Download Probability

-
Expected values
Trees, tables and Venn diagrams
Factorials, combinations and permutations
Conditional probability
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This standard has the highest failure rate of
all standards in statistics.
Last year approximately 75% of those who
sat the old probability standard from KBHS
did not achieve.
A standard with very simple principles and
mathematical processes.
Commonly misunderstood questions and
incorrect methods used.
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1. Repeated practice (hundreds of times)
▪ By familiarising yourself with questions you can ‘feel’ the
correct method to use.
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2. Understanding the question
▪ You are incredibly likely to use the wrong methods if you
do not carefully READ the question several times.
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3. Common sense
▪ Like most things in statistics, the most simple way to
approach a question is to forget about the mathematics
and simply consider the problem.
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The number 1 skill you will need to develop
throughout this topic is to select the correct
representation for a situation.
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This will generally be one of 3 things- a tree
diagram, a Venn diagram or a table.
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The situation will dictate which one makes
most sense.
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Expected values are estimates of a mean
outcome from the set of all possibly
outcomes which are weighted by probability.
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We take the value of each event outcome and
multiply it by the outcome probability. The
sum of these results gives us the expected
value.
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Expected value (µ):
𝐸(𝑋) =
(𝑥𝑖 𝑝𝑖 )
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Variance and standard deviation are two very
common measures of spread. The standard
deviation is the square root of the variance.
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VAR(X) = SD(X)2 or (σ2)
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And
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σ =
𝑉𝐴𝑅
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The variance of a random variable is the
expected squared distance of each individual
x from its mean.
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These distances are weighted as with E(X).
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𝑉𝐴𝑅 𝑋 = 𝐸 𝑋 2 − 𝐸(𝑋)2
x
1
2
3
p
0.5
0.25
0.25
E(X):
Each individual x is weighted:
1 × 0.5 + 2 × 0.25 + 3 × 0.25
0.5
+
0.5
+
0.75
E(X) = 1.75
E(X²):
Each x value is squared and weighted
12 × 0.5 + 22 × 0.25 + (32 × 0.25)
1 × 0.5 + 4 × 0.25 + (9 × 0.25)
0.5
+
1
+
2.25
E(X2) = 3.75
VAR(X):
The expected value of X is squared and subtracted from the expected value of X²
VAR(X) = 3.75 − 1.752
VAR(X) = 0.6875
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When given a linear function ‘y’ that acts
upon a random variable ‘x’ we can also find
expected values and variance of that
function- E(Y) and VAR(Y).
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From working through a simple example we
can see how the function affects E(X) and
VAR(X).
x
1
2
3
p
0.5
0.25
0.25
E(X):
Each individual x is weighted:
1 × 0.5 + 2 × 0.25 + 3 × 0.25
0.5
+
0.5
+
0.75
E(X) = 1.75
E(X²):
Each x value is squared and weighted
12 × 0.5 + 22 × 0.25 + (32 × 0.25)
1 × 0.5 + 4 × 0.25 + (9 × 0.25)
0.5
+
1
+
2.25
E(X2) = 3.75
VAR(X):
The expected value of X is squared and subtracted from the expected value of X²
VAR(X) = 3.75 − 1.752
VAR(X) = 0.6875
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Now lets say a function of y=2x+3 is applied
to this events outcome.
x
1
2
3
p
0.5
0.25
0.25
y
5
7
9
p
0.5
0.25
0.25
Earlier we worked
out that:
E(X) = 1.75
VAR(X) = 0.6875
y
5
7
9
p
0.5
0.25
0.25
E(Y):
Each individual y is weighted:
(5 x 0.5) + (7 x 0.25) + (9 x 0.25)
2.5 + 1.75 +
2.25
= 6.5
VAR(Y):
E(Y²) – E(Y) ²
45 – 6.5²
45 – 42.25
= 2.75
E(Y²):
Each individual y² is weighted:
( 5² x 0.5) + (7² x 0.25) + (9² x 0.25)
(25 x 0.5) + (49 x 0.25) + (81 x 0.25)
12.5 + 12.25 + 20.25
= 45
For reference:
E(X) = 1.75
VAR(X) = 0.6875
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The linear function is y=2x+3
E(X) = 1.75
E(Y) = 6.5
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By putting 1.75 in for x we find that:
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E(aX + b) = a x E(X) + b
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▪ or
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E(Y) = aE(X) + b
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The linear function is y=2x+3
VAR(X) = 0.6875
VAR(Y) = 2.75
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VAR(Y) = a² x VAR(X)
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The relationship is difficult to see at first, but
makes sense we consider the reasons.
