University of Texas SDS 321 Midterm Exam

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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 [Show More]

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