# Achievement Standard 2 by accinent

VIEWS: 7 PAGES: 16

• pg 1
```									   Achievement Standard 2.6 (90289)
(Simulate probability situations, and apply the normal distribution)

Level 2               Internal               2 credits

Key words: simulation, probability, normal distribution, outcome, relative frequency, AND, OR,
expected value, trial, mean, theta (  ), mu (  ), standard deviation, bell shape curve, symmetrical,
certain, standard normal distribution, Z score, theoretical probability, tree diagram, conditional,
independent, replacement, model, inverse normal, x bar,

Achieve: (Design and use a simulation method to
explore a situation involving probabilities AND Use the
normal distribution model to find probabilities in
straightforward problems.)
Basic Probability
Fraction, decimal or percentage
A probability must only be expressed as a fraction, decimal or percentage

One
The greatest that a probability can be is 1
0                                               1

IMPOSSIBLE                                        Certain

Probability + NOT = 1
As 1 is the maximum probability, probabilities must add to 1
E.g. if P(Rain) = 0.23 the P(not Rain) = 1 – 0.23 = 0.77

Equally likely outcomes
Look at shape of object to see that all events have an equal chance i.e. coin or dice.
A' s
P( A ) 
outcomes
Long run relative frequency (exercise 26.1 is all practical)
Similar to equally likely outcomes but based on a practical
E.g. Drop toast experiment
Buttered side up     10
Buttered side down   12
TOTAL                22
P(Up) = 10/22 = 5/11
This might be a good point to recap simplification of fractions

1
AND/OR
AND = 
1                 1                             1 1 1
E.g. if P(Ace) =  and P(Tails) =    then P(Ace AND Tails) =  
6                 2                             6 2 12
This might be a good point to recap multiplication of fractions

OR = +
4                    4                           4   4   8   2
E.g. if P(Jack) =   and P(Queen) =        then P(Jack OR Queen) =         
52                   52                          52 52 52 13
This might be a good point to recap addition of fractions
Page 274 Ex 26.2

Expected Value/Number (achieve and merit work)
There are two types of questions, the first are obvious.
i.e. if a coin is tossed 70 times, we would expect 35 heads and 35 tails.

Expected number of = probability  trials

The second type of question involve tables
Expected values from probability tables
E.g. 1) We know that the probability of a fair die is 1/6 but what would be the average value of each
roll be?

Number        1         2              3               4          5             6
Probability   1/6       1/6            1/6             1/6        1/6           1/6
1/6       2/6            3/6             4/6        5/6           6/6
1 2 3 4 5 6          21
Expected value =                       3.5 i.e. The mean average of all the numbers of a large
6 6 6 6 6 6          6
trial would be 3.5.

E.g. 2) A biased dice is known to produce the following results

Number        1        2         3      4         5         6
Probability 0.01     0.25       0.1   0.11      0.03       0.5
Expected value = sum of (number multiplied by probability)
0.01      0.5       0.3   0.44      0.15        3
Total = 4.4

In other words, the average number that we expect to obtain is 4.4
i.e. if we roll the dice 300 times adding the score each time then divide by 300 we would expect the

The following exercise needs knowledge of Venn Diagrams. This is not needed at level 2 but can
do no harm being taught, as it is required at level 3
Page 292 Ex 26.6

2
Normal Distribution
Do the following none calculator test with the students in silence
1) Solve (2x + 3)(x – 4) = 0                                           (Answer = –3/2 & 4)
2) Find distance between (2 , 3) and (5 , 7)                            (Answer = 5)
3) What is the product of the gradients of two perpendicular lines? (Answer = –1)
4) For the line y = 3x + 4, what is the gradient?                       (Answer = 3)
2
5) Factorise x – x – 6                                                  (Answer = (x + 2)(x – 3))
6) Expand (4x2)3                                                        (Answer = 64x6)
7) Find mid point of (8 , 17) and (10 , 21)                            (Answer = (9 , 19))
8) Gradient of line connecting (4 , 9) and (6 , 17)                     (Answer = 4)
1/2
9) Value of 9                                                           (Answer = 3)
                A
10) Make r the subject of A =  r2                                      Answer : r     
                
                  
Now:
1) Find mean average (  )
2) Find standard deviation (  ) (the calculator will do this for you)
3) Draw a histogram of class results with each bar being a mark out of 10 e.g.

