1) Test to see if 
can serve as a probability distribution function
for x = 1, 2, or 3. If yes, make a table of values, find
Random Variables & Probability Distributions |
Discrete and Continuous Random Variables
Random Variable is a quantity that assumes a numerical value as a result of an experiment.
ex: the number of heads obtained in 10 flips of a coin, the heights of certain plants in an
experimental garden or the salaries of employees at a given company.
Discrete Random Variable can assume only certain distinct values.
ex: the number of boys in a family of 6 children.
Continuous Random Variable can assume a continuum of values over a given interval.
ex: the weight of coffee in a certain size jar.
These 2 types of random variables result in 2 types of probability distributions.
Discrete distributions are based on counting successes rather than measuring data
like a continuous distribution does. In other words, the data values are frequencies
instead of measures.
For example, if we're counting the number of customers in a Metro Store who buy
the ice cream on special, we would label a "buy" as a success and we'd
build our probability distribution from the integer frequencies recorded.
However, if we're measuring the heights of grade 9 students in Montreal,
our data would be a range of real numbers between the shortest and the tallest student.
Probability Distributions
A Probability Distribution is a function which assigns probabilities to the values
of a random variable resulting from an experiment. So, it is a listing of all possible
outcomes of an experiment and the corresponding probabilities.
We make a table of x and P(x)
Note: the symbol S (sigma) means summation. So S f (x) = 1 means that if we
add up all the probabilities listed, we will get exactly 1.
.
Conditions on a Probability Distribution Function
For f (x) to serve as a probability distribution, 2 conditions must be met.
1) 0 [ f(x) [ 1 for all x
the probabilty of any outcome is greater than or equal to 0 and less than or equal to 1
2) Sf(x) = 1 for all x
The sum of all the probabilities is exactly 1.
Example 1:
The table shows the distribution for the number of points
rolled on a balanced die. (1 dice cube)
| points | probability |
| 1 | 1/6 |
| 2 | 1/6 |
| 3 | 1/6 |
| 4 | 1/6 |
| 5 | 1/6 |
| 6 | 1/6 |
Note that sum of all the probabilities = 1 and each one falls between 0 and 1.
.
The Mean (µ) and the Variance ( r 2) of a Probability Distribution
The mean µ of a probability distribution = S xP(x)
We take the sum of each random variable value times its probability.
The variance (square of the st. dev.) r 2 = S[(x – µ)2 P(x)] for a probability distribution.
We take the sum of the square of the difference between the value (x) and the mean
times the probability of x for each x-value in the distribution.
.
For the distribution in the table above,
l = 1(1/6) + 2(1/6) + 3(1/6) + 4(1/6) + 5(1/6) + 6(1/6) = 21/6 or 3.5
r 2 = (1 – 3.5)²(1/6) + (2 – 3.5)²(1/6) + (3 – 3.5)²(1/6) + (4 – 3.5)²(1/6) + ... = 2.92
the standard deviation is the square root of the variance = 1.71.
.
Practice
1) Test to see if can serve as a probability distribution function
for x = 1, 2, or 3. If yes, make a table of values, find
.
2) Test to see if 
can serve as a probability distribution function
for x = 1, 2, 3 or 4. If yes, make a table of values, find l and find r with formulas.
.
3) Using the data in the table from #2, find the probability that x = 2 or 3.
.
4) Test if the data in this table is a probability distribution. If yes, find l and r .
| value | probability |
| 0 | 0.10 |
| 1 | 0.40 |
| 2 | 0.30 |
| 3 | 0.10 |
| 4 | 0.10 |
.
Solutions
1) 
can serve as a probability distribution function.
| points | probability |
| 1 | 4/15 |
| 2 | 5/15 |
| 3 | 6/15 |
| total | 15/15 = 1 |
l = 1(4/15) + 2(5/15) + 3(6/15) = 32/15 = 2.133
=
[(1 – 2.133)²(4/15) + (2 – 2.133)²(5/15) + (3 – 2.133)²(6/15)]1 / 2 = 0.806
.
2) 
can serve as a probability distribution function for x = 1, 2, 3 or 4.
| value | probability |
| 1 | 1/30 |
| 2 | 4/30 |
| 3 | 9/30 |
| 4 | 16/30 |
| total | 30/30 = 1 |
l = 1(1/30) + 2(4/30) + 3(9/30) + 4(16/30) = 99/30 = 3.3
=
[(1 – 3.3)²(1/30) + (2 – 3.3)²(4/30) + (3 – 3.3)²(9/30) + (4 – 3.3)²(16/30)]1 / 2 = 0.83066
.
3) From the table, P(x = 2) = 4/30 and P(x = 3) = 9/30, so P(x = 2 or 3) = 13/30 or 0.4333.
.
4) Yes, the data in the table is a probability distribution since all probabilities
are between 0 and 1 and the sum of the probabilities = 1.
l = 0(.10) + 1(.40) + 2(.30) + 3(.10) + 4(.10) = 1.7
=
[(0 – 1.7)²(.10) + (1 – 1.7)²(.40) + (2 – 1.7)²(.30) + (3 – 1.7)²(.10) + (4 – 1.7)²(.10)]1 / 2 = 0.795
.
.
.
(all content of the MathRoom Lessons © Tammy the Tutor; 2002 - ).