SCHOOL OF ENGINEERING

ENGINEERING MATHEMATICS 2B

SCEE08010


Exam Date: 10/05/2019 From and To: 14:30-16:00 Exam Diet: May 2019
Please read full instructions before commencing writing

Exam paper information
 This exam paper consists of TWO sections.
 SECTION A: TEN multiple choice questions. Candidates should attempt ALL TEN
questions. Answers should be given on the multiple choice question sheet provided.
 SECTION B: TWO long form questions. Candidates should attempt BOTH questions.

Special instructions
 Do not remove this exam paper from the exam hall. Your exam paper must be
returned with your script book.
 This is an open book exam. Students are permitted to bring the following textbooks:
Glyn James: Modern Engineering Mathematics and Glyn James: Advanced Modern
Engineering Mathematics. Students are also permitted two A4 sides of their own
notes.
 Students should assume reasonable values for any data not given in a question nor
available on a datasheet, and should make any such assumptions clear on their script.
 Students in any doubt as to the interpretation of the wording of a question, should
make their own decision, and should state it clearly on their script.
 Please write your name in the space indicated at the right hand side on the front cover
of the answer book. Also enter your examination number in the appropriate space on
the front cover.
 Write ONLY your examination number on any extra sheets or worksheets used and
firmly attach these to the answer book(s).
 This examination will be marked anonymously.

Special items
 MCQ sheet, pencil, rubber, Data Sheet (2 pages)