frequency

1    2 3 . . . etc.
Marks out of 10

If there were enough people doing the test this should follow a normal distribution curve or bell
shape. This is unlikely due to the small numbers involved. If not a perfect bell shape, more classes
results are required. The more we collect, the closer it will be to a bell shape.

What would a ‘cleverer’ class’s graph look like?
What would a ‘less able’ class’s graph look like?
What would the graph be like if the questions were harder?
What would the graph look like if the questions were easier?

3
Casio Graphic Calculator: Averages from Discrete Data
(This is not required for 2.6, but some students might be interested)
To enter data
1) STAT mode (second in on left the top row)
2) Enter data into list 1 column i.e. 12, 13, 14 (press Exe after each entry)
3) F2 (CALC)
4) F6 (SET) make sure the first two lines of the menu is set up to
1Var XList :List1
1Var Freq :1 (The 1 indicates that there is one of each number in List1. If this is to be a frequency
5) EXE

To find various statistics
1) Once you have followed the instructions above and entered a set of data
2) EXIT or F2 (CALC) to obtain bottom menu with arrow above F6 and CALC above F2
3) F1 (1VAR)
Using the arrow down key you can see various types of statistics including mean, mode, median,
maximum, minimum, sum of data, number of pieces of data and quartiles. Take care with mode as
if there is no mode it chooses the maximum value for the mode or if there is more than one mode it
chooses the highest mode value.

(If you are calculating averages and it has an error, check that you have followed instruction 4 from
entering data correctly)

To clear previously entered data
1) In STAT mode
2) EXIT or F2 (CALC) to obtain bottom menu with arrow above F6 and CALC above F2
3) F6 (arrow right)
4) To delete single piece of data, use arrow key to highlight piece of data and then press F3
(DEL) or
5) To delete all of row use arrow key to highlight any piece of data in row and then press F4
(DEL-A), F1 (YES)
Graphically the mean and deviations look like:

0.5%         2%     13.5%      34%     34%         13.5%   2%      0.5%
Xa                 
likely/probable
-      68%  + 

very likely/very probable
 - 2              95%              + 2

almost certain
 - 3                        99%                             + 3
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Notice that the normal distribution is symmetrical i.e. 50% either side of mean.
Probability questions can now be asked using the bell shape curve i.e. on the above graph the
probability that the result is bigger than Xa is 97.5%

With the results that we have for our test result out of ten explain what the standard deviation and
mean average mean i.e. if the mean is 6 and the standard deviation is 1.5 then we expect to find:
34% of the students with results between 6 and 7.5,
34% of the students with results between 6 and 4.5
47.5% (34 + 13.5) of the students with results between 7 and 10 etc.
(with such a small amount of data we are unlikely to obtain realistic results)

All normal distributions have different means (  ) and standard deviations (  ), but all have 68%,
95%, 99% pattern around mean.

X

Example:
In a test, the mean is 70 and the standard deviation is 5. Assume normal distribution.
From the above data the following graph can be drawn

0.5%     2%         13.5%        34%     34%    13.5%    2%        0.5%
55          60           65         70    75       80         85

From the above, certain probabilities can be stated as long as it involves the values that lie on
multiples of standard deviations away from the mean.
i.e.
0.5% of the students scored more than 85.
16% of the students scored less than 65.
50% of the students scored more than 70.
68% of the students scored between 65 and 75 i.e. it is likely that a student’s result is between
these two values.
95% of the students scored between 60 and 80 i.e. it is very likely that a student’s result is between
these two values.
99% of the students scored between 55 and 85 i.e. it is almost certain that a student’s result is
between these two values. Alternatively, if I chose a student at random there is only a 1% chance
that his score is outside of these two values.
