In this section, we will delve into the concept of probability distributions, which are fundamental in understanding how probabilities are assigned to different outcomes in a random experiment. This topic is crucial for anyone looking to analyze data and make informed decisions based on statistical methods.

1. Random Variable: A variable that takes on different values based on the outcome of a random event.
2. Probability Distribution: A function that describes the likelihood of different outcomes for a random variable.
3. Discrete vs. Continuous Distributions: Discrete distributions deal with countable outcomes, while continuous distributions deal with outcomes that can take any value within a range.

1. Binomial Distribution: Describes the number of successes in a fixed number of independent Bernoulli trials (e.g., flipping a coin).
2. Poisson Distribution: Describes the number of events occurring within a fixed interval of time or space.

1. Normal Distribution: Describes data that clusters around a mean. It's symmetric and bell-shaped.
2. Exponential Distribution: Describes the time between events in a Poisson process.

Definition: The binomial distribution models the number of successes in a fixed number of independent trials, each with the same probability of success.

Formula: \[ P(X = k) = \binom{n}{k} p^k (1-p)^{n-k} \] Where:

• \( n \) = number of trials
• \( k \) = number of successes
• \( p \) = probability of success in a single trial
• \( \binom{n}{k} \) = binomial coefficient

Example: Suppose you flip a fair coin 10 times. What is the probability of getting exactly 6 heads?

Solution: \[ P(X = 6) = \binom{10}{6} (0.5)^6 (0.5)^4 \] \[ P(X = 6) = \frac{10!}{6!(10-6)!} (0.5)^{10} \] \[ P(X = 6) = 210 \times (0.5)^{10} \] \[ P(X = 6) = 210 \times 0.0009765625 \] \[ P(X = 6) \approx 0.205 \]

Definition: The normal distribution is a continuous probability distribution characterized by a symmetric, bell-shaped curve.

Formula: \[ f(x) = \frac{1}{\sqrt{2\pi\sigma^2}} e^{-\frac{(x-\mu)^2}{2\sigma^2}} \] Where:

• \( \mu \) = mean
• \( \sigma \) = standard deviation

Example: Suppose the heights of adult men are normally distributed with a mean of 70 inches and a standard deviation of 3 inches. What is the probability that a randomly selected man is taller than 73 inches?

Solution: First, convert the height to a z-score: \[ z = \frac{X - \mu}{\sigma} \] \[ z = \frac{73 - 70}{3} \] \[ z = 1 \]

Using standard normal distribution tables or a calculator, find the probability corresponding to \( z = 1 \): \[ P(Z > 1) = 1 - P(Z \leq 1) \] \[ P(Z > 1) = 1 - 0.8413 \] \[ P(Z > 1) = 0.1587 \]

So, the probability that a randomly selected man is taller than 73 inches is approximately 0.1587.

Problem: A factory produces light bulbs, and the probability that a bulb is defective is 0.02. If a sample of 100 bulbs is taken, what is the probability that exactly 3 bulbs are defective?

Solution: \[ P(X = 3) = \binom{100}{3} (0.02)^3 (0.98)^{97} \] \[ P(X = 3) = \frac{100!}{3!(100-3)!} (0.02)^3 (0.98)^{97} \] \[ P(X = 3) \approx 0.180 \]

Problem: The weights of apples in a farm are normally distributed with a mean of 150 grams and a standard deviation of 20 grams. What is the probability that a randomly selected apple weighs less than 130 grams?

Solution: First, convert the weight to a z-score: \[ z = \frac{130 - 150}{20} \] \[ z = -1 \]

Using standard normal distribution tables or a calculator, find the probability corresponding to \( z = -1 \): \[ P(Z < -1) = 0.1587 \]

So, the probability that a randomly selected apple weighs less than 130 grams is approximately 0.1587.

In this section, we covered the basics of probability distributions, including both discrete and continuous types. We explored the binomial and normal distributions in detail, providing formulas, examples, and practical exercises to reinforce the concepts. Understanding these distributions is crucial for analyzing data and making informed decisions based on statistical methods.