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Remember, the variance measures the
squared distance, it is influenced by the
square of the co-efficient in the function (a²).
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Also remember, variance measures spread.
By adding a constant we change all the
outcome values but their relative difference is
the same (the difference between the
numbers). Constants have no effect.
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If we have two independent variables that
are combined we establish several
relationships. Most of these are based on
theories outside the scope of this course so
the theory behind these are not required
though you do need to be able to compute
them.
If variables are not independent they cannot
be combined.
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If we have two random variables (in this case
labelled X and Y) that are to be added we find
that:
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E(T) = E(X+ Y) = E(X) + E(Y)
▪ And
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VAR(T) = VAR(X+Y) = VAR(X) + VAR(Y)
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The mean weight of a can is estimated to be
30 grams with a standard deviation of 3
grams. The mean weight of beans in the can
is estimated as 470 grams with a standard
deviation of 6 grams. Find the mean total
weight of a can of beans along with its
variance.
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E(X) = 30 grams
SD(X) = 3 grams
VAR(X) = 9 grams
E(T) = E(X) + E(Y)
E(T) = 30 + 470
E(T) = 500 grams
VAR(T) = VAR(X) + VAR(Y)
E(Y) = 470 grams
VAR(T) = 9 + 36
SD(Y) = 6 grams
VAR(Y) = 36 grams VAR(T) = 45 grams
SD(T) = 6.7 grams
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The difference between two variables is a
little bit different. As we expect:
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E(T) = E(X –Y) = E(X) – E(Y)
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But…
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VAR(T) = VAR(X –Y) = VAR(X) + VAR(Y)
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This is because of the squared term in
functions of variables.
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Consider:
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T = X –Y
VAR(T) = (+1)²VAR(X) (-1)²VAR(Y)
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▪ so
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VAR(T) = (+1)VAR(X) (+1)VAR(Y)
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The average diameter of the inside of a nut is
4.5mm with a variance of 0.005mm. The
average diameter of a bolt is 4.4mm with a
variance of 0.004mm. Find the average gap
between a nut and a bolt.
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E(X) = 4.5 mm
E(Y) = 4.4 mm
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VAR(X) = 0.005 mm
VAR(Y) = 0.004 mm
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E(T) = E(X –Y) = E(X) – E(Y)
E(T) = 4.5 – 4.4 = 0.1 mm
VAR(T) = VAR(X –Y) = VAR(X) + VAR(Y)
VAR(T) = 0.005 + 0.004 = 0.009 mm
SD(T) = √0.009
SD(T) = 0.094 mm
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Combining all of the information we have
learned so far we can see that:
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Z = aX + bY
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E(Z) = E(aX + bY) = aE(X) + bE(Y)
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VAR(Z) = VAR(aX + bY) = a²VAR(X) + b²VAR(Y)
▪ Assuming x and y are independent.
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A truck carts a mixture of two different types
of metal. When loading, the truck received 3
scoops of small metal and 4 scoops of large
metal. From experience, the driver knows the
average weight of a scoop of small metal is
500kgs and the average weight of a scoop of
large metal is 400kgs. If the variance of
weights are 200kgs for small metal and
250kgs for large metal, estimate the trucks
weight and variance.
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Z = 3X + 4Y
▪ The total load is 3 small scoops (X) and 4 large scoops (Y).
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E(Z) = 3(500) + 4(600)
E(Z) = 3900 kgs
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VAR(Z) = 3² x 200 + 4² x 250
VAR(Z) = 5800 kgs
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SD(Z) = 76.16 kgs
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