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Mathematical Biostatistics Boot Camp 1 Week 1 Quiz 1. (1 point) What is P (A ∪ B) always equal to? a. 1 − P (A C ∩ BC ) b. 1 − P (AC ∪ B C ) c. 1 − P (AC )P (B C ) d. P (AC ∩ B C ) 1. a. Solution: P (A ∪ B) = 1 − P (A ∪ B)C = 1 − P (AC ∩ B C ) 2. (1 point) Which of the following are always true about P ( √ It is smallerPthan or equal to Pn i=1 P (Ei ) It is equal to P (Ei ). It is smaller than maxi P (Ei ). n i=1 √ √ . Sn i=1 Ei )? (Check all that apply.) It is larger than or equal to min P (E ). It is larger than or equal to max P (E ). i i i i It is smaller than mini P (Ei ). 3. (1 point) Consider inuenza epidemics for two parent heterosexual families. Suppose that the probability is 17% that at least one of the parents has contracted the disease. The probability that the father has contracted inuenza is 12% while the probability that both the mother and father have contracted the disease is 6%. What is the probability that the mother has contracted inuenza? a. 12% b. 6% c. 5% d. 25% e. 11% f. 17% 3. e. Solution: Let's have following events: A : father has the u, and B : mother has the u We know the following: P (A ∪ B) = 0.17, P (A) = 0.12, and P (A ∩ B) = 0.06. From the following formula: P (A ∪ B) = P (A) + P (B) − P (A ∩ B) we can derive: P (B) = P (A ∪ B) + P (A ∩ B) − P (A) = 0.17 + 0.06 − 0.12 = 0.11 Page 1 of 6 4. (1 point) A random variable, X is uniform, so that it's density is f (x) = 1 for 0 ≤ x ≤ 1. What is it's 75th percentile? Express your answer to two decimal places. a. 0.25 b. 0.50 c. 0.75 d. 0.10 e. 0.90 c. 4. Solution: The cumulative distribution function for f (x) is: x Z F (x) = Z x f (t) dt = 0 0 x 1 dt = [t]0 = x − 0 = x We want to nd value of x75 such that F (x75 ) = 0.75. We know that F (x75 ) = x75 , so x75 = 0.75. 5. (1 point) A Pareto density is for 1 < x < ∞? a. x1 − 1 b. 1 − c. 1 − d. x13 e. x1 1 x2 for 1 < x < ∞. What is the distribution function associated with this density 1 x 1 x3 5. Solution: Z x F (x) = 1 b. x 1 1 1 1 1 1 1 dt = − = = − =1− t2 t 1 t x 1 x x Page 2 of 6 6. (1 point) a. b. c. d. What is the quantile p from the density e−x (1 + e−x )−2 ? log ((1 − p) /p) p/ (1 − p) (1 − p) /p 1/ (1 + ex ) e. log (p/ (1 − p)) 6. e. Solution: First, we need to nd cumulative distribution function: x e−t −∞ (1 + e−t ) Z F (x) = 2 dt, using substitution u(t) = 1 + e−t and du = −e−t dt we get: Z u(x) F (x) = u(−∞) −1 du = u2 Z 1+e−x ∞ 1+e−x −1 1 1 ex e−x ex du = = = = u2 u ∞ 1 + e−x e−x (ex + 1) ex + 1 Now we want to nd a value of xp such that F (xp ) = p. F (xp ) = p exp =p exp + 1 exp = p (exp + 1) exp = pexp + p exp − pexp = p exp (1 − p) = p p 1−p p xp = log 1−p exp = Page 3 of 6 7. (1 point) of c? a. b. c. d. e. f. g. Suppose that a density is of the form cxk for some constant c > 1 and 0 < x < 1. What is the value k+2 1 k 2 1 k+1 k−1 k k+1 7. Solution: Z 1 xk+1 cx dx = c k+1 k 0 1 = 0 g. c k+1 1 c c x = (1 − 0) = must be equal to 1, so: 0 k+1 k+1 k+1 c = 1 ⇒ c = k + 1. k+1 8. (1 point) Suppose that the time in days until hospital discharge for a certain patient population follows a density f (x) = 12 e−x/2 for x > 0. What is the median discharge time in days? a. 1.0 b. 1.4 c. 1.8 d. 2.2 e. 2.6 8. b. Solution: First, we need to nd cumulative distribution function F (x): Z x h ix h i0 1 −t/2 e dt = −e−t/2 = e−t/2 = e0 − e−x/2 = 1 − e−x/2 0 x 0 2 Now we need to nd the value of x50 such that F (x50 ) = 0.5: F (x) = F (x50 ) = 0.5 1 − e−x50 /2 = 0.5 e−x50 /2 = 0.5 1 2 1 x50 /2 = − log 2 −x50 /2 = log x50 /2 = log (2) x50 = 2 log (2) ≈ 1.386 R code: qexp ( 0 . 5 , 1 / 2) Page 4 of 6 9. (1 point) Consider the density given by 2xe−x for x > 0. What is the median? a. 1.03 2 b. 0.83 c. 0.79 d. 0.24 e. 0.15 9. b. Solution: First, we need to nd cumulative distribution function F (x): x Z 2 2te−t dt, F (x) = 0 using substitution u(t) = −t2 and du = −2tdt we get: Z u(x) u Z e du = − F (x) = − u(0) 0 −x2 2 0 eu du = [eu ]−x2 = e0 − e−x = 1 − e−x 2 Median is the value of x50 such that F (x50 ) = 0.5. F (x50 ) = 0.5 2 1 − e−x50 = 0.5 2 e−x50 = 0.5 −x250 1 = log 2 x250 = log (2) p x50 = log (2) ≈ 0.833 Page 5 of 6 10. (1 point) Suppose h(x) is such that 0 < h(x) < ∞ for x = 1, 2, . . . , I . Then c · h(x) is a valid PMF (probability mass function) when c is equal to what? a. b. c. 1 PI 1+ PI x=1 x=1 h(x) h(x) P I x=1 −1 h(x) d. h(I) c. 10. Solution: We know that 0 < c·h(x) < ∞. The second condition to function to be a valid PMF is, that I X P x c·h(x) = 1. c · h(x) = 1 x=1 c I X h(x) = 1 x=1 c= I X !−1 h(x) x=1 Question: 1 2 3 4 5 6 7 8 9 10 Total Points: 1 1 1 1 1 1 1 1 1 1 10 Score: Page 6 of 6