Random Variables and Probability Distributions Notes

Random Variables

Discrete random variables are random variables whose possible values can be listed.
- Examples:
  - The number that comes up on the roll of a die.
  - The number of siblings a randomly chosen person has.

Continuous random variables are random variables that can take on any value in an interval.
- Examples:
  - The height of a randomly chosen college student.
  - The amount of electricity used to light a randomly chosen classroom.

Specifies the probability for each possible value of the random variable.
Properties:
- 0 ≤ P(X) ≤ 1 for every possible X.
- The sum of all of the probabilities must equal one.

Example 1: Not a probability distribution because the probability of 3 is not between 0 and 1.
Example 2: A probability distribution because all of the probabilities are between 0 and 1, and they all add up to 1.
Example 3: Not a probability distribution because the probability of 1 is not between 0 and 1.

1,000 parking spaces:
- 142 covered long-term spaces: $2/hour
- 378 covered short-term spaces: $4.50/hour
- 423 uncovered long-term spaces: $1.50/hour
- 57 uncovered short-term spaces: $4/hour
Let X represent the hourly parking fee for a randomly selected space. Find the probability distribution of X.
- Possible values of X: $1.50, $2, $4, $4.50
- Probabilities:
  - P(1.50) = \frac{423}{1000} = 0.423
  - P(2) = \frac{142}{1000} = 0.142
  - P(4) = \frac{57}{1000} = 0.057
  - P(4.50) = \frac{378}{1000} = 0.378

Example: Probability distribution and histogram for the number of boys in a family of five children.

Let X be the number of heads that come up. Find the probability distribution of X.
Four equally likely outcomes: HH, HT, TH, TT
- Possible values for X: 0, 1, 2
- Probabilities:
  - P(X = 0) = \frac{1}{4} = 0.25
  - P(X = 1) = \frac{2}{4} = 0.5
  - P(X = 2) = \frac{1}{4} = 0.25

The mean of a random variable provides a measure of center for the probability distribution of a random variable.
To find the mean of a discrete random variable, multiply each possible value by its probability, then add the products.
- \mu = \sum X \cdot P(X)

Let X represent the number of defective pixels on a randomly chosen monitor.
Probability distribution of X:
- P(0) = 0.2
- P(1) = 0.5
- P(2) = 0.2
- P(3) = 0.1
Find the mean number of defective pixels:
- \mu = (0 \cdot 0.2) + (1 \cdot 0.5) + (2 \cdot 0.2) + (3 \cdot 0.1) = 1.2

Probability 0.4 of a $30,000,000 loss
Probability 0.5 of a $20,000,000 profit
Probability 0.1 of a $40,000,000 profit
Let X represent the profit. Find the probability distribution of the profit and the expected value of the profit.
Probability distribution of X:
- P(-30) = 0.4
- P(20) = 0.5
- P(40) = 0.1
E(X) = (-30 \cdot 0.4) + (20 \cdot 0.5) + (40 \cdot 0.1) = 2
There is an expected gain of $2,000,000.

\mu = 1.2
\sigma^2 = (0^2 \cdot 0.2) + (1^2 \cdot 0.5) + (2^2 \cdot 0.2) + (3^2 \cdot 0.1) - 1.2^2 = 0.76
\sigma = \sqrt{0.76} = 0.872

Probability distribution of X (number of occupants):
- P(1) = 0.7
- P(2) = 0.15
- P(3) = 0.1
- P(4) = 0.03
- P(5) = 0.02
Find P(2): P(2) = 0.15
Find P(more than 3): P(4) + P(5) = 0.03 + 0.02 = 0.05
Find the probability that a car has only one occupant: P(1) = 0.7
Find the probability that a car has fewer than four occupants: P(1) + P(2) + P(3) = 0.7 + 0.15 + 0.1 = 0.95
Compute the mean:
- \mu = (1 \cdot 0.7) + (2 \cdot 0.15) + (3 \cdot 0.1) + (4 \cdot 0.03) + (5 \cdot 0.02) = 1.52
Compute the standard deviation:
- \sigma^2 = ((1 - 1.52)^2 \cdot 0.7) + ((2 - 1.52)^2 \cdot 0.15) + ((3 - 1.52)^2 \cdot 0.1) + ((4 - 1.52)^2 \cdot 0.03) + ((5 - 1.52)^2 \cdot 0.02) = 0.8696
- \sigma = \sqrt{0.8696} = 0.933