Transcript continuous - Walton High

First we need to understand
the variables.
A random variable is a value of an outcome such as
counting the number of heads when flipping a coin,
which is an example of a discrete random variable.
These variables can also be in interval form like the
range of the mid 50% of SAT scores in which they are
known as continuous random variables.
More on Discrete Random
Variables
• X has a countable number of
possible values
• Has to take in account whether
or not it is asking for less or
greater than and less than or
equal equal to
• X> 2 is different from X≥ 2
• The probability of X> 2 is .6
• Probability of X≥ 2 is .9
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4
.1
.3
.4
.2
Probability of Discrete Random
Variables
• The probability has to be between 0
and 1
• Sum of probabilities is 1
• P1 + P2 + … + Pk = 1
• Ex from last slide: .1 + .3 + .4 + .2 = 1
If simpler terms don't work for you...
Another Example
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2
3
4
5
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7
.05
.3
.2
.15
.02
.05
.23
Example continued:
• Let us say that we wanted to know X = 4, X < 4, and X ≤
4
• X = 4’s probability would be .15 since we want to know
what is the probability of 4 and nothing else
• X < 4 is equivalent to X ≤3 since it is everything under 4.
The probability of it would be .55 since 1’s probability of
.05 + 2’s probability of .3 + 3’s probability of .2 = .55
• X≤ 4 is everything below 4 including 4 itself. So in
addition of having .05 + .3 + .2, we include the value of 4
into the probability. The probability of X≤4 is .7
Probability Histograms
• Histograms can show
probability distribution and
distribution of data
• The most likely is 3 because it
has the highest probability with
.4
• The least likely is 1 which has
a probability of .1
• It is easy to compare different
distributions with histograms
Continuous Random Variables
• It takes all the numbers in an interval
• It ignores signs like ≤ or ≥ and treats them
just like < or >
• Think of it like a spinner.
• It doesn’t land exactly on one
point, but instead there are many
numbers that are in between
the different numbers
Uniform Distribution and
Continuous Random Variables
• We want all outcomes to
be equally likely
• Can’t assign a probability
to an individual outcome
• We assign probabilities to
areas under a density
curve
• Use ranges for probability
• like P(0 < X < .5) = .5
• P (0 < x < .2) = .2
• P(.5 < x ≤ .8) = .3
ALL Individual outcomes have a
probability of 0
• Since it is continuous, you can’t get 1
exact point
• Say it was .5, .5 has no length on the
uniform distribution.
• It only has length if it is < or >
Normal Distributions as Probability
Distributions
• Normal distributions are probability
distribution
• N(μ, δ ) notation for Normal distribution
• μ is the mean
• δ is the standard deviation
• Formula for standardizing
Normal Distributions as Probability
Distributions continued
• To find the probability of <x, follow the z
score formula and change from z score
into probability
• To find >x, do the z score formula, convert
z score into probability and then find the
complement of it ( 1 – p ) and that is the
probability that it will be above
Example
• A radar unit is used to measure speeds of
cars on a motorway. The speeds are
normally distributed with a mean of 90
km/hr and a standard deviation of 10
km/hr. What is the probability that a car
picked at random is travelling at more than
100 km/hr?
Example continued
• N(μ, δ ) = N(90, 10) Car goes on average 90
kilometers/hour with a standard deviation of
10kilometers/hour
• P( X > 100 ) Trying to find probability of car going faster
than 100kilometers/hour
• Z = (100 – 90)/ 10
• Z=1
• 1 = .8413
• Because we are trying to find greater than 100, we have
to do 1 - .8413 which is .1587
The mean of discrete random
variables
X: (a value of a discrete
random variable)
P(x): the probability of
the x value occuring
To find the mean
multiply X by P(x) for
each pair and add the
products together.
The mean is
represented by
(1x.10) + (2x.30)...
Too many letters >.<
Example continued
• You can also do it with a calculator
• 2nd, Vars, normalcdf( ,
• For lower you would enter 100 since we are trying to find greater
than 100
• Upper you can basically put it as an insanely high number like
9999999999999999999999999999999999
• μ would be 90 since that is the average speed of the cars
• δ you would put 10 since that is the standard deviation that was
given
• Then paste it and hit enter and you should .1586552596 which is
slightly different than using the z-score table, but that is due to round
off error.
Simplify it further
• Input X in L1 and P(x)
in L2
• Click stat, go right to
calc, select 1-Var
Stats
• type (L1,L2)
• ENTER
is your mean for the
discrete random
variables
Finding the variance of a discrete
random variable
The variance of discrete variables can be
found by subtracting the mean of the data
set from each individual variable and
squaring the difference. Then multiply the
result by the probability of that X value.
Repeat for each variable and add each
answer to eachother.
((x1-mean)^2)xP1 + ((x2-mean)^2)xP2....
Or...
Just get the variance from squaring the
standard deviation from 1-Var Stats
(L1,L2)
Stardard deviation of discrete
variables?
If you don't know
how to get the
standard deviation
from the variance
Just square root
the variance...
Textbook Version
The LAW of large numbers
Meaning:
the average of the
values of X observed in
many trials must
approach the mean.
Rules for means
Rule one: If X is a random variable and a and b are fixed numbers, then the
mean of a plus b times x= a+ b times the mean of x.
The face I made when I attempted to understand this formula:
Rule two: If x and y are two independent random variables, then the mean of
X and Y is equal to the sum of the mean of X and the sum of the mea of Y
The Rules of Variances
This is how we prepared
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