Binomial and Normal Distributions
Complete Notes & Formulae for Twelfth Grade (Precalculus)
1. Binomial Distribution
Conditions:
A binomial experiment must satisfy these four conditions:
1. Fixed number of trials (n)
2. Two possible outcomes: success or failure
3. Independent trials: outcome of one doesn't affect another
4. Constant probability (p): probability of success stays the same
Probability Formula:
\[ P(X = x) = \binom{n}{x} p^x (1-p)^{n-x} \]
\[ = \frac{n!}{x!(n-x)!} \cdot p^x \cdot q^{n-x} \]
where:
• \( n \) = number of trials
• \( x \) = number of successes
• \( p \) = probability of success on each trial
• \( q = 1 - p \) = probability of failure
Example:
A coin is flipped 5 times. Find the probability of getting exactly 3 heads.
Given: \( n = 5 \), \( x = 3 \), \( p = 0.5 \)
\( P(X = 3) = \binom{5}{3} (0.5)^3 (0.5)^2 \)
\( = 10 \times 0.125 \times 0.25 \)
\( = 0.3125 \) or 31.25%
2. Binomial Distribution Statistics
Formulas:
Mean (Expected Value):
\[ \mu = np \]
Variance:
\[ \sigma^2 = npq = np(1-p) \]
Standard Deviation:
\[ \sigma = \sqrt{npq} = \sqrt{np(1-p)} \]
Example:
A basketball player makes 70% of free throws. She takes 20 shots. Find mean, variance, and standard deviation.
Given: \( n = 20 \), \( p = 0.7 \), \( q = 0.3 \)
Mean: \( \mu = 20(0.7) = 14 \) shots
Variance: \( \sigma^2 = 20(0.7)(0.3) = 4.2 \)
Standard Deviation: \( \sigma = \sqrt{4.2} \approx 2.05 \) shots
3. Normal Distribution
Properties:
The normal distribution is a continuous, symmetric, bell-shaped distribution
• Notation: \( X \sim N(\mu, \sigma) \)
• Mean, median, and mode are equal
• Total area under curve = 1
• 68% of data within 1 standard deviation
• 95% of data within 2 standard deviations
• 99.7% of data within 3 standard deviations
4. Z-Scores (Standard Scores)
Definition:
A z-score tells you how many standard deviations a value is from the mean
\[ z = \frac{x - \mu}{\sigma} \]
where:
• \( x \) = raw score (data value)
• \( \mu \) = mean of distribution
• \( \sigma \) = 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: mean = 75, SD = 10. Find z-score for a score of 90.
\( z = \frac{90 - 75}{10} = \frac{15}{10} = 1.5 \)
The score is 1.5 standard deviations above the mean
5. Find Values from Z-Scores
Formula:
To find the raw score (x) from a z-score, rearrange the z-score formula:
\[ x = \mu + z\sigma \]
Example:
Heights: mean = 170 cm, SD = 8 cm. What height has a z-score of -1.5?
\( x = 170 + (-1.5)(8) \)
\( x = 170 - 12 \)
\( x = 158 \) cm
6. Standard Normal Distribution
Definition:
A standard normal distribution has mean = 0 and standard deviation = 1
\[ Z \sim N(0, 1) \]
Finding Probabilities:
• Use z-table or calculator to find areas under the curve
• \( P(Z < z) \) = area to the left of z
• \( P(Z > z) = 1 - P(Z < z) \)
• \( P(a < Z < b) = P(Z < b) - P(Z < a) \)
7. Distribution of Sample Means
Formulas:
If samples of size n are drawn from a population with mean \( \mu \) and standard deviation \( \sigma \):
Mean of sample means:
\[ \mu_{\bar{x}} = \mu \]
Standard deviation of sample means (Standard Error):
\[ \sigma_{\bar{x}} = \frac{\sigma}{\sqrt{n}} \]
Z-score for sample mean:
\[ z = \frac{\bar{x} - \mu}{\sigma/\sqrt{n}} \]
8. Central Limit Theorem (CLT)
Statement:
The Central Limit Theorem states that the distribution of sample means approaches a normal distribution as the sample size increases, regardless of the population distribution
Key Points:
• Works for ANY population distribution (even non-normal)
• Usually, \( n \geq 30 \) is considered "large enough"
• If population is normal, sample means are normal for ANY n
• Larger samples → better approximation to normal
Application:
With large samples, we can use normal distribution to find probabilities about sample means:
\[ \bar{X} \sim N\left(\mu, \frac{\sigma}{\sqrt{n}}\right) \]
9. Normal Approximation to Binomial
When to Use:
When n is large, binomial distribution can be approximated by normal distribution if:
\[ np \geq 10 \quad \text{and} \quad nq \geq 10 \]
Parameters:
Use normal distribution with:
Mean: \( \mu = np \)
Standard Deviation: \( \sigma = \sqrt{npq} \)
Continuity Correction:
Since binomial is discrete and normal is continuous, apply continuity correction:
• \( P(X = k) \approx P(k - 0.5 < X < k + 0.5) \)
• \( P(X \leq k) \approx P(X < k + 0.5) \)
• \( P(X \geq k) \approx P(X > k - 0.5) \)
• \( P(X < k) \approx P(X < k - 0.5) \)
Example:
60% of students pass. Out of 100 students, find P(X ≥ 65 pass).
Check: \( np = 60 \geq 10 \) and \( nq = 40 \geq 10 \) ✓
Use: \( \mu = 60 \), \( \sigma = \sqrt{100(0.6)(0.4)} = 4.9 \)
With continuity correction: \( P(X \geq 65) \approx P(X > 64.5) \)
Z-score: \( z = \frac{64.5 - 60}{4.9} \approx 0.92 \)
Use z-table to find probability
10. Quick Reference Summary
Binomial Distribution:
\( P(X=x) = \binom{n}{x} p^x (1-p)^{n-x} \)
\( \mu = np \), \( \sigma = \sqrt{npq} \)
Normal Distribution:
Z-score: \( z = \frac{x-\mu}{\sigma} \)
Find x: \( x = \mu + z\sigma \)
Sample Means:
\( \mu_{\bar{x}} = \mu \), \( \sigma_{\bar{x}} = \frac{\sigma}{\sqrt{n}} \)
Normal Approximation:
Use when: \( np \geq 10 \) and \( nq \geq 10 \)
Apply continuity correction (±0.5)
📚 Study Tips
✓ Binomial: fixed trials, two outcomes, independent, constant probability
✓ Z-score tells you standard deviations from the mean
✓ Central Limit Theorem works for ANY population distribution
✓ Use normal approximation to binomial when np and nq ≥ 10
✓ Always apply continuity correction when approximating discrete with continuous