Page 350 Ex 30.1 (OK Q’s if time)
Page 353 Ex 30.2 (good Q’s)

The Standard Normal Distribution
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The normal distribution graphs up to this point can only be analysed using values that are multiples
of the standard deviation. This is rather limited.

The standard normal distribution is a perfect normal distribution with a mean of 0 and a standard
deviation of 1. Instead of using X for certain values, we use Z.

The probability (area) of Z (+ve only) being between the mean and a certain value can be found on
page 516. These values are as decimals i.e. they will not be bigger than 1 (0.5 as only half the
graph is used in the table)

e.g.
P(0 < Z < 1.43) = 0.4236 or 42.36%
P(0 < Z < 0.7) = 0.2580
P(–0.7 < Z < 0) = 0.2580
P(0 < Z < 2.163) = 0.4846 + 0.0001 = 0.4847

P(Z > 1.57) = 0.5 – 0.4418 = 0.0582 = 5.82%

P(Z < 1.593) = 0.5 + 0.4445 = 0.9445 = 94.45%

P(Z > 2.35) = 0.5 – 0.4906 = 0.0094 = 0.94%

Page 358 Ex 30.3 Q1a–f, 2a–f

Converting to the Standard Normal Distribution
To be able to work with data we need to convert normal distribution to a standard normal
distribution i.e. probability that data lies between mean and any point. This is given in table on page
516

All we need to do to convert a particular value (X) to see at what point it is on a standard normal
X
distribution we use the formula for Z value/score Z          (on formula sheet). We will use this

later.

E.g.1) If year 11 boys had a mean height of 1.67m and a standard deviation of 9cm. Find the
probability that a person chosen at random is bigger than 1.81m. Assume data is normally
distributed

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Question is asking i.e. P(X > 1.81) notice that this is not a Z.
First using above formula convert X score to Z score
1.81  1.67
Z                1.5555555 use 1.556
0.09

Therefore the probability can now be written as P(Z>1.556)
Draw diagram (stress importance of this step)

0.4401

0
1.556

From table: 1.556 gives 0.4401 (remember this value is the area between the mean and the stated
value. We do not want this. We need the ‘other’ area)
Therefore P(Z > 1.556) = 0.5 – 0.4401 = 0.0599 = 5.99%

E.g. 2) Party bags of M and M‘s claim they have 255 in them. The company know that the mean
average is 259 and has a standard of 2.

What is the probability that a bag chosen at random has less than 255 M and M’s in them?

Question is asking i.e. P(X < 255) notice that this is not a Z.
First using above formula convert X score to Z score
255  259
Z              2
2
Therefore the probability can now be written as P(Z<-2)
Draw diagram (stress importance of this step)

0.47725

Therefore P(Z > –2) = 0.5 – 0.47725 = 0.02275 = 2.275%
Page 359 Ex 30.4 (using formula and practical problems, Achieve and Merit questions)

7
Casio Graphic Calculator: Practical problems using Normal Distribution
Using the calculator at this stage will hinder your understanding of this topic, but if you are only
after achieve ……
1) STAT mode
2) DIST (F5)
3) NORM (F1)
4) Ncd (F2)
5) Enter the lower value of X
6) Enter the Upper value of X
7) Enter the standard deviation
8) Enter mean
9) EXE

E.g. 1) If year 11 boys had a mean height of 1.67m and a standard deviation of 9cm. Find the
probability that a person chosen at random is between 1.7 and 1.81m. Assume data is
normally distributed
For step 5) above use 1.7
For step 6) above use 1.81
For step 7) above use 0.09
For step 8) above use 1.67

E.g. 2) If year 11 boys had a mean height of 1.67m and a standard deviation of 9cm. Find the
probability that a person chosen at random is bigger than 1.81m. Assume data is normally
distributed. Note that this is the same question in the above notes.
