Transcript PowerPoint - Paradigm Lost

Probability
Normal Distribution
What is Normal Distribution?
Any event can have at least one possible
outcome.
A trial is a single event. An experiment consists
of the same trial being performed repeatedly
under the same conditions.
If an experiment is performed with enough
trials, the populations of each possible outcome
this
This
is aistypical
the
can be distributed according to different
symmetrical
Poisson
patterns.
Gaussian,
Distribution.
or Normal
Note
Distribution.
the lack of
Learn
symmetry.
this! notice the numbers. We’ll deal with them later
How it works...
•The Normal Distribution is characterised by
grouped continuous data.
•The typical graph is a histogram of the populations
of each
grouped
For
example,
if werange of possible outcome values
returned to the era of
norm-standardised
testing for NCEA, then
the distribution of test
scores, as percentages,
would look something
To
likepass,
this: you would need a score of 50% or greater.
Notice about 50% of all candidates achieved this
that 50% pass-rate is quite
important.
•For any Normally-Distributed
data, the central peak is the
mean, μ.
•AND
•50% of all data is <μ;
•which means 50% of all data is
a quick bit of
revision...
•the standard deviation now comes into its own.
•Recall:
•For a set of continuous data, the mean, μ, is a
measure of central tendancy - it is one value that
represents the peak data value population.
•The majority of the data does not equal μ.
•The standard deviation, σ, is analogous to the
mean difference of every data value from μ.
and now, back to the graph and it’s
numbers...
features of Normal Distribution...
the x-axis is
asymptotic
95% of all data
lie within 2σ of μ
the peak is the mean, μ.
50% of the data lie either
side of μ.
68% of all data lie
within 1σ of μ
99% of all data lie
within 3σ of μ
the distribution is
symmetrical about
μ
summary:
the x-axis is asymptotic
the peak is the mean, μ.
the distribution is symmetrical about μ; 50% of the
data lie either side of μ.
68% of all data lie within 1σ of μ
}
95% of all data lie within 2σ of μ
these
percentages are
rounded
99% of all data lie within 3σ of μ
this distribution lets us calculate the probability
that any outcome will be within a specified range
of values
Ah, the wonder that is
Z
u
•The Normal Distribution of outcome frequencies is
defined in terms of how many standard deviations
either side of the mean contain a specified range of
outcome values.
•In order to calculate the probability that the
outcome of a random event X will lie within a
specified multiple of σ either side of μ, we use an
intermediate Random Variable,
X - μ Z.
Z=
σ
For this relationship to hold true, for Z, σ = 1 and μ = 0; hence, for
a Normally distributed population, the range is from -3σ to +3σ
Z, PQR, and You
•Probability calculations using Z give the likelihood that an
outcome will be within a specified multiple of σ from the mean.
There are three models used:
P(t) = the probability that an outcome t
is any value of X up to a defined
P(Z<t) ≡ P(μ<Z<t) + 0.5
multiple of σ beyond μ
Q(t) = the probability that an outcome t
is any value of X between μ and
a defined multiple of σ
P(μ<Z<t)
R(t) = the probability that an outcome t
is any value of X greater than
a defined multiple of σ below μ
P(Z>t)
Solving PQR
Problems
1. Enter
RUN
mode. carefully.
Read the problem
2. OPTN
Draw
a
diagram
sketch
the
Bell
3. F6
Curve,
and
use
this
to
identify
the
4. F3
problem
as
P,
Q
or
R
5. F6
ThisYou
gives
the
F-menu
for
PQR.
could now use the Z probability
1. Choose
the
function
(P,
Q
or
R)
tables to calculate P, Q or R, or use
appropriate
to
your
problem;
a Graphic Calculator such as the
Casio fx-9750G Plus
2. enter the value of t, and EXE.
Calculating Z from Real
Data
The PQR function assumes a perfectly-symmetrical distribution
about μ. Real survey distributions are rarely perfect.
For any set of real data, we can calculate μ and σ, and therefore
Z.
For example, if μ=33 and σ=8, then to find P(X<20):
SO... use
X
μ
Z= σ
to calculate Z, and
then use PQR.
[ ]
[ ]
20-μ
P(X<20) = P(Z <
)
σ
20-33
= P(Z <
) Now, use the R
8
function, and
= P(Z < -1.625) subtract the
result from 1.
continuity
• Normally-distributed data is often continuous.
• If asked to calculate probability for continuous
data above a value q, apply the principle of
measurement error, and take 0.5 the basic unit
above the stated value.
• This is because any value in the range q-0.5 to
q+0.5 will be recorded as q.
Inverse Normal
This is the reverse process to finding the
probability.
Given the probability that an event’s outcome will
lie within a defined range, we can rearrange the Z
X=Zσ+
equation to give
μ
But...
k
we cannot define X, as it represents the
entire range of values of all possible
outcomes.
What the equation will give us is the
is thevalue
upperk.or lower limit of the range of X
that is included in the P calculations
an example...
•withif X is a normally-distributed variable
•is k?σ=4, μ=25, and p(X<k) = 0.982, what
Use the
PQRway...
model,
and sketch
The
using
along
graphic
calculator,
for a
bell curve to identify the regions
example the trusty Casio fx9750G;
being included in the p range.
Use the ND table to find the value
Z:
•ofMODE:
STATS
Z range is from -1 to +1, so find
•0.982
F5 ➜ F1 ➜ F3
- 0.5 = 0.482
•Area = probability, as a decimal
σGives
= Z = 2.097
μ=
So, k = 4 x 2.097 + 25 = 33.388
EXECUTE
Combinations of
Variance
• The real world is rarely a simple place.
However, apparently complex
relationships can be rationalised to form
straightforward equations.
• In addition to functional relationships that
involve a single Random Variable, there
can be interactions between two or more
independent random variables, X and Y.
Sum of Random
Variables
• For each Random
Variable there is a calculable
variance - this is true irrespective of the number of
possible outcomes.
• Sums of Random Variables occur when we want
the likelihood of a specific pair of outcomes (T)
from two independent events;
• X+Y=T
• Simply,
• VAR(T) = VAR(X + Y) = VAR(X) + VAR(Y)
• and
• VAR(X - Y) = VAR(X) + VAR(Y)
Got it? Whether
adding or subtracting,
you always add the
independent
Variances!
Linear Combinations
of Random Variables
•We know that for a single Random Variable X, the
linear function is
•E(aX + b) = aE(X) + b,
and VAR(aX + b) =
2
a VAR(X)
NOTE:
•If we introduce a second Random Variable
Y,this only
holds true if X
and Y are
•E(aX + bY) = aE(X) + bE(Y)
independent
•so
•VAR(aX + bY) = a VAR(X) + b VAR(Y)
2
2
an example...
•A hydroponic lettuce grower has her weekly costs expressed by two
random variables - the number of plants X, and the liquid fertiliser
concentrate costs Y. Both variables are independent. The standard
deviation of X is 150, and the standard deviation of Y is 4 litres. Each
lettuce costs $0.50 to irrigate and each litre of concentrate costs $10.
Find the standard deviation of her costs.
•X: σ = 150, so VAR(X) = 150 = 22500
•Y: σ = 4, so VAR(Y) = 4 = 16
•VAR(0.5X + 10Y) = 0.5 VAR(X) + 10 VAR(Y)
•
= (0.25 x 22500) + (100 x 16)
•
= 5625 + 1600
•
= 7225
• so, σ = 7225 = $85
2
2
2
½
2