Midterm
SDS321
Spring 2015
You may use a single (2 sides) page of notes, and you may use a calculator.
This exam consists of five questions, containing multiple sub-questions. The assigned
points are noted next to e
...
Midterm
SDS321
Spring 2015
You may use a single (2 sides) page of notes, and you may use a calculator.
This exam consists of five questions, containing multiple sub-questions. The assigned
points are noted next to each question; the total number of points is 25. You have 75
minutes to answer the questions.
Please answer all problems in the space provided on the exam. Use extra pages if
needed. Of course, please put your name on extra pages.
Read each question carefully, show your work and clearly present your answers. Note,
the exam is printed two-sided - please don’t forget the problems on the even pages!
Good Luck!
Name:
UTeid:
1
Question 1 (3 points)
(a) (1 point) How many solutions are there to the equation x1 + x2 = 5, where
x1 and x2 are non-negative integers?
6. This can either be solved using a stars-and-bars type approach, or by
direct enumeration of options ((0,5),(1,4),(2,3),(3,2),(4,1),(5,0)).
1 point for correct answer.
• If they used only positive numbers, give no marks but don’t drop marks
on the second part for only using positive numbers.
• If they only included half the pairs – e.g. counted (1,4) and (4,1) as the
same – give no marks but don’t drop marks on the second part if they
do the same.
(b) (2 points) How many solutions are there to the equation x1 + x2 + 2x3 = 5,
where x1, x2 and x3 are non-negative integers? Hint: what values can x3
take?
12. x3 can be 0,1 or 2. If x3 = 0, there are 6 options as above. If x3 = 1,
then x1 + x2 = 3, so 4 options. If x3 = 2, then x1 + x2 = 1, so 2 options.
2 points for correct answer. If the answer is incorrect but they showed
working:
• If the answer is correct except tQuestion 2 (3 points) Let X ∼ N(70, 100) be a normal random variable with expectation µ = 70 and standard deviation σ = 10. For the following problems,
please express your probability in terms of Φ(z) and then evaluate it using the
standard normal table provided on the last page of this exam script. Recall that
Φ(z) = P(Z ≤ z), where Z ∼ N(0, 1).
(a) (1 point) Find P(X < 60).
P(X < 60) = P
Z < 60 − 70
10
= P(Z < −1) = 1 − P(Z ≤ 1)
= 1 − Φ(1) = 1 − 0.8413 = 0.1587
• 0.5 marks for translating into Z form (either P(Z ≤ 1) or Φ(1) notation
are fine, it doesn’t matter if they use “<” or “≤ notation). 0.5 marks
for right answer. More/fewer sig figs are fine.
• If they use an incorrect value for µ or σ in calculating Z, give 0 marks
here but if they repeat the mistake in later parts, give marks.
(b) (1 point) Find P(X ∈ [50, 60])
P(X ∈ [50, 60]) =P(X ≤ 60) − P(X ≤ 50)
P(X ≤ 50) =P
Z ≤
50 − 70
10
= 1 − P(Z ≤ 2) = 1 − 0.9772 = 0.0228
P(X ∈ [50, 60]) =0.1587 − 0.0228 = 0.1359
• 0.5 marks for translating into Z form, 0.5 marks for right answer. They
can either translate the whole thing into Z form, or just the P(Z < 50)
part. Again, it doesn’t matter if they use “<” or “≤ notation, and
more/fewer sig figs are fine.
• If they made a mistake in part (a) and used that value here, but everything else is correct, give the mark.
• If they used an incorrect value for µ or σ here, and used the same incorrect value in part (a), give the mark assuming everything is consistent
with the values used.
(c) (1 point) Find P(X ∈ [50, 60] ∪ [80, 90]). Hint: No further calculation necessary. By symmetry, this is twice the answer in (b), i.e. 0.2718. 1 point for
2*answer in (b). Or, they can calculate it using tables – in which case, 0.5
marks for Z notation, 0.5 marks for correct answer (with the guidelines from
part (b) for mistakes carried forward).
3
Question 3 (5 points) A continuous random variable X has probability density function.
fX(x) = c e−|x|/2 =
c e−x/2 x ≥ 0
c ex/2 x < 0
(1)
(a) (1 point) Find c.
1 = R ∞
−∞ f(x)dx = 2 · c
R ∞
0
e
−x/2dx = 2c[−2e
−x/2
]
∞
0 = 4c ⇒ c = 1/4.
1 point; 0.5 points for setting up correct integral relationship (either R ∞
−∞ f(x)dx =
1, or R ∞
0
f(x)dx = 0.5); 0.5 points for solving integral correctly.
(b) (2 points) Find the CDF of X, i.e. FX(x) = P(X ≤ x) for x ≥ 0 and x < 0.
If you were unable to solve part a, you may give your answer in terms of c.
For x < 0, FX(x) = R x
−∞ f(t)dt = c
R x
−∞ e
t/2dt = c[2e
t/2
]
x
−∞ = 2cex/2 =
0.5e
x/2
For x > 0, FX(x) = R x
−∞ f(t)dt = c
R 0
−∞ e
x/2dt + c
R x
0
e
−t/2dt = 0.5 +
c[−2e
−t/2
]
x
0 = 0.5 − 2ce−x/2 + 2c = 1 − 0.5e
−x/2
• 1 mark for each case: 0.5 marks for setting up the integral, 0.5 marks
for solving the integral.
• For the seco
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