For step 5) above use 1.81
For step 6) above use any large number i.e. 50 as this number is a lot bigger than the mean
For step 7) above use 0.09
For step 8) above use 1.67
Compared to the answer obtained using tables, this is virtually the same. I would round the

Practicals
Use of random numbers to:-
1) Choose a number that has a 60% chance. i.e. 1,2,3,4,5,6 is ok and 7,8,9,0 is not ok
2) Choose a number that has a 1 in 6 chance i.e. 1 is ok and 2,3,4,5,6. Take care of using
Random6 to choose a number between 1 and 6.

Monty Hall Problem
Not a good example for working out statistics but good to use as a demonstration of how a practical
activity can show if something works or not.

You have three doors, you pick one, and Monte shows you which one of the other doors is the
wrong choice. Do you keep your choice or do you swap.

For a computer simulation go to http://www.cut-the-knot.com/hall.shtml

There is much discussion as to which is the correct answer but the following convinces me:

8
The best explanation I can give is as follows. There are three possible places the prize can be, just
for arguments sake lets say we always start by picking door number 1. There are two possible
choices for the second question switch or do not switch. If we do not switch here are the outcomes:

Prize is                 Monty
I pick                       Result
behind                   opens

1           1        2 or 3       Stay on 1, Win

2           1            3        Stay on 1, Lose

3           1            2        Stay on 1, Lose

I will win 1/3rd of the time by sticking with my original choice. But if I do choose to switch:

Prize is       I       Monty
Result
behind:      pick      opens

1          1         2        Switch to 3, Lose

2          1         3         Switch to 2, Win

3          1         2         Switch to 3, Win

So, if I switch, I will win 2/3rds of the time.
In the original problem, you have three doors, you pick one, and Monte shows you which of the
other doors is the wrong choice. The key: he is showing you all but one of the remaining doors, and
he will only show you ones which are wrong. The best bet is obfuscated by the fact that there're only
two other doors - by showing you only one wrong answer, as that's all that are available without
totally   giving     it   away,    the     magnitude     of   the   new     information     gets    lost.
Think of this instead: I have a standard deck of poker cards. The winner is the player with the ace of
spades. You get to pick one, face down, and do not get to look at it yet. I have 51 cards left. I thumb
through them and show you 50 of them that are not the AofS. Now, is the one you picked the AofS
(you picked 1 card randomly out of 52), or is it the one I picked out of the remaining 51 cards (I get
to look, and get to pick any of the 51 remaining cards)? I'd expect you to get the AofS correctly on a
single pull, oh... about 1 out of 52 tries. I would expect me to have (and hence pick) the AofS about
51 out of 52 times. Put another way: if you managed to get the AofS on the first pull, I obviously
have to go with something else; otherwise, I am gonna pick the AofS from what is left, and win.
Now, let us get in line with the Monte Hall problem - after we both do our picks, you are betting on
who has the AofS. You can either take the card you picked, or switch and take the one I picked. Go
with          my           choice            -        it       is        the          right          bet.
I have explained this to others before, and some have said, "It isn't the same problem." It IS the
same problem, precisely. It just has the proportion set-up so the result is obvious. Reduce the
number of cards from 52 to 3, and the correct choice is the same.

Below is a convincing argument but not correct
There are 24 possibilities:
12 possibilities of switching
Prize is behind door      Contestant chooses Monty Opens             Switch to    Result
1                       1               2                      3          L
1                       1               3                      2          L
2                       1               3                      2         W
3                       1               2                      3         W
2                       2               1                      3          L
2                       2               3                      2          L
9
1                        2                    3              1           W
3                        2                    1              3           W
3                        3                    1              2           L
3                        3                    2              1           L
2                        3                    1              2           W
1                        3                    2              1           W
The above table shows that if you switch you will win 6 out of 12 times or 50%

Prize is behind door  Contestant chooses Monty Opens                     Result
1                      1                  2                      W
1                      1                  3                      W
2                      1                  3                      L
3                      1                  2                      L
2                      2                  1                      W
2                      2                  3                      W
1                      2                  3                      L
3                      2                  1                      L
3                      3                  1                      W
3                      3                  2                      W
2                      3                  1                      L
1                      3                  2                      L
The above table shows that if you do not switch you will win 6 out of 12 times or 50%

Greedy Pig
Everyone stand up. Teacher rolls a dice. If the number is a two, you are out. Any other number
you add it to your score. You can sit down at any point.

