Probability Distributions Overview

Study Notes

Probability Distributions

A probability distribution is a function that describes the likelihood of obtaining each possible value of a random variable.
The sum of all probabilities in a probability distribution must equal 1.

Example 1

In example (a), the sum of probabilities is 10/9, which is not equal to 1, so it is not a probability distribution.
In example (b), the sum of probabilities is 1, so it is a probability distribution.

Calculating Probability

The probability of obtaining a specific value (x) in a binomial distribution is calculated by using the binomial probability formula:
P(x)=(nx)px(1−p)n−xP(x) = \binom{n}{x} p^{x} (1 - p)^{n-x}P(x)=(xn)px(1−p)n−x, where:
- n: number of trials
- x: number of successes
- p: probability of success on a single trial

Mean, Variance, and Standard Deviation

Mean: The expected value of a random variable.
Variance: A measure of the spread of the distribution.
Standard Deviation: The square root of the variance, also a measure of spread.

Calculating Mean, Variance, and Standard Deviation

The mean of a discrete probability distribution can be calculated using the formula: μ=∑iXiP(Xi)\mu = \sum_{i} X_{i}P(X_{i})μ=∑iXiP(Xi), where XiX_{i}Xi is the value of the random variable and P(Xi)P(X_{i})P(Xi) is the probability of obtaining that value.
The variance of a discrete probability distribution can be calculated using the formula: σ2=∑i(Xi−μ)2P(Xi)\sigma^{2} = \sum_{i} (X_{i} - \mu)^{2} P(X_{i})σ2=∑i(Xi−μ)2P(Xi).
The standard deviation of a discrete probability distribution is the square root of the variance: σ=σ2\sigma = \sqrt{\sigma^{2}}σ=σ2.

Probability Distributions

A probability distribution is valid only when the sum of probabilities equals 1.
Problem 1 (a) is not a probability distribution because the sum of all probabilities is 109\frac{10}{9}910, which is not equal to 1.
Problem 1 (b) is a probability distribution because the sum of all probabilities is 1.

Binomial Distribution

The binomial distribution is a discrete probability distribution that describes the probability of obtaining a specific number of successes in a sequence of n independent trials, each of which can have only two possible outcomes (success or failure).
In Problem 1 Part B, a graduate of XY University has a 75% chance of getting a job in their field, making it a binomial distribution.
The probability of exactly 4 out of 7 graduates getting hired is approximately 0.1711, calculated using the formula: P(X=k)=(nk)pk(1−p)n−kP(X=k) = \binom{n}{k} p^{k} (1 - p)^{n-k}P(X=k)=(kn)pk(1−p)n−k, where n is the number of trials, k is the number of successes, and p is the probability of success.

Mean, Variance, and Standard Deviation

In Problem 2, Flower World determines the probabilities for the number of flower arrangements they deliver each day, this can be represented as a probability distribution.
The mean of the distribution is calculated as the sum of the product of each value and its corresponding probability, which is 7.6 in this case.
The variance is calculated as the sum of the squared differences between each value and the mean, weighted by the probability of that value, resulting in a variance of 1.16.
The standard deviation is the square root of the variance, and is approximately 1.13 in this case.

Description

This quiz covers the fundamentals of probability distributions, including their definitions and properties. It explains how to calculate probabilities, mean, variance, and standard deviation in the context of binomial distributions. Test your understanding of these essential concepts in probability theory.