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The Normal Distribution

CSEC Additional Mathematics Essential Knowledge: The normal distribution is the most important probability distribution in statistics. It describes data that clusters around a mean, forming the classic “bell curve.” Understanding the normal distribution is crucial for analyzing real-world data, making predictions, and solving probability problems in CSEC Add Maths.

Key Concept: The normal distribution is a continuous probability distribution characterized by its symmetric bell-shaped curve. It is completely defined by two parameters: the mean (μ) which determines the center, and the standard deviation (σ) which determines the spread.

Part 1: Understanding the Normal Distribution

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Properties of the Normal Curve

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The Bell Curve

The normal distribution curve, often called the “bell curve,” has these key characteristics:

Interactive Normal Distribution Curve
💡 Try adjusting the mean (μ) and standard deviation (σ) to see how they affect the curve!
  • Symmetric: The left and right sides are mirror images
  • Unimodal: Single peak at the mean (μ)
  • Asymptotic: Approaches but never touches the x-axis
  • Mean = Median = Mode: All measures of center coincide
  • Total area under curve = 1: Represents 100% probability
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Mean and Standard Deviation Comparison

The normal distribution \(N(\mu, \sigma^2)\) is defined by:

\[\text{Mean } (\mu) \quad \text{and} \quad \text{Standard Deviation } (\sigma)\]
Effect of Different Means and Standard Deviations
💡 Observe how changing the mean shifts the curve, while changing the standard deviation changes its spread!

Effect of changing parameters:

  • Changing μ shifts the curve left or right
  • Changing σ makes the curve wider (larger σ) or narrower (smaller σ)
📝 Example 1: Recognizing Normal Distributions

Which of these are likely to follow a normal distribution?

(a) Heights of adult males in a country

(b) Scores on a well-designed exam

(c) Annual income of all people in a country

1
Heights: Usually normally distributed – most people are around average height, fewer very short or very tall
2
Exam scores: Often normally distributed if test is well-designed – most scores around average, fewer very high or very low
3
Income: Typically NOT normally distributed – skewed right (more people with lower incomes, few with very high incomes)

Part 2: The Empirical Rule (68-95-99.7 Rule)

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The 68-95-99.7 Rule

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Empirical Rule Definition

For any normal distribution:

Visualizing the Empirical Rule
💡 Click the buttons to see how data is distributed within 1, 2, and 3 standard deviations from the mean!
\[P(\mu – \sigma \leq X \leq \mu + \sigma) \approx 0.68\] \[P(\mu – 2\sigma \leq X \leq \mu + 2\sigma) \approx 0.95\] \[P(\mu – 3\sigma \leq X \leq \mu + 3\sigma) \approx 0.997\]
📝 Example 2: Applying the Empirical Rule

The heights of adult women are normally distributed with mean 162 cm and standard deviation 6 cm.

(a) What percentage of women are between 156 cm and 168 cm tall?

(b) What percentage are between 150 cm and 174 cm tall?

1
For part (a): 156 = 162 – 6 = μ – σ, 168 = 162 + 6 = μ + σ
2
Within 1σ of μ → 68% of women
3
For part (b): 150 = 162 – 12 = μ – 2σ, 174 = 162 + 12 = μ + 2σ
4
Within 2σ of μ → 95% of women
📝 Example 3: Finding Percentages with Empirical Rule

IQ scores are normally distributed with μ = 100 and σ = 15. What percentage of people have IQ between 85 and 115?

IQ Distribution with Empirical Rule Applied
💡 The shaded area shows that 68% of people have IQ scores between 85 and 115 (μ ± σ)
1
Calculate boundaries: 85 = 100 – 15 = μ – σ, 115 = 100 + 15 = μ + σ
2
Apply empirical rule: Within 1σ of μ = 68%
3
Answer: 68% of people have IQ between 85 and 115

Part 3: Standard Normal Distribution and Z-scores

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Converting to Standard Normal

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Z-score Formula

A z-score measures how many standard deviations a value is from the mean:

\[z = \frac{x – \mu}{\sigma}\]

Where:

  • \(z\) = z-score (standard score)
  • \(x\) = raw score
  • \(\mu\) = mean of distribution
  • \(\sigma\) = standard deviation of distribution
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Interpreting Z-scores
Standard Normal Distribution with Z-scores
💡 The standard normal distribution has μ = 0 and σ = 1. Z-scores tell you how many standard deviations a value is from the mean!

