Math 6 SL Probability Distributions Practice Test Mark Scheme

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1 Math 6 SL Probability Distributions Practice Test Mark Scheme. (a) Note: Award A for vertical line to right of mean, A for shading to right of their vertical line. AA N (b) evidence of recognizing symmetry e.g. 05 is one standard deviation above the mean so d is one standard deviation below the mean, shading the corresponding part, = 00 d d = 95 A N (c) evidence of using complement e.g. 0., p P(d < X < 05) = 0.6 A N [6]. (a) (i) valid approach e.g. np, 5 5 E(X) = A N (ii) evidence of appropriate approach involving binomial e.g. X ~ B 5, 5 recognizing that Mark needs to answer or more questions correctly e.g. P(X ) valid approach e.g. P(X ), P(X = ) + P(X =4) + P(X = 5) P(pass) = A N M (b) (i) evidence of summing probabilities to e.g (a + b) = some simplification that clearly leads to required answer e.g a + b = 4a + b = 0.4 AG N0 A IB Questionbank Maths SL

2 (ii) correct substitution into the formula for expected value e.g. 0(0.67) + (0.05) (0.04) some simplification e.g a + 4b (0.04) = correct equation e.g. a + 5b = 0.75 evidence of solving A a = 0.05, b = 0.0 AA N4 (c) attempt to find probability Bill passes e.g. P(Y ) correct value 0.9 A Bill (is more likely to pass) A N0 [7]. (a) E(X) = A N (b) evidence of appropriate approach involving binomial 0 e.g. (0.), (0.) (0.) 7, X ~ B(0, 0.) P(X = ) = 0.0 A N (c) METHOD P(X ) = (= ) evidence of using the complement (seen anywhere) e.g. any probability, P(X > ) = P(X ) P(X > ) = 0. A N METHOD recognizing that P(X > ) = P(X 4) e.g. summing probabilities from X = 4 to X = 0 correct expression or values 0 0 e.g. (0.) 0 r (0.) r r 4 r P(X > ) = 0. A N [6] IB Questionbank Maths SL

3 4. (a) evidence of approach.7 e.g. finding 0.4, using correct working.7 e.g =, graph σ =. A N (b) (i) evidence of attempting to find P(X < 5.4) e.g. using z =.7 P(X < 5.4) = 0.95 A N (ii) evidence of recognizing symmetry e.g. b = 4.4, using z =.7 b = 6.6 A N [7] 5. (a) evidence of using binomial probability 7 e.g. P(X = ) = (0.) (0.) 5 P(X = ) = 0.5 A N (b) METHOD evidence of using the complement M e.g. (P(X )) P(X ) = 0.6 P(X ) = 0.6 A N METHOD evidence of attempting to sum probabilities e.g. P( heads) + P( heads) P(7 heads), correct values for each probability e.g P(X ) = 0.6 A N M [5] 6. Note: Candidates may be using tables in this question, which leads to a variety of values. Accept reasonable answers that are consistent with working shown. W ~ N(.5, 0. ) (a) (i) z =.67 (accept.67) P(W < ) = (accept answers between and 0.045) A N (ii) z = IB Questionbank Maths SL

4 (iii) P(W >.) = 0.59 A N.5 kg AA Note: Award A for a vertical line to left of mean and shading to left, A for vertical line to right of mean and shading to right. (iv) Evidence of appropriate calculation M eg ( ), P = AG N0 Note: The final value may vary depending on what level of accuracy is used. Accept their value in subsequent parts. (b) (i) X ~ B(0, ) Evidence of calculation eg P(X = 0) = ( ) 0 P(X = 0) = ( sf) A N (ii) METHOD Recognizing X ~ B(0, ) (may be seen in (i)) M P(X 6) = (or P(X = ) P(X = 6)) evidence of using the complement eg P(X 7) = P(X 6), P(X 7) = P(X < 7) P(X 7) = 0.67 A N METHOD Recognizing X ~ B(0, ) (may be seen in (i)) For adding terms from P(X = 7) to P(X = 0) P(X 7) = = 0.67 A N N [] 7. X ~ N(, ), P(X < ) = 0., P(X > ) = 0. P(X < ) = 0.9 Attempt to set up equations 0.46,. AA IB Questionbank Maths SL 4

5 = 0.46 =. 5 =.6 =.5, = 4.99 AA N4 [6]. (a) (i) a b 0.5 (ii) (a) 0.4 (A) (b) (or ) 0.5 ( sf) (N) 6 (b) (i) Sketch of normal curve (ii) c (A) 4 [0] (a) P(M 50) = P (M < 50) = P Z 0 = P(Z <.) = 0.90 = 0.09 (accept to 0.090) P(M 50) = 0.09 (G) (b) P(Z <.96) = 0.05 = Z.96 (0) = < M < a = 5, b = 69 5 < M < 69 (G) Note: Award (G) if only one of the end points is correct. [5] IB Questionbank Maths SL 5

6 0. Note: Where accuracy is not specified, accept answers with greater than sf accuracy, provided they are correct as far as sf (a) z = 9.5 =.00 P (Z > ) = () = 0.4 = 0.57 = 0.59 ( sf) = 5.9% P (H > 97) = 0.59 (G) = 5.9% (b) Finding the 99 th percentile (a) = 0.99 => a =.7 (accept.) => 99% of heights under (9.5) = = 0 ( sf) 99% of heights under 09.6 = 0 cm ( sf) (G) Height of standard doorway = = 7 cm 4 [7] (a) Z = = P(Z <.4) = P(Z <.4) = 0.99 = 0.00 P(W < 5) = 0.00 (G) (b) P(Z < a) = 0.05 P(Z < a) = a = =.96 µ = (0.50) = = 5.9 = 6.0 ( sf) (AG) =.00 P(Z <.00) = P(Z <.00) = = IB Questionbank Maths SL 6

7 µ = 5.9 (G) mean = 6.0 ( sf) (AG) (c) Clearly, by symmetry µ = Z = = =.96σ σ = 0.55 kg (d) On average, cost saving bag cement saving bag = 0.5 kg = 0.5(0.0) = $0.40 To save $5000 takes 5000 = 500 bags 0.40 []. (a) Area A = 0. (b) EITHER Since p (X ) = p (X ), then and are symmetrically disposed around the (R) mean. Thus mean = = 0 Notes: If a candidate says simply by symmetry = 0 with no further explanation award [ marks] (M, A, R). As a full explanation is requested award an additional for saying since p(x < ) = p(x > ) and another for saying that the normal curve is symmetric. p (X ) = 0. p p p (X ) = 0. p p Z = 0. Z = 0.9 Z = 0. Z = 0.9 So = = 0 5 IB Questionbank Maths SL 7

8 (c) 0 = 0.9 Note: Award for 0, for standardizing, and for 0.9. =. (or.) = or.. =.56 ( sf) (AG) 5 Note: Working backwards from =.56 to show it leads the given data should receive a maximum of [ marks] if done correctly. (d) p (X ) = p 0 Z (or.56).56 Note: Award for standardizing and for = p (Z ) (or 0.64 or 0.64) = (0.6407) = 0.79 ( sf) 5 [6]. (a) p (4 heads) = 7 = 70 = ( sf) (b) p ( heads) = 56 = ( sf) (c) p (5 heads) = p ( heads) (by symmetry) p ( or 4 or 5 heads) = p (4) + p () = ( sf) [6] IB Questionbank Maths SL

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