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.

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**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.

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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.

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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.

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**Practice
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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 l and find r with formulas.

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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.

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3) Using the data in the table from #2, find the probability that *x = 2* or *3*.

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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 |

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**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

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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

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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.

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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

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*(all content of the MathRoom Lessons **© Tammy the Tutor; 2002 - ).*