functions 𝑓𝑋 (𝑥) and 𝑓𝑌 (𝑦) (10 Points)
b) Compute the probability given as 𝑃(0.25 < 𝑋 < 0.5| 𝑌 = 0.5) (10 Points)
ANSWER 1:
a) Marginal distribution for the random variable x can be obtained from:
+∞
1
1
𝑥(1 + 3𝑦 2 )
𝑥𝑦
3𝑦 3
𝑥
𝑓𝑋 (𝑥) = ∫ 𝑓𝑋𝑌 (𝑥, 𝑦) 𝑑𝑦 = ∫
𝑑𝑦 =
+
| =
4
4 4×3 0 2
−∞
0
Similarly, marginal distribution for the random variable y can be obtained from:
+∞
2
2
𝑥(1 + 3𝑦 2 )
1 4 × 3𝑦 2
1 + 3𝑦 2
𝑓𝑌 (𝑦) = ∫ 𝑓𝑋𝑌 (𝑥, 𝑦) 𝑑𝑥 = ∫
𝑑𝑥 = +
| =
4
2
4×2 0
2
−∞
0
b) The required probability
0.5
𝑃(0.25 < 𝑋 < 0.5| 𝑌 = 0.5) = ∫ 𝑓𝑋|𝑦 (𝑥|𝑌 = 0.5) 𝑑𝑥
0.25
So the conditional probability function must be obtained as follow
𝑓𝑋𝑌 (𝑥, 𝑦)
𝑓𝑋|𝑦 (𝑥) =
=
𝑓𝑌 (𝑦)
𝑥(1+3𝑦 2 )
4
1+3𝑦 2
=
2
𝑥
2
Thus,
0.5
0.5
0.25
0.25
0.5
𝑥
𝑥2
𝑃(0.25 < 𝑋 < 0.5| 𝑌 = 0.5) = ∫ 𝑓𝑋|𝑦 (𝑥|𝑌 = 0.5) 𝑑𝑥 = ∫ 𝑑𝑥 = |
= 0.0468
2
4 0.25
QUESTION 2: To determine the life of a machine component, a sample of 10 machine
components are taken from a production line, and the life of the components are found to be
[2.00 2.5 1.85 2.25 1.95 1.85 2.10 2.05 2.15 1.90].
Based on the sampling data, find a 95% confidence interval for the mean life of the machine
component. (30 Points)
ANSWER 2: since the population variance is unknown and n<30, t-distribution is used:
̅ − 𝐭𝛂
̅ + 𝐭𝛂
𝐗
̅ <𝝁 <𝐗
̅
,𝐧−𝟏 𝐒𝑿
,𝐧−𝟏 𝐒𝑿
𝟐
𝟐
=0.05; SD=n-1=10-1=9; t 0.0𝟐𝟓;𝟏𝟗 = 2.262 is read from the t-table.
The mean and standard deviation of the samples can be computed from the sample given as:
2
𝟐
𝑆
0.202
̅
X = 𝟐. 𝟎𝟔 ve 𝑆𝑋̅ = √ 𝑛 = √ 9 = 𝟎. 𝟎𝟔𝟕 →
𝟐. 𝟎𝟔 − 2.262 × 𝟎. 𝟎𝟔𝟕 < 𝜇 < 𝟐. 𝟎𝟔 + 2.262 × 𝟎. 𝟎𝟔𝟕
1.908 < 𝜇 < 2.211
QUESTION 3: The diameter of a shaft used in a motor is supposed to be 20 mm. If the shaft
diameter is either too small or too large, the motor does not perform properly. Therefore, the
manufacturer takes a sample of 25 shafts, and finds to have sample mean 20.01 mm and sample
variance 0.01. With the significant level of  = 0.05, can we say that the shaft conforms the
specification? (30 Points)
ANSWER 3:
Step 1: 𝐇𝟎 : 𝛍 = 𝟐𝟎
𝐇𝟏 : 𝛍 ≠ 𝟐𝟎
Step 2: since n=25<30 we need to use t-table. The test is two-sided test thus 𝐭 𝜶,𝒏−𝟏 =
𝟐
𝐭 𝟎.𝟎𝟐𝟓,𝟐𝟒 = ±𝟐. 𝟎𝟔𝟒
Step 3: The test statistics:
𝐭𝐡 =
̅−𝛍
𝐗
𝐒𝑿
̅
=
𝟐𝟎.𝟎𝟏−𝟐𝟎
√0.01/𝟐𝟓
= 𝟎. 𝟓
Step 4: compare the test statistics with the t-value as:
| 𝐭 𝐡 = 𝟎. 𝟓| < |𝐭 𝟎.𝟎𝟐𝟓,𝟐𝟒 = 𝟐. 𝟎𝟔𝟒| thus 𝐇𝟎 can be accepted.
