Download Assessment Schedule – KOHIA 2014 (Statistics) BOARD GAMES

Assessment Schedule – KOHIA 2014 (Statistics) BOARD GAMES: AS91585
Achievement
Achievement with Merit
Achievement with Excellence
Apply probability concepts in solving problems involves:
selecting and using methods
demonstrating knowledge of concepts and terms
communicating using appropriate representations.
Apply probability concepts, using relational
thinking, in solving problems involves:
selecting and carrying out a logical sequence
of steps
connecting different concepts or
representations
demonstrating understanding of concepts
and also relating findings to a context or
communicating thinking using appropriate
statements.
Apply probability concepts, using extended abstract
thinking, in solving problems involves:
devising a strategy to investigate or solve a problem
identifying relevant concepts in context
developing a chain of logical reasoning
making a statistical generalisation
and also where appropriate, using contextual
knowledge to reflect on the answer.
Evidence Statement
One
Expected Coverage
(a)
Probability of landing on the place =
Achievement (u)
23+30+37
200
=
90
200
Merit (r)
Excellence (t)
Probability correctly
calculated.
= 0.45 or 45%
Allow for estimation
between 40% - 50%
(b)
Consider the four combinations: (2,5), (3,4), (4,3) and (5,2).
1
4
4
P(2,5) = x =
P(3,4) =
P(4,3) =
P(5,2) =
36
2
36
3
36
4
36
x
x
x
36
3
36
2
36
1
36
=
=
=
1296
6
1296
6
1296
Any rounding.
4
1296
+
6
1296
+
6
1296
+
4
1296
=
20
1296
= 1.54%
The most likely die roll combination is 7, because it has the most outcomes (1,6), (2,5), (3,4),
(4,3), (5,2), (6,1) as compared to any other combination. The distribution of probabilities is
symmetric about the number 7. OR The mean result or expected value of one throw is 7.
AS91585 Kohia 2014
Correct probability
calculated, with some
working.
Any rounding.
1296
4
Using the addition rule, P(landing on ‘Take a chance’) =
(c)
Calculation of two
relevant probabilities
of (2,5), (3,4), (4,3)
or (5,2).
Calculates about 6
turns.
Calculates about 6
turns with justification
and one assumption
given.
Calculates about
5 turns with
logical reasoning.
Therefore, given there are 40 spaces on the board, 40 spaces ÷ 7 moves = 5.71
This approximates to about 6 turns to go around the board. Assumptions are that the
probability of going to jail or being sent elsewhere by Take a Chance or Community card are
small and don’t have much effect on the number of turns.
Since people get another throw if they roll doubles, the expected value of one turn is more than
7.
The expected move is 7 whether doubles is rolled or not.
If going to jail is counted as equivalent to a move of zero, the expected number of spaces
5
1
5
1
1
5
moved in a turn is × 7 + × × 14 + × × × 21 = 8.264
6
6
6
6
6
6
40÷8.264 = 4.840
The small chance of being moved by a card or going to jail would make the number of turns
slightly higher than 4.84, but not much.
The expected number of turns to go around the board is about 5.
(d)
P(rolling 3 doubles | go to jail)
=
=
Calculates P(rolling
three doubles), any
rounding.
P(rolling 3 doubles and going to jail)
6
1
( )3 ÷ [
36
40
𝑃(go to jail)
8
1
+
40
x
15
6
25
36
232
+ ( )3 ] =
Correct calculation
with one error, any
rounding.
= 0.1077 = 10.77%
N
N1
N2
A3
A4
M5
M6
E7
E8
No relevant
evidence.
Making
progress.
1 of u
2 of u
3 of u
1 of r
2 of r
1t
2 of t
Two
Expected Coverage
Achievement (u)
(a)
P(S∩C’) =
AS91585 Kohia 2014
Yes
No
No
2
1
3
20
5
25
Coin
Coin
4/25
Spinner
Yes
18
4
22
Correct answer
with working, any
rounding.
Yes
No
Spinner
Yes
18
4
22
No
2
1
3
20
5
25
Tree, table or Venn
diagram
Probability found from
incorrect tree, table or
Venn Diagram.
Merit (r)
Excellence (t)
(b)
P(Not used spinner ∩ Not used coin) = 1/25 = 0.04.
As experimental probability is greater than zero at 0.04, the events are not mutually exclusive. The
sample size is small at n = 25 but if even one person in the population has used neither spinner nor coin
toss that is evidence that the two events are not mutually exclusive.
