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🎯⭐ INTERACTIVE LESSON

Normal Distributions

Learn step-by-step with interactive practice!

Normal Distributions - Complete Interactive Lesson

Part 1: The Normal Curve

📊 The Normal Distribution

Part 1 of 7 — Bell Curves and the Empirical Rule


The Normal Distribution

The normal distribution is the most important distribution in statistics. It is:

  • Symmetric and bell-shaped
  • Described by two parameters: mean μ\mu and standard deviation σ\sigma
  • Notation: X∼N(μ,σ)X \sim N(\mu, \sigma)

The Empirical Rule (68-95-99.7)

For any normal distribution:

RangePercentage
μ±1σ\mu \pm 1\sigma68% of data
μ±2σ\mu \pm 2\sigma95% of data
μ±3σ\mu \pm 3\sigma99.7% of data

Example: IQ scores follow N(100,15)N(100, 15)

  • 68% of scores between 100±15=[85,115]100 \pm 15 = [85, 115]
  • 95% between 100±30=[70,130]100 \pm 30 = [70, 130]
  • 99.7% between 100±45=[55,145]100 \pm 45 = [55, 145]

🔑 The Empirical Rule gives a quick estimate for normal data. For exact probabilities, use z-scores.

Normal Distribution Check 🎯

Empirical Rule Calculations 🧮

Adult male heights follow N(70,3)N(70, 3) inches.

1) What percentage of men are between 67 and 73 inches tall?

2) What percentage are shorter than 64 inches? (Hint: 64 = 70 − 2(3))

3) Between what two heights do the middle 99.7% of men fall? Give the upper bound.

Part 2: Z-Scores

📏 Z-Scores and the Standard Normal

Part 2 of 7 — Standardizing Values


The Z-Score Formula

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

z=x−μσz = \frac{x - \mu}{\sigma}

Z-ScoreInterpretation
z=0z = 0At the mean
z=1z = 1One SD above the mean
z=−2z = -2Two SDs below the mean

The Standard Normal Distribution

When we standardize: Z∼N(0,1)Z \sim N(0, 1)

This allows us to use one table (or calculator) for all normal distributions.

Example: Heights ∼N(70,3)\sim N(70, 3). A person is 76 inches tall. z=76−703=2z = \frac{76 - 70}{3} = 2 They are 2 standard deviations above the mean.


Using the Z-Table

The z-table gives P(Z≤z)P(Z \leq z) — the area to the left of zz.

To FindMethod
P(Z≤z)P(Z \leq z)Read directly from table
P(Z≥z)P(Z \geq z)1−P(Z≤z)1 - P(Z \leq z)
P(a≤Z≤b)P(a \leq Z \leq b)P(Z≤b)−P(Z≤a)P(Z \leq b) - P(Z \leq a)

🔑 Always sketch the normal curve, shade the region, then calculate.

Z-Score Practice 🎯

Z-Score Calculations 🧮

ACT scores follow N(21,5)N(21, 5).

1) Find the z-score for a student who scored 31. (Give as a whole number)

2) Find the z-score for a student who scored 16.

3) A student has z=−0.4z = -0.4. What was their ACT score?

Part 3: Normal Calculations

🔢 Normal Probability Calculations

Part 3 of 7 — Finding Areas and Percentiles


Forward Problems: X → Z → Probability

Given a value xx, find the probability:

  1. Compute z=x−μσz = \frac{x - \mu}{\sigma}
  2. Look up P(Z≤z)P(Z \leq z) in the z-table
  3. Adjust for the direction (left tail, right tail, between)

Example: Scores ∼N(500,100)\sim N(500, 100). Find P(X>650)P(X > 650).

  • z=650−500100=1.5z = \frac{650 - 500}{100} = 1.5
  • P(Z≤1.5)=0.9332P(Z \leq 1.5) = 0.9332
  • P(X>650)=1−0.9332=0.0668=6.68%P(X > 650) = 1 - 0.9332 = 0.0668 = 6.68\%

Backward Problems: Probability → Z → X

Given a percentile, find the value:

  1. Find the z-score from the table that matches the given probability
  2. Solve for x=μ+zσx = \mu + z\sigma

Example: What score is at the 90th percentile if μ=500,σ=100\mu = 500, \sigma = 100?

  • 90th percentile → z=1.28z = 1.28 (from table: P(Z≤1.28)=0.8997≈0.90P(Z \leq 1.28) = 0.8997 \approx 0.90)
  • x=500+1.28(100)=628x = 500 + 1.28(100) = 628

🔑 "Top 10%" = 90th percentile. "Bottom 25%" = 25th percentile.

Normal Calculations 🎯

Finding Percentiles 🧮

Baby weights at birth follow N(7.5,1.2)N(7.5, 1.2) lbs.

1) What z-score corresponds to a baby weighing 9.9 lbs?

2) Using P(Z≤2)=0.9772P(Z \leq 2) = 0.9772, what percent of babies weigh less than 9.9 lbs? (Express as a number, e.g., 97.72)

3) The 84th percentile has z≈1z \approx 1. What is the 84th percentile weight? (in lbs, one decimal)

Part 4: Assessing Normality

📈 Assessing Normality

Part 4 of 7 — Is a Normal Model Reasonable?


Why Check?

A normal curve can model a distribution of data only if the data have a roughly normal shape. Later, inference procedures for means also ask whether the sample data could reasonably come from a normal population.


Check the Shape with a Graph

Make a histogram, dotplot, stemplot or boxplot of the data and look for:

Supports a normal modelArgues against a normal model
One main peak (unimodal)Two or more distinct peaks
Roughly symmetricStrong skew in either direction
No outliersOutliers, gaps or clusters

🔑 No real data set is perfectly normal. We look for "close enough": roughly symmetric and unimodal with no outliers.