Convenor of Board of Examiners: Mr S Warrington
External Examiner: Professor M Cartmell
Section A
Answer all multiple-choice questions on the machine marking form provided. You may use the
script book provided for any rough working.
Question 1
The are 3 children in a family. Assuming that the probability of a child been born a boy or
a girl is exactly the same, what is the probability that all children in this family are girls,
given that the first born child is a girl?
A. 0.5
B. 0.75
C. 0.33
D. 0.25
E. 0.66 (5)
Question 2
Let X be a continuous random variable with probability density function fX(x) = λe
−λx
for x ≥ 0. If Y = X2 what is the probability density function fY (y) of Y ?
A. fY (y) = 1− e−λ
√
y
B. fY (y) =
λ
2y
1
2 e−λ
√
y
C. fY (y) = 1− e−λy2
D. fY (y) =
λ
2y
− 1
2 e−λ
√
y
E. fY (y) = 1− λ2y−
1
2 e−λ
√
y (5)
Question 3
Let X and Y be random variables with means µX = µY = 0 and standard deviations
σX = 2 and σY = 10. Define the random variable Z = 2X + Y . If the covariance between
X and Z is Cov(X,Z) = −4, then what is the variance Var(X − Y )?
A. 128
B. 86
C. 116
D. 99
E. 22 (5)
Question 4
Let X be a random variable with Poisson distribution with expectation λ and further let
Y = (X + 2)(X + 1). What is the value of the expectation of Y ?
A. λ2 + 3λ+ 1
B. λ2 + 3λ+ 2
C. λ2 + 4λ+ 2
D. 3λ2 + 3λ+ 2
E. 4λ2 + 4λ+ 2
SCEE08010 Engineering Mathematics 2B - May 2019
(5)
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Question 5
Suppose you are taking a multiple-choice test with 5 choices for each question. In answering
a question on this test, the probability that you know the answer is p. If you don’t know the
answer, you can choose one at random. What is the probability that you knew the answer
to a question, given that you answered it correctly?
(5)
A. 5p4p+1
B. p+ 15(1− p)
C. 15(1− p)
D. p4p+1
E. p5p+1
Question 6
Suppose that the weight in pounds of a British adult can be represented by a normal random
variable with mean 150 lbs and variance 900 lb2. A lift containing a sign “maximum capacity
12 people” can safely carry 2000 lbs. The probability that 12 British people will overload
the lift is close to
(5)
A. 0.5499
B. 0.9729
C. 0.0271
D. less than 0.0001
E. more than 0.9999
Question 7
A machine is used to manufacture gears for the transmission system of a car. After pro-
cessing, gears are classified based on their characteristics in “Excellent”, “Good” and “Ac-
ceptable”. Once mounted, “Excellent” gears have a 1% probability to fail, “Good” and
“Acceptable” gears fail in 5% and 10% of the cases respectively. An engineer runs a test
and finds out that, out of 1000 gears, 300 were “Excellent”, 500 were “Good” and 200 were
“Acceptable”. How many of the 1000 gears do you expect to fail?
A. 131
B. 48
C. 90
D. 11
E. 910 (5)
Question 8
Nova Rockets Ltd, a company that specialises in re-usable space rockets, developed a rocket
composed of 8 first-stage boosters. A booster is re-usable if, after launch, it lands intact
on earth. After extensive tests, engineers at Nova Rockets estimate that each booster has
a 60% probability of landing intact. The company is planning two sequential launches and
has a total of 12 boosters. Given that (a) each launch requires 8 boosters and (b) there
SCEE08010 Engineering Mathematics 2B - May 2019
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is no time to build new boosters for the second launch, you have been asked to calculate
the probability that Nova Rocket will be able to launch the second rocket. Which of the
following R statements gives you the correct answer?
A. > 1 - pbinom(4,8,0.6)
[1] 0.4511456
B. > 1 - pbinom(3,8,0.6)
[1] 0.8263296
C. > dbinom(4,8,0.6)
[1] 0.2322432
D. > pbinom(4,8,0.6)
[1] 0.9718061
E. > 1 - pbinom(4,8,0.6)
[1] 0.02819386
(5)
Question 9
As an engineer at McCarrie Ltd, a fast-food chain, you have been asked to check and report
on the ice-cream dispensing machine in a restaurant in Glasgow. On site you obtained 15
cups of ice cream and weighed them. Their average weight was 70 g and the standard
deviation of the samples was 8.6 g. Using Student t-distribution you found that:
> qt(0.95, df = 13:16)
[1] 1.770933 1.761310 1.753050 1.745884
> qt(0.975, df = 13:16)
[1] 2.160369 2.144787 2.131450 2.119905
Which of the following is the 95% confidence interval for the weight of the ice cream cups
you will report?
A. 65.12 < µ < 72.19
B. 66.54 < µ < 76.51
C. 65.24 < µ < 74.76
D. 64.09 < µ < 76.30
E. 66.54 < µ < 72.50 (5)
Question 10
Edison, a car manufacturer specialised in electric vehicles, is testing a new prototype, Model
