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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