# biosta - P(H X P(E = 1/3 X 1/3 = 0.111 III a X ~ b(15,0.10...

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II A) Had a low rating on academic performance the probability of this 1 in 3 = 0.333 b) was rated excellent in terms of their nutritional status the probability of this occurring is 1 in 3 = 0.333 c) Had low academic performance and an excellent nutritional status the probability of this is P(L) + P(E) = 1/3 + 1/3 = 0.667 d) Had low academic performance given that he or she had excellent nutritional status The probability is P(L) X P(E) = 1/3 X 1/3 = 0.111 e) Had high academic performance or excellent nutritional status
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Unformatted text preview: P(H) X P(E) = 1/3 X 1/3 = 0.111 III a) X ~ b(15,0.10) b) Pr(X=0) = 0.2059 c) Pr(X= 1) = 0.3432 d) Pr (X≤ 1) = Pr(X=0) + Pr(X=1) = 0.2059 + 0.3432 = 0.5491 e) Pr(X ≥ 2) = 1 x Pr(X ≤ 1) = 1 – 0.5491 = 0.4509 No. of Successes (x) 1 2 3 4 5 Pr( X= x) 0.2059 0.3432 0.2669 0.1285 0.0428 0.0105 Pr(X≤x) 0.2059 0.5490 0.8159 0.9444 0.9873 0.9978 Pr(X>x) 0.7941 0.4510 0.1841 0.0556 0.0127 0.0022...
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