Z. The car is brought to cruising speed and its position (in metres) is tracked at intervals
of 1 s. The test is a success if the car reached a speed of, at least 110 km/h. You have been
tasked to determining whether this is the case. To do so you collect the time vector t (in
seconds) and the position d (in metres) of the car. As you know there is a linear relationship
between the distance d and the time t, you set out to find the speed the car reached by
running a linear regression (model d = aˆt+ bˆ) with the following code:
> r<-lm(d~t)
> summary(r)
Call:
SCEE08010 Engineering Mathematics 2B - May 2019
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lm(formula = d ~ t)
Residuals:
Min 1Q Median 3Q Max
-0.225621 -0.082933 -0.003759 0.083387 0.287022
Coefficients:
Estimate Std. Error t value Pr(>|t|)
(Intercept) 2.127e+02 2.072e-02 10263 <2e-16 ***
t 3.330e+01 3.563e-04 93467 <2e-16 ***
---
Signif. codes: 0 ‘***’ 0.001 ’**’ 0.01 ’*’ 0.05 ’.’ 0.1 ’ ’ 1
Residual standard error: 0.1028 on 98 degrees of freedom
Multiple R-squared: 1,Adjusted R-squared: 1
F-statistic: 8.736e+09 on 1 and 98 DF, p-value: < 2.2e-16
You then obtained the following diagnostic plots
500 1000 1500 2000 2500 3000 3500
−
0.
2
−
0.
1
0.
0
0.
1
0.
2
0.
3
Fitted values
R
es
id
ua
ls
Residuals vs Fitted
70
7349
−2 −1 0 1 2
−
2
−
1
0
1
2
3
Theoretical Quantiles
St
an
da
rd
ize
d
re
sid
ua
ls
Normal Q−Q
70
73 49
500 1000 1500 2000 2500 3000 3500
0.
0
0.
5
1.
0
1.
5
Fitted values
St
an
da
rd
ize
d
re
sid
ua
ls
Scale−Location
70
7349
0.00 0.01 0.02 0.03 0.04
−
2
−
1
0
1
2
3
Leverage
St
an
da
rd
ize
d
re
sid
ua
ls
Cook's distance
Residuals vs Leverage
70
12
73
Which of the following statements is true?
A. The diagnostic plots allow to rule out that the residuals are homoscedastic. Since
there are no issues with the remaining plots and the regression is significant at
the 0.1% level the regression line, which explains 87% of the variation in the data,
is
t = 207.2 + 356x.
B. The diagnostic plots clearly show that the noise is not homoscedastic, therefore
no additional conclusion can be drawn.
C. The diagnostic plots seem to indicate that the noise is homoscedastic. Since there
are no issues with the remaining plots and the regression is statistically significant
SCEE08010 Engineering Mathematics 2B - May 2019
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(below the 0.1% level) the test was successful as the car achieved a speed of almost
120 km/h.
D. The diagnostic plots seem to indicate that the noise is homoscedastic. However the
“Residuals vs Leverage” plot shows that there is a major issue with the residuals
and therefore no additional conclusion can be drawn.
E. The diagnostic plots show that the residuals are heteroscedastic. Since there are
no issues with the remaining plots and the regression is significant at the 0.1%
level the regression line, which explains 87% of the variation in the data, is
t = 8.7780 + 2.3501x. (5)
SCEE08010 Engineering Mathematics 2B - May 2019
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Section B
Answer all questions in the script book provided, marking clearly the number of the question.
Question 1
(a) For f(x, y, z) = x2y + y2z + z2x find ∇f at the point (3,2,0) and the derivative there
along the direction of the vector u = iˆ+ 2jˆ+ 2kˆ. (5)
(b) Use Green’s theorem to calculate the work of F = y2iˆ + x2jˆ on a path formed by a
triangle with vertices at (0, 0), (0, 1) and (−1, 0) in the anticlockwise direction. (10)
(c) Verify your answer to the above part by computing the line integral directly. (10)
SCEE08010 Engineering Mathematics 2B - May 2019
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Question 2
(a) Show using vector calculus arguments that a conservative field that is also solenoidal
has a harmonic potential field. (5)
(b) Formulate an iterated surface integral for the flux of F(x, y, z) = x2iˆ + xzjˆ + 3zkˆ out
of the sphere x2 + y2 + z2 = 4, in spherical coordinates. You are not required to solve
this integral. (10)
(c) Using the divergence theorem compute the flux out of the sphere by calculating a
volume integral. (10)
END OF EXAM
SCEE08010 Engineering Mathematics 2B - May 2019
11.2 REVIEW OF BASIC PROBABILITY THEORY 911
11.2.5 Sample measures
It is conventional to denote a random variable by an upper-case letter (X, say), and an
actual observed value of it by the corresponding lower-case letter (x, say). An observed
value x will be one of the set of possible values (sample space) for the random variable,
which for a discrete random variable may be written as a list of the form {v1, v2, v3, . . . }.