Key z-score values:

  • \(z = 0\): Exactly at the mean
  • \(z = 1\): 1 standard deviation above mean
  • \(z = -1\): 1 standard deviation below mean
  • \(z = 2\): 2 standard deviations above mean
  • \(z = -2\): 2 standard deviations below mean
📝 Example 4: Calculating Z-scores

Test scores are normally distributed with μ = 75 and σ = 10. Calculate the z-score for:

(a) A score of 85

(b) A score of 60

1
For 85: \(z = \frac{85 – 75}{10} = \frac{10}{10} = 1\)
2
Interpretation: Score of 85 is 1 standard deviation above the mean
3
For 60: \(z = \frac{60 – 75}{10} = \frac{-15}{10} = -1.5\)
4
Interpretation: Score of 60 is 1.5 standard deviations below the mean
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Standard Normal Distribution Table Visualization

The standard normal distribution has μ = 0 and σ = 1. We use z-tables to find probabilities:

Using Z-table to Find Probabilities
z = 0.0

Current Probability: P(Z ≤ 0.0) = 0.5000

💡 Drag the slider to see how different z-scores correspond to different probabilities in the standard normal distribution!

Part 4: Calculating Probabilities with Normal Distribution

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Finding Probabilities using Z-scores

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Step-by-Step Process
1
Identify μ and σ from problem
2
Convert raw score(s) to z-score(s): \(z = \frac{x – \mu}{\sigma}\)
3
Sketch normal curve and shade required area
4
Use z-table to find probabilities
5
Perform calculations to find required probability
📝 Example 5: Probability Less Than a Value

Test scores are normally distributed with μ = 75, σ = 10. Find the probability a randomly selected score is less than 85.

Probability that X < 85 (z = 1)
💡 The shaded area represents P(X < 85) = 0.8413 or 84.13%
1
Convert to z-score: \(z = \frac{85 – 75}{10} = 1\)
2
We want: \(P(X < 85) = P(Z < 1)\)
3
Look up z = 1 in table: \(P(Z < 1) = 0.8413\)
4
Answer: Probability = 0.8413 or 84.13%
📝 Example 6: Probability Between Two Values

Heights of men are normally distributed with μ = 175 cm, σ = 8 cm. Find the percentage of men between 170 cm and 185 cm tall.

Probability that 170 < X < 185
💡 The shaded area represents P(170 < X < 185) = 0.6284 or 62.84%
1
Convert both to z-scores: \(z_1 = \frac{170 – 175}{8} = -0.625\), \(z_2 = \frac{185 – 175}{8} = 1.25\)
2
We want: \(P(170 < X < 185) = P(-0.625 < Z < 1.25)\)
3
Find probabilities: \(P(Z < 1.25) = 0.8944\), \(P(Z < -0.625) = 0.2660\) (from table)
4
Calculate: \(0.8944 – 0.2660 = 0.6284\)
5
Answer: 62.84% of men are between 170 cm and 185 cm tall
📝 Example 7: Probability Greater Than a Value

IQ scores have μ = 100, σ = 15. What percentage of people have IQ above 130?

Probability that X > 130 (z = 2)
💡 The shaded area represents P(X > 130) = 0.0228 or 2.28%
1
Convert to z-score: \(z = \frac{130 – 100}{15} = 2\)
2
We want: \(P(X > 130) = P(Z > 2)\)
3
Find \(P(Z < 2)\) from table: 0.9772
4
Use complement: \(P(Z > 2) = 1 – P(Z < 2) = 1 - 0.9772 = 0.0228\)
5
Answer: 2.28% of people have IQ above 130

Part 5: Inverse Normal Distribution

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Finding Values Given Probabilities

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Finding X from Probability

Sometimes we know the probability and need to find the corresponding x-value:

\[x = \mu + z\sigma\]

Where \(z\) is found from the z-table using the given probability.

Inverse Normal: Finding X from Probability
90th Percentile

Corresponding z-score: z = 1.28

Corresponding x-value (μ=75, σ=10): x = 87.8

💡 Adjust the percentile slider to see how different percentiles correspond to different z-scores and x-values!
📝 Example 8: Finding Score Given Percentile

Test scores are normally distributed with μ = 75, σ = 10. What score represents the 90th percentile?

1
90th percentile means: \(P(Z < z) = 0.90\)
2
Find z from table: z ≈ 1.28 (since \(P(Z < 1.28) ≈ 0.8997\))
3
Convert to x: \(x = \mu + z\sigma = 75 + 1.28(10) = 75 + 12.8 = 87.8\)
4
Answer: The 90th percentile score is approximately 88
📝 Example 9: Finding Cutoff Score

An exam has normally distributed scores with μ = 65, σ = 12. If the top 15% get an A, what is the cutoff score for an A?

