Binomial and Normal Distributions
Complete Notes & Formulae for Eleventh Grade (Algebra 2)
1. Find Probabilities Using the Binomial Distribution
What is a Binomial Distribution?
A probability distribution for experiments with EXACTLY TWO outcomes (success or failure)
Requirements (BINS):
• Binary - Only two possible outcomes
• Independent - Trials are independent
• Number - Fixed number of trials (n)
• Same - Same probability for each trial (p)
Binomial Probability Formula:
\[ P(X = x) = \binom{n}{x} p^x (1-p)^{n-x} \]
where:
• \( n \) = number of trials
• \( x \) = number of successes
• \( p \) = probability of success on one trial
• \( 1-p \) = probability of failure (also written as \( q \))
• \( \binom{n}{x} = \frac{n!}{x!(n-x)!} \) = combination formula
Mean and Standard Deviation:
\[ \mu = np \]
\[ \sigma = \sqrt{np(1-p)} \]
Example:
A coin is flipped 5 times. Find P(exactly 3 heads).
Given: \( n = 5, x = 3, p = 0.5 \)
\( P(X=3) = \binom{5}{3}(0.5)^3(0.5)^{2} \)
\( = \frac{5!}{3!2!} \times 0.125 \times 0.25 \)
\( = 10 \times 0.03125 = 0.3125 \)
Answer: 0.3125 or 31.25%
2. Find Probabilities Using the Normal Distribution
Normal Distribution (Bell Curve):
A continuous probability distribution that is symmetric and bell-shaped
\[ X \sim N(\mu, \sigma) \]
Read as: "X follows a normal distribution with mean μ and standard deviation σ"
Properties:
• Symmetric about the mean
• Mean = Median = Mode
• Total area under curve = 1
• Defined by two parameters: μ (mean) and σ (standard deviation)
Empirical Rule (68-95-99.7 Rule):
• About 68% of data falls within 1 standard deviation of mean
• About 95% of data falls within 2 standard deviations of mean
• About 99.7% of data falls within 3 standard deviations of mean
3. Find Z-Values (Z-Scores)
Z-Score Formula:
The z-score tells you how many standard deviations a value is from the mean
\[ z = \frac{x - \mu}{\sigma} \]
where:
• \( x \) = data value
• \( \mu \) = mean
• \( \sigma \) = standard deviation
Interpretation:
• \( z > 0 \): Value is ABOVE the mean
• \( z < 0 \): Value is BELOW the mean
• \( z = 0 \): Value EQUALS the mean
Example:
Test scores: μ = 75, σ = 10. Find z-score for x = 90.
\( z = \frac{90 - 75}{10} = \frac{15}{10} = 1.5 \)
Answer: z = 1.5 (score is 1.5 standard deviations above mean)
4. Standard Normal Distribution
Definition:
A normal distribution with mean = 0 and standard deviation = 1
\[ Z \sim N(0, 1) \]
Using Z-Table:
• Z-table gives area (probability) to the LEFT of a z-value
• P(Z < z) is read directly from table
• P(Z > z) = 1 - P(Z < z)
• P(a < Z < b) = P(Z < b) - P(Z < a)
Finding Probabilities:
Steps:
1. Convert x-value to z-score using \( z = \frac{x - \mu}{\sigma} \)
2. Look up z-score in z-table (or use calculator)
3. Interpret probability based on question
5. Find Values of Normal Variables
Inverse Normal (Working Backwards):
Sometimes you know the probability and need to find the value
\[ x = \mu + z\sigma \]
Steps:
1. Find z-score from z-table that corresponds to given probability
2. Use formula \( x = \mu + z\sigma \) to find x-value
Example:
Scores: μ = 500, σ = 100. Find the score that 90% of people score below.
Step 1: P(Z < z) = 0.90 → z ≈ 1.28 (from z-table)
Step 2: \( x = 500 + (1.28)(100) = 500 + 128 = 628 \)
Answer: 628
6. Distributions of Sample Means (Central Limit Theorem)
Central Limit Theorem (CLT):
For large sample sizes, the distribution of sample means is approximately normal, regardless of the population distribution
\[ \bar{X} \sim N\left(\mu, \frac{\sigma}{\sqrt{n}}\right) \]
Key Points:
• Mean of sample means = population mean (μ)
• Standard deviation of sample means = \( \frac{\sigma}{\sqrt{n}} \) (called standard error)
• CLT applies when \( n \geq 30 \) (or population is already normal)
Standard Error Formula:
\[ \sigma_{\bar{x}} = \frac{\sigma}{\sqrt{n}} \]
The standard error measures the variability of sample means
Z-Score for Sample Means:
\[ z = \frac{\bar{x} - \mu}{\sigma/\sqrt{n}} \]
where:
• \( \bar{x} \) = sample mean
• \( \mu \) = population mean
• \( \sigma \) = population standard deviation
• \( n \) = sample size
Example:
Population: μ = 100, σ = 20. Sample size n = 64. Find P(sample mean > 105).
Step 1: Standard error = \( \frac{20}{\sqrt{64}} = \frac{20}{8} = 2.5 \)
Step 2: \( z = \frac{105 - 100}{2.5} = \frac{5}{2.5} = 2.0 \)
Step 3: P(Z > 2.0) = 1 - P(Z < 2.0) = 1 - 0.9772 = 0.0228
Answer: 0.0228 or 2.28%
7. Binomial vs Normal Distribution
| Feature | Binomial | Normal |
|---|---|---|
| Type | Discrete | Continuous |
| Values | Counting (0, 1, 2, 3...) | Any real number |
| Parameters | n (trials), p (probability) | μ (mean), σ (std dev) |
| Shape | Can be skewed | Always symmetric |
| Use When | Fixed trials, success/failure | Continuous measurements |
Normal Approximation to Binomial:
When \( np \geq 10 \) and \( n(1-p) \geq 10 \), binomial can be approximated by normal with:
• \( \mu = np \)
• \( \sigma = \sqrt{np(1-p)} \)
8. Quick Reference Summary
Key Formulas:
Binomial Probability:
\[ P(X=x) = \binom{n}{x}p^x(1-p)^{n-x} \]
Binomial Mean & SD:
\[ \mu = np, \quad \sigma = \sqrt{np(1-p)} \]
Z-Score:
\[ z = \frac{x - \mu}{\sigma} \]
Inverse Normal:
\[ x = \mu + z\sigma \]
Standard Error:
\[ \sigma_{\bar{x}} = \frac{\sigma}{\sqrt{n}} \]
Z-Score for Sample Means:
\[ z = \frac{\bar{x} - \mu}{\sigma/\sqrt{n}} \]
📚 Study Tips
✓ Use binomial for counting successes in fixed trials
✓ Use normal for continuous measurements or large samples
✓ Always convert to z-scores when working with normal distribution
✓ Central Limit Theorem requires n ≥ 30 for non-normal populations
✓ Remember: larger samples have smaller standard error