z .00 .01 .02 .03 .04 .05 .06 .07 .08 .09
.0 .5000 .5040 .5080 .5120 .5160 .5199 .5239 .5279 .5319 .5359
.1 .5398 .5438 .5478 .5517 .5557 .5596 .5636 .5675 .5714 .5753
.2 .5793 .5832 .5871 .5910 .5948 .5987 .6026 .6064 .6103 .6141
.3 .6179 .6217 .6255 .6293 .6331 .6368 .6406 .6443 .6480 .6517
.4 .6554 .6591 .6628 .6664 .6700 .6736 .6772 .6808 .6844 .6879
.5 .6915 .6950 .6985 .7019 .7054 .7088 .7123 .7157 .7190 .7224
.6 .7257 .7291 .7324 .7357 .7389 .7422 .7454 .7486 .7517 .7549
.7 .7580 .7611 .7642 .7673 .7704 .7734 .7764 .7794 .7823 .7852
.8 .7881 .7910 .7939 .7967 .7995 .8023 .8051 .8078 .8106 .8133
.9 .8159 .8186 .8212 .8238 .8264 .8289 .8315 .8340 .8365 .8389
1.0 .8413 .8438 .8461 .8485 .8508 .8531 .8554 .8577 .8599 .8621
1.1 .8643 .8665 .8686 .8708 .8729 .8749 .8770 .8790 .8810 .8830
1.2 .8849 .8869 .8888 .8907 .8925 .8944 .8962 .8980 .8997 .9015
1.3 .9032 .9049 .9066 .9082 .9099 .9115 .9131 .9147 .9162 .9177
1.4 .9192 .9207 .9222 .9236 .9251 .9265 .9279 .9292 .9306 .9319
1.5 .9332 .9345 .9357 .9370 .9382 .9394 .9406 .9418 .9429 .9441
1.6 .9452 .9463 .9474 .9484 .9495 .9505 .9515 .9525 .9535 .9545
1.7 .9554 .9564 .9573 .9582 .9591 .9599 .9608 .9616 .9625 .9633
1.8 .9641 .9649 .9656 .9664 .9671 .9678 .9686 .9693 .9699 .9706
1.9 .9713 .9719 .9726 .9732 .9738 .9744 .9750 .9756 .9761 .9767
2.0 .9772 .9778 .9783 .9788 .9793 .9798 .9803 .9808 .9812 .9817
2.1 .9821 .9826 .9830 .9834 .9838 .9842 .9846 .9850 .9854 .9857
2.2 .9861 .9864 .9868 .9871 .9875 .9878 .9881 .9884 .9887 .9890
2.3 .9893 .9896 .9898 .9901 .9904 .9906 .9909 .9911 .9913 .9916
2.4 .9918 .9920 .9922 .9925 .9927 .9929 .9931 .9932 .9934 .9936
2.5 .9938 .9940 .9941 .9943 .9945 .9946 .9948 .9949 .9951 .9952
2.6 .9953 .9955 .9956 .9957 .9959 .9960 .9961 .9962 .9963 .9964
2.7 .9965 .9966 .9967 .9968 .9969 .9970 .9971 .9972 .9973 .9974
2.8 .9974 .9975 .9976 .9977 .9977 .9978 .9979 .9979 .9980 .9981
2.9 .9981 .9982 .9982 .9983 .9984 .9984 .9985 .9985 .9986 .9986
3.0 .9987 .9987 .9987 .9988 .9988 .9989 .9989 .9989 .9990 .9990
3.1 .9990 .9991 .9991 .9991 .9992 .9992 .9992 .9992 .9993 .9993
3.2 .9993 .9993 .9994 .9994 .9994 .9994 .9994 .9995 .9995 .9995
3.3 .9995 .9995 .9995 .9996 .9996 .9996 .9996 .9996 .9996 .9997
3.4 .9997 .9997 .9997 .9997 .9997 .9997 .9997 .9997 .9997 .9998
z 1.282 1.645 1.960 2.326 2.576 3.090 3.291 3.891 4.417
Φ (z) .90 .95 .975 .99 .995 .999 .9995 .999 95 .999 995
2[1− Φ (z)] .20 .10 .05 .02 .01 .002 .001 .000 1 .000 01
Figure 11.2 Table of
the standard normal
cumulative distribution
function Φ(z).
920 APPLIED PROBABILITY AND STATISTICS
The measured lifetimes of a sample of 20 electronic components gave an average of
1250 h, with a sample standard deviation of 96 h. Assuming that the lifetime has a normal
distribution, find a 95% confidence interval for the mean lifetime of the population, and
test the hypothesis that the mean is 1300 h.
Solution The appropriate figure from the t table is t0.025,19 = 2.093, so the 95% confidence interval is
(1250 ± 2.093(96)/!20) = (1205, 1295)
The claim that the mean lifetime is 1300 h is therefore rejected at the 5% significance
level. The same conclusion is reached by evaluating
which exceeds t0.025,19 in magnitude.
Example 11.5
n α = 0.10 α = 0.05 α = 0.025 α = 0.01 α = 0.005 n
1 3.078 6.314 12.706 31.821 63.657 1
2 1.886 2.920 4.303 6.965 9.925 2
3 1.638 2.353 3.182 4.541 5.841 3
4 1.533 2.132 2.776 3.747 4.604 4
5 1.476 2.015 2.571 3.365 4.032 5
6 1.440 1.943 2.447 3.143 3.707 6
7 1.415 1.895 2.365 2.998 3.499 7
8 1.397 1.860 2.306 2.896 3.355 8
9 1.383 1.833 2.262 2.821 3.250 9
10 1.372 1.812 2.228 2.764 3.169 10
11 1.363 1.796 2.201 2.718 3.106 11
12 1.356 1.782 2.179 2.681 3.055 12
13 1.350 1.771 2.160 2.650 3.012 13
14 1.345 1.761 2.145 2.624 2.977 14
15 1.341 1.753 2.131 2.602 2.947 15
16 1.337 1.746 2.120 2.583 2.921 16
17 1.333 1.740 2.110 2.567 2.898 17
18 1.330 1.734 2.101 2.552 2.878 18
19 1.328 1.729 2.093 2.539 2.861 19
20 1.325 1.725 2.086 2.528 2.845 20
21 1.323 1.721 2.080 2.518 2.831 21
22 1.321 1.717 2.074 2.508 2.819 22
23 1.319 1.714 2.069 2.500 2.807 23
24 1.318 1.711 2.064 2.492 2.797 24
25 1.316 1.708 2.060 2.485 2.787 25
26 1.315 1.706 2.056 2.479 2.779 26
27 1.314 1.703 2.052 2.473 2.771 27
28 1.313 1.701 2.048 2.467 2.763 28
29 1.311 1.699 2.045 2.462 2.756 29
∞ 1.282 1.645 1.960 2.326 2.576 ∞
Figure 11.8 Table
of the t distribution tα,n.
(Based on Table 12
of Biometrika Tables
for Statisticians,
Volume 1. Cambridge
University Press, 1954.
By permission of the
Biometrika trustees.)
Tn =
1250 1300–
96 /!20
-----------------------------
= 2.33–