1
Top 15% means: \(P(Z > z) = 0.15\) or \(P(Z < z) = 0.85\)
2
Find z from table: z ≈ 1.04 (since \(P(Z < 1.04) ≈ 0.8508\))
3
Convert to x: \(x = 65 + 1.04(12) = 65 + 12.48 = 77.48\)
4
Answer: Students need about 77.5 or higher to get an A

Quiz: Test Your Understanding

Normal Distribution Quiz
Question 1: The weights of apples are normally distributed with μ = 150g, σ = 20g. What percentage of apples weigh between 130g and 170g?
Answer:
130g = 150 – 20 = μ – σ
170g = 150 + 20 = μ + σ
Within 1σ of μ → 68% (using empirical rule)
Approximately 68% of apples weigh between 130g and 170g.
Question 2: Test scores are normally distributed with μ = 70, σ = 8. Find the z-score for a score of 86.
Answer:
\(z = \frac{x – \mu}{\sigma} = \frac{86 – 70}{8} = \frac{16}{8} = 2\)
A score of 86 is 2 standard deviations above the mean.
Question 3: The heights of women are normally distributed with μ = 162 cm, σ = 6 cm. What is the probability a randomly selected woman is taller than 174 cm?
Answer:
\(z = \frac{174 – 162}{6} = \frac{12}{6} = 2\)
\(P(Z > 2) = 1 – P(Z < 2) = 1 - 0.9772 = 0.0228\)
Probability = 0.0228 or 2.28%
Question 4: IQ scores have μ = 100, σ = 15. What is the 75th percentile IQ score?
Answer:
For 75th percentile: \(P(Z < z) = 0.75\)
From z-table: z ≈ 0.674 (since \(P(Z < 0.67) = 0.7486\), \(P(Z < 0.68) = 0.7517\))
\(x = \mu + z\sigma = 100 + 0.674(15) = 100 + 10.11 = 110.11\)
The 75th percentile IQ is approximately 110.
Question 5: A factory produces bolts with normally distributed lengths, μ = 5 cm, σ = 0.2 cm. Bolts shorter than 4.6 cm or longer than 5.4 cm are rejected. What percentage are accepted?
Answer:
4.6 cm = 5 – 0.4 = μ – 2σ
5.4 cm = 5 + 0.4 = μ + 2σ
Within 2σ of μ = 95% (empirical rule)
So 95% are accepted, 5% are rejected.

🎯 Key Concepts Summary

  • Normal Distribution: Bell-shaped, symmetric curve defined by μ (mean) and σ (standard deviation)
  • Empirical Rule (68-95-99.7):
    • 68% within 1σ of μ
    • 95% within 2σ of μ
    • 99.7% within 3σ of μ
  • Z-score: \(z = \frac{x – \mu}{\sigma}\) measures how many σ a value is from μ
  • Standard Normal Distribution: μ = 0, σ = 1. Use z-table to find probabilities
  • Finding Probabilities:
    • Convert x to z
    • Use z-table to find \(P(Z < z)\)
    • For \(P(Z > z)\), use \(1 – P(Z < z)\)
    • For \(P(a < Z < b)\), calculate \(P(Z < b) - P(Z < a)\)
  • Inverse Normal: Given probability, find z from table, then \(x = \mu + z\sigma\)
  • Common Applications:
    • Test scores and grading
    • Quality control in manufacturing
    • Biological measurements (heights, weights)
    • Psychological test scores (IQ, personality tests)
  • CSEC Exam Strategy:
    • Always sketch the normal curve and shade the required area
    • Clearly show z-score calculations
    • Use the z-table provided in the exam
    • For word problems, identify μ and σ first
    • Check if answer makes sense (probabilities between 0 and 1)

CSEC Exam Strategy: Normal distribution questions frequently appear in Paper 2. Common question types: (1) Calculate probabilities using z-scores, (2) Apply empirical rule, (3) Find percentiles/quartiles, (4) Solve real-world application problems. Always show your working: write the z-score formula, substitute values, show table lookup, and state final probability/answer. Remember to use the z-table provided in the exam formula sheet.

Real-World Applications

Grading on a curve (test score distributions)
Quality control in manufacturing (product dimensions)
Medical measurements (blood pressure, cholesterol levels)
Psychological testing (IQ scores, personality traits)
Financial modeling (stock returns, risk assessment)
Biological studies (animal sizes, plant growth)
Sports statistics (player performance metrics)

Pro Tip: When using the z-table, remember it gives \(P(Z < z)\). For \(P(Z > z)\), subtract from 1. For negative z-values, use symmetry: \(P(Z < -z) = 1 - P(Z < z)\). Always sketch the normal curve to visualize the area you need!

Common Mistakes to Avoid: 1. Forgetting to convert to z-scores before using the table
2. Confusing \(P(Z < z)\) with \(P(Z > z)\)
3. Using raw scores instead of z-scores with the standard normal table
4. Not checking if answer is reasonable (probability between 0 and 1)
5. Forgetting that total area under curve = 1 (100% probability)
6. Misreading the z-table (combining row and column correctly)

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