Reaction time
Work in pairs. One person drops the ruler and the other person catches it. Collect classes’ results.
From the results, much statistics can be calculated. I.e. Probability person’s reaction time is bigger
than…., smaller than… Excellence question could be what value are 60% of students better then.

Weetbix packet
In every Weetbix packet there is a playing card. Either a Jack, Queen, King or Ace of hearts. How
many packets are needed to get a full set? A good use of random numbers here to simulate the
choosing of cards.

Mini Lotto
Choose two numbers from 1 – 6. How many random numbers from 1 – 6 are required to win.
Simulate with dice
Theoretical probability can be found using trees
Once mean and standard deviation are found, questions using normal distribution can be asked i.e.
Find the probability that a student wins after 3 goes.

Homework
Sara has maths on 4 days each week and every day she has homework. Her teacher checks the
class’s homework on only 1 of those days. This day is chosen at random by the teacher. Students
who have not done their homework for that day get a detention. Sara says she did her homework
on 32 out of the 40 days on which she had maths last term.
Describe a simulation (probability experiment) that Sara could use to predict the probability that:
i)     she had not done her homework
ii)    the teacher checked

10
Casio Graphic Calculator: Generation of random numbers
1) Run mode
2) OPTN
3) F6 (arrow right)
4) PROB (F3)
5) Ran# (F4)
6) EXE
Take care with trying to generate only certain random numbers. E.g. Pressing 6Ran# will
only give numbers between 0 and 5.999999, not between 1 and 6. A way round this is
choose a number one bigger then you require e.g. to generate numbers from 1 to 8 press
9Ran# and ignore any 0’s that appear a the start of the number.

MERIT: (Use simulations and theoretical probability
techniques to solve problems AND Solve problems
using the normal distribution model)
Probability Trees
Probability trees are used to show with a picture what is going on mathematically. (I always use
them!)

Although both spinners are spun simultaneously, they can be treated as being spun one after the
other

Independent
Green
Green                       Yellow

Blue
Red

Spinner A                          Spinner B
P(Red) = ½                        P(Yellow) = 2/3
P(Blue) = 1/3                     P(Green) = 1/3
P(Green) = 1/6

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P(Y) = 2/3

P(R) = ½
P(G) = 1/3
                          +
P(Y) = 2/3

P(B) = 1/3                                                           =
P(G) = 1/3
1
P(G) = 1/6

P(Y) = 2/3

P(G) = 1/3

Using the above tree diagram, any question can now be easily asked. This is a good time to use
AND =  & OR = +. Describe the situations using the words AND & OR, this well indicate whether

1 2 2 1
P(Red and Yellow) =         
2 3 6 3

1 1        1 17
P(not the same colour) = 1 -    = 1    
6 3       18 18

 1 1 1 1 1 2 1 1 2      7
P(only one green) = P(RG or BG or GY) =        =     =
 2 3 3 3 6 3   6 9 18  18
Page 281 Ex 26.3

Probability without Replacement (Conditional)
Events cannot happen at the same time as the result of the first event affects the second event.

12
5 balls are placed in a bag (3 black & 2 white). A ball is chosen and put to one side. A second ball
is chosen and placed to one side. This is the same as choosing two balls at once.