Step 5: At the confidence level of 95%, we can say that the mean diameter of the machine
component is equal to 20 mm; thus it conform the specifications.
QUESTION 4: The following table presents data on the fatigue strength value of mild steel
corresponding to the number of cycles. Representative data follow, with x: the number of cycles
and y: wear volume.
y
350
320
300
280
270
x
10000 20000 30000 40000 50000
(a) Fit the simple linear regression model using least squares (15 Points).
(b) Comment on how good the model is (10 Points).
(c) Predict fatigue strength corresponding to x =25000 cycles (5 Points).
ANSWER 3:
Fatigue strength (y)
Number of cycles (x)
xy
x2
y2
350
10000
3500000
100000000
122500
320
20000
6400000
4E+08
102400
300
30000
9000000
9E+08
90000
280
40000
11200000
1.6E+09
78400
270
50000
13500000
2.5E+09
72900
a) The coefficient of a linear model can be found from:
∑ 𝐘 = 𝐧𝐚 + 𝐛 ∑ 𝐗 and ∑(𝐗𝐘) = 𝐚 ∑ 𝐗 + 𝐛 ∑ 𝐗 𝟐
𝟏𝟓𝟐𝟎 = 𝟓𝒂 + 𝟏𝟓𝟎𝟎𝟎𝟎𝒃 and 𝟒𝟑𝟔𝟎𝟎𝟎𝟎𝟎 = 𝟏𝟓𝟎𝟎𝟎𝟎𝒂 + 𝟓. 𝟓𝑬 + 𝟎𝟗𝒃
After solving these two equations: 𝐚 = 𝟑𝟔𝟒 and 𝐛 = −𝟎. 𝟎𝟎𝟐,
thus the model is 𝐲̂ = 𝟑𝟔𝟒 − 𝟎. 𝟎𝟎𝟐𝐱.
b) The goodness of fit can be determined from the following expression:
𝑹𝟐 = 𝟎. 𝟗𝟕 indicates that the model represents the data quite well.
c) for x=25000 , the predicted response is
𝐲̂ = 𝟑𝟔𝟒 − 𝟎. 𝟎𝟎𝟐𝐱 = 𝟑𝟔𝟒 − 𝟎. 𝟎𝟐 × 𝟐𝟓𝟎𝟎𝟎 = 𝟑𝟏𝟒
Sum
1520
150000
43600000
5.5E+09
466200
QUESTION 5:
A class with 45 students takes statistics lecture and the mean and standard deviation of the final
grade of the class is 45 and 15, respectively. Find out what grade a student takes if his/her final
grade is 65.
Class level
Perfect
Excellence
Very good
Good
Upper
intermediate
Intermediate
Weak
Bad
Mean of the class
grade
Lower Upper
bound bound
80,00
100,00
70,00
79,99
62.50
69,99
57.50
62.49
52.50
57.49
47.50
42.50
0
52.49
47.49
42.49
Letter grade depending on the lower bound of T-value
AA
BA
BB
CB
CC
DC
DD
57
59
61
63
65
52
54
56
58
60
47
49
51
53
55
42
44
46
48
50
37
39
41
43
45
32
34
36
38
40
27
29
31
33
35
67
69
71
62
64
66
57
59
61
52
54
56
47
49
51
42
44
47
37
40
43
ANSWER 5:
First, t-score for the student must be calculated from
𝑇=
𝑋 − 𝑋̅
65 − 45
× 10 + 50 =
× 10 + 50 = 63.33
𝑆
15
Since the mean of the class grade is 45, the level of the class is Weak. Thus, the score is above
the lower limit of BB, thus the student takes letter grade of BB.