Identification of relevant
theory related to
mutually exclusive
events.
Explanation of why
events are not mutually
exclusive with
justification.
(c)
P(Player two wins) =
P(wins first turn) + P(wins third turn)
1 1
3 5
2
1
= ( x ) + ( x x )=
2 2
4 6
5
2
P(Player one wins) = P(wins second turn) + P(Wins third turn)
3 1
3 5 3
1
=( x ) +( x x )=
4 6
4 6 5
2
As it is equally likely for either Player one or Player two to win, Oliver could omit this section of the game
and this have no effect on the long-term outcomes of the game, given it is based on luck as opposed to
skill.
He could choose the starter more quickly eg he could have the players flip a coin which could be a
quicker, although potentially a less exciting way to start the game.
Tree, table or Venn
diagram with three
probabilities correctly
identified
Tree, table or Venn
diagram with all relevant
probabilities correctly
identified, or correct
working allow one
calculation error
N
N1
N2
A3
A4
M5
M6
E7
E8
No relevant
evidence.
Making
progress.
1 of u
2 of u
3 of u
1 of r
2 of r
1 of t with
minor error
1 of t
Three
Expected Coverage
Achievement (u)
Merit (r)
(a)
P(both lose) = (1/3 x ½ x 2/3 x ¾ x 5/6 x 7/8) = 35/576 =
0.0625
Probability of one player losing correctly
calculated.
Correctly calculates
probability.
CAO
(b)
P(Primula wins) = P(P wins first round)
Correctly calculates probability of one
winning outcome.
Correctly calculates
probability.
OR P(P wins the second round)
OR P(P wins third round)
= (⅓x ½) + (⅓x ½ x ⅔x ¼)+ (⅓x ½ x ⅔ x ¾ x 5/6 x ⅛)
= 1/6 + 1/36 + 5/576 = 13/64 = 0.2031
AS91585 Kohia 2014
CAO
Correct statement in
context about
equally likely
probabilities,
supported by
working.
AND
Advice given to
Oliver consistent
with equally likely
probabilities and
sensible advice
given for the game.
Excellence (t)
(c)
P(Kwins∩Kfirst) =½(⅔+⅓ x ½ x ⅓ + ⅓ x ½ x ⅔ x ¾ x1/6)
= 53/144 = 0.3681
Correctly calculates probability of one
winning outcome.
Correctly calculates
probability of
P(Kwins∩Kfirst) OR
P(Kwins∩Ksecond) OR
P(kwins)
P(Kwins∩Ksecond) =½( ½×⅔ + ½×⅓×¾×⅓ +
½×⅓×¾×⅔×⅞ x1/6) = 223/1152 = 0.1936
P(kwins)=0.3681+0.1936 = 0.5617
Correctly calculates probability of
P(k first│K wins)
CAO
CAO
P(k first│K wins) = P(kwins∩kfirst)│K wins)
= 0.3681/0.5617 = 0.6553
(d)
Model distribution of probabilities out of 100 if
probabilities are as described at start of question with
Katrin going first:
Number of arrows
for a win
1
2
3
Primula
Calculates number or proportion of
expected wins for 1, 2 and 3 arrows for
either Primula or Katrin.
Katrin
18
3
0
67
6
1
Calculates number or
proportion of expected
wins for 1, 2 and 3
arrows for both Primula
and Katrin, AND notes
difference in
probabilities for model
and observations.
Calculates number or proportion of expected
wins for 1, 2 and 3 arrows for both Primula and
Katrin, AND notes difference in probabilities for
model and observations.
The size of difference between probabilities is
noted and explanations for possible reasons are
given wrt to sample size and sampling variability
and/or change in model probabilities.
In the observed games Primula is winning more often
than the model predicts. Even though 100 is a relatively
small sample size for investigating a probability, Primula
is winning 67% more often than predicted, which is a
bigger difference than would be expected due to
sampling variability. Katrin’s and Primula’s estimates of
their probability of winning may need to be adjusted.
N
N1
N2
A3
A4
M5
M6
E7
E8
No relevant
evidence.
Making
progress.
1 of u
2 of u
3 of u
1 of r
2 of r
1 of t
2 of t
Judgement Statement
Score range
AS91585 Kohia 2014
Not Achieved
Achievement
Achievement
with Merit
Achievement
with Excellence
0–7
8 – 12
13 – 18
19 – 24