Survey
* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
Assessment Schedule Statistics and Modelling: Use probability distribution models to solve straightforward problems (90646) Achievement Criteria Qn No. Use probability distribution models to solve straightforward problems. 1 Evidence Code Binomial Distribution P(X < 4; n = 10, π = 0.15) = 0.9901 A Judgement Accept alternative rounding. CAO Sufficiency Achievement: 2 x Code A ACHIEVEMENT 2 Poisson Distribution = 24 per hour No repeated = 4 per ten minute interval distributions. P(X 1) = 1 – P(X = 0) = 1 – 0.0183 A CAO A CAO = 0.9817 3a Normal Distribution P( 50 < X < 60) = P( -0.976 < Z < 1.463) = 0.7638 (0.76369 GC) Achievement Criteria Qn No. ACHIEVEMENT WITH MERIT 3b 4 Use probability distribution models to solve problems Evidence Inverse Normal P(X < k) = 0.025 Z = -1.96 k = 45.96 Mbps Judgement Sufficiency Accept alternative rounding for all answers. Achievement with Merit: AM Normal Distribution P(X > 4) = P(Z > 1.333) = 0.0913 (0.0912 GC) Binomial Distribution P(X = 0; n = 4, π = 0.0913) = 0.6818 (0.6821 GC) 5a Code A AM Sum of two normally distributed independent random variable T=X+Y E(T) = 150 + 80 = 230 min σ(T) = 222 152 26.627 P(X < 240) = P(Z < 0.376) = 0.6465 (0.6464 GC) AM Statistics and Modelling 90646 Assessment Schedule Page 2 of 3 Achievement plus 2 x Code M or 3 x Code M Achievement Qn Criteria No. ACHIEVEMENT WITH EXCELLENCE 5b Use and justify probability distribution models to solve complex problems. Evidence Code Linear combination of independent random variables E (C ) $0.25 150 $0.40 80 Require evidence of linear combination used. = 69.5 σ(C)= .25 22 .4 15 = 8.139 P(X < 55) = P(Z < -1.782) = 0.0373 (0.0374 GC) 2 6 2 2 Sufficiency Excellence: Merit plus 2 Poisson distribution Two assumptions from: faults occur at random Each fault is independent Faults cannot occur simultaneously The probability of a fault is proportional to the length of the time interval. Inverse Poisson P(X = 0) = 1 – 0.98 = 0.02 e 0.02 λ = 3.912 P(X< 5) = 0.7987 Judgement 2 x Code E AME E Or equivalent. Or equivalent. Must have TWO assumptions. AME Statistics and Modelling 90646 Assessment Schedule Page 3 of 3