P(B) = 2/4

P(B) = 3/5
P(W) = 2/4

P(W) = 2/5         P(B) = 3/4

P(W) = 1/4

3 2  6   3
P(B and then W) =        
5 4 20 10

 3 2 2 1  6   2 8     2
P(both the same colour) = P(BB or WW) =               
 5 4 5 4   20 20  20 5
 3 2      6 14 7
P(not both Black) = 1 -     1       
5 4       20 20 10
Page 286 Ex 26.4

More than one z-value
Page 358 Ex 30.3 Q1ghi, 2ghi

Excellence: (Interpret the outcome of modelling a
probability situation)
Inverse Normal
Inverse normal problems: This is the reverse of the previous work and as such the tables are used
in reverse i.e. probabilities are known and you are asked to find the Z value.

E.g. 1) Find the value of Z if P(0 < Z < k) = 23.89% (this is an easy example compared to the
exercises) You still need to draw diagrams.

0      k

23.89% = 0.2389 (from table this gives a Z score of 0.64).

13
If we had known mean and standard deviation of data we could work out X value i.e. using the
X
rearrangement of Z       (on formula sheet) i.e. X  Z   (not on formula sheet)


I.e. in the above example if the mean was 56 and the standard deviation was 3, X = 3(0.64) + 56 =
57.92

E.g. 2)
Find k if P(Z > k) = 0.2576
Draw diagram                                   0.2576

0     k

0.5 – 0.2576 = 0.2424

Look up 0.2424 in body of table.
0.65 gives 0.2422
0.651 gives 0.2425 (this is closer) therefore answer = 0.651
Page 364 Ex 30.5 not Q8(finding Z score given probability from diagrams and values of
probability)

Casio Graphic Calculator: Inverse Normal Distribution
Using the calculator at this stage will hinder your understanding of this topic and excellence exam
questions will require more than just an answer to obtain full marks.
1) STAT mode (2)
2) DIST (F5)
3) NORM (F1)
4) InvN (F3)
5) Enter the area (this is the area from the LHS to the point in question)
6) Enter standard deviation
7) Enter mean
8) CALC (F1)
E.g. Using a mean of 56 and a standard deviation of 3, find the value of Z if P(0 < Z < k) =
23.89%
We need the value from the left so the area is 0.5 + 0.2389 = 0.7389
1) Enter area of 0.7389
2) Enter 3 for standard deviation
3) Enter 56 for mean
4) CALC (F1)
The answer is 57.919 (same as example above)

14
Practical problems
The Year 11 girls have a mean height of 1.59m and a standard deviation of 7 cm. Jo is 1.69m but is
not sure of the minimum height that is required to join the basketball team. She knows to keep the
numbers down the manager just lets the tallest 15% of students join. Can she join?

15%

1.59

Area between 1.59 and x is 0.35 (0.5 – 0.15)
In body of table, area of 0.35 gives Z score of 1.036 or 1.037

Substitute either value of Z into X  Z  
X = 0.07(1.036) + 1.59
This gives 1.66m so yes she can play.
Page 364 Ex 30.5 Q8
Page 366 Ex 30.6 (practical inverse problems)

For the exam students are required to know the difference between theoretical and experimental
probability.

Theoretical probability: use the shape or information you know about events to predict the
probability of outcomes occurring. E.g. the probability of obtaining heads on a coin and 2 on a dice
1 1 1
is  
2 6 12

Experimental probability: use the results of experiments to predict the probability of outcomes
occurring. E.g. To find the experimental probability of obtaining heads on a coin we flick the coin
numerous times and note the number of times the event happened.

TEST
1) Go to the following website for past papers: http://www.nzqa.govt.nz/ncea/ and then type in
the number 90289
2) Use the following link: www.studyit.org.nz
3) See Achieve mind map below

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Expected    Relative
AND
number    frequency
OR

Basic
On formula                  One value                                                                Not
sheet.                    of z only.                                Probability

X                                         Fraction,             Easy probability

Standard normal  z                                             decimal or                  trees
       
percentage
2.6
Achieve
(Normal)
Normal
distribution
(likely/probable
= 68%  1 ),
&
Bell shape
curve
Design and use
a simulation
(very likely/very             (almost
probable =                 certain =                          Dice or
95%  2 )               99%  3 )                         random
numbers etc

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