The AP exam checks normality with graphs and the empirical rule; normal probability plots are not part of the course.

Normality Assessment 🎯

Check with the Empirical Rule

Compare the actual percent of data within 1, 2 and 3 standard deviations of the mean with 68%, 95% and 99.7%.

Example: 40 test scores have xˉ=50\bar{x} = 50 and s=8s = 8.

IntervalCountPercentNormal model
50±850 \pm 8 (42 to 58)2972.5%68%
50±1650 \pm 16 (34 to 66)3895%95%
50±2450 \pm 24 (26 to 74)40100%99.7%

The percents are close to 68–95–99.7, so a normal model is reasonable.


The Sample Data Condition (Preview of Unit 4)

For t-procedures, the population distribution should be approximately normal or n≥30n \geq 30. With a small sample, check a graph of the sample data: it should be free from strong skewness and outliers.

Empirical Rule Check 🧮

A sample of 50 values has xˉ=20\bar{x} = 20 and s=4s = 4. There are 33 values between 16 and 24, 48 values between 12 and 28, and all 50 between 8 and 32.

1) Percent within 1 SD of the mean?

2) Percent within 2 SDs?

3) Percent within 3 SDs?

Sample Data Condition 🔍

Part 5: Intervals, Percentiles and Comparing Relative Position

➕ Intervals, Percentiles and Comparing Relative Position

Part 5 of 7 — Working with Areas in Both Directions


Setting Up the Inequality

StatementArea to the left of the cutoff
Lowest pp% are below xax_ap/100p/100
Highest pp% are above xbx_b1−p/1001 - p/100
Middle pp% lie between xax_a and xbx_bSplit the remaining area evenly: each tail gets 12(100−p)\frac{1}{2}(100 - p)%

Middle Intervals

Example: Battery life is N(500,40)N(500, 40) hours. Find the middle 90%.

  • Tails: 10% outside, so 5% in each tail.
  • z=±1.645z = \pm 1.645 (the 5th and 95th percentiles of ZZ).
  • Interval: 500±1.645(40)=434.2500 \pm 1.645(40) = 434.2 to 565.8565.8 hours.

The Highest pp%

Example: Which battery life marks the top 15%?

  • 85% of the area is to the left: z≈1.04z \approx 1.04.
  • x=500+1.04(40)=541.6x = 500 + 1.04(40) = 541.6 hours.

🔑 Always translate "top", "bottom", "middle" into the area to the LEFT before using a table or invNorm.

Intervals and Cutoffs 🎯

Comparing Relative Position

Percentiles (or z-scores) let you compare values from different normal distributions.

Example: Ana scores 82 on a test that is N(74,6)N(74, 6). Ben scores 88 on a test that is N(80,10)N(80, 10).

Studentz-scorePercentile
Ana(82−74)/6≈1.33(82 - 74)/6 \approx 1.33P(Z<1.33)≈0.908P(Z < 1.33) \approx 0.908, about 91st
Ben(88−80)/10=0.80(88 - 80)/10 = 0.80P(Z<0.80)≈0.788P(Z < 0.80) \approx 0.788, about 79th

Ben's raw score is higher, but Ana did better relative to her group.

⚠️ When smaller is better (race times, golf scores), a more negative z-score is the better performance.

Relative Position 📊

Intervals and Percentiles 🧮

Exam scores are N(500,100)N(500, 100).

1) Lower cutoff of the middle 95% of scores? (use z=1.96z = 1.96)

2) Score at the 30th percentile? (use z=−0.52z = -0.52, nearest whole number)

3) A student scores 650. At what percentile is that score? (nearest whole number)

Part 6: Problem-Solving Workshop

🛠️ Normal Distribution Workshop

Part 6 of 7 — Comprehensive Practice


Strategy for Normal Distribution Problems

  1. Identify μ\mu and σ\sigma from the problem
  2. Sketch the curve and shade the desired region
  3. Standardize using z=(x−μ)/σz = (x - \mu)/\sigma
  4. Use the table or calculator to find probabilities
  5. For percentiles: work backward from probability to z to x

Worked Example

Problem: A machine fills cereal boxes with μ=368\mu = 368 g and σ=4\sigma = 4 g. What proportion of boxes have less than 360 g?

Solution:

  1. z=(360−368)/4=−8/4=−2z = (360 - 368)/4 = -8/4 = -2
  2. P(Z≤−2)=0.0228P(Z \leq -2) = 0.0228
  3. About 2.28%2.28\% of boxes are underfilled.

Follow-up: What weight is exceeded by 90% of boxes?

  1. "Exceeded by 90%" means 10th percentile (10% are below)
  2. z=−1.28z = -1.28 (from table)
  3. x=368+(−1.28)(4)=368−5.12=362.88x = 368 + (-1.28)(4) = 368 - 5.12 = 362.88 g

Workshop Problems 🎯

Part 7: Review & Applications

📋 Normal Distribution Review

Part 7 of 7 — Summary and Applications


Key Formulas

FormulaWhen to Use
z=x−μσz = \frac{x - \mu}{\sigma}Convert any value to standard normal
x=μ+zσx = \mu + z\sigmaConvert from z-score back to original units
P(a≤Z≤b)=P(Z≤b)−P(Z≤a)P(a \leq Z \leq b) = P(Z \leq b) - P(Z \leq a)Find probability between two values
P(X≤xa)=p/100P(X \leq x_a) = p/100xax_a is the ppth percentile

Common z-Values to Know

Confidence Levelz*
90%1.645
95%1.960
99%2.576

Final Review 🎯

Mastery Review 🔽