University:
California State University, Northridge
Course:
MATH 440A | Mathematical Statistics I
Academic year:

2010
Views:

69

Pages:

2
Author:

Rory Rhodes

Given that test is positive Positive Pres. a Abs. c Total a+c PPV = a a+c Sensitivity = P(test+|disease) = 9/10 = 90% Present 9 Absent 1 Total 10 NPV = P(no disease|test-) = 99/100 = 99% Specificity = P(test-|no disease) = 99/100 = 99% PPV = P(disease|test+) = 9/10 = 90% Positive Negative Total Yes 18 2 20 No 8 72 80 Total 26 74 100 PPV = P(con|test+) = 18/26 NPV = P(no con|test-) = 72/80 M = person is male T = correctly identifies tap water T and M are independent => P(T) = P(T|M) P(T) = 60/100 = 0.60 P(T|M) = 21/35 = 3/5 = 0.60 T & M are independent!! P(T) ≠ P(T|B) B = person is a bottled water drinker P(T|B) = 24/80 = 3/10 = 0.30 ≠ 0.60 = P(T) so B & T are not independent H1 = heads on the first flip H2 = heads on the second flip H3 = heads on the third flip H4 = heads on the fourth flip P(H1 and H2 and H3 and H4) = P(H1) * P(H2) * P(H3) * P(H4) = 1/2 * 1/2 * 1/2 * 1/2 = 1/16 P1 = person 1 doesn't have health insurance P2 = person 2 doesn't have health insurance P3 = person 3 doesn't have health insurance ... Pn = person n doesn't have health insurance P(P1 or P2 or P3 or ... or Pn) = 1 - P(P1 and P2 and P3 and ... and Pn) (assuming independence) = 1 - (1/3)^n = 1 - 0.0000059049 = 0.9999940951 Probability that at least one of them has health insurance = 1 - P(nobody has health insurance) W = winning the game D = playing during the day P(W|D) = 4/8 = 0.5 P(W|D) = 6/21 = 0.2857 not equal -> so not independent But they are really close!