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

1 2 3 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

1 2 3 4 5 6 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 manynumbers 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

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

The mean of discrete random variables

(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

最终的 27 Total Brutal Hours Spent Developing this PowerPoint between our group members.

This is how we prepared

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