Solved Exercises on Normal Distributions in Grade 10

Master normal distributions: mean, standard deviation, z-scores, and probability calculations through these 5 detailed exercises.

Solution: Exercises 1 to 3
1 Standard Normal Distribution
Exercise 1
The heights of adult males in a city are normally distributed with mean μ = 70 inches and standard deviation σ = 3 inches. Find the probability that a randomly selected male is taller than 75 inches.
Definition:

Normal Distribution: Continuous probability distribution characterized by mean μ and standard deviation σ

Standard Normal (Z): N(0,1) distribution with mean 0 and standard deviation 1

Z-score Formula: Z = (X - μ)/σ, converts any normal to standard normal

Normal distribution method:
  1. Identify mean (μ) and standard deviation (σ)
  2. Convert raw score to z-score: Z = (X - μ)/σ
  3. Use standard normal table or calculator to find probability
  4. Interpret the result in context
Given
μ = 70, σ = 3, X > 75
Z-score
Z = (75-70)/3 = 1.67
Probability
P(Z > 1.67)
Step 1: Identify parameters

Mean (μ) = 70 inches, Standard deviation (σ) = 3 inches

Step 2: Calculate z-score

Z = (X - μ)/σ = (75 - 70)/3 = 5/3 = 1.67

Step 3: Find P(Z > 1.67)

P(Z > 1.67) = 1 - P(Z ≤ 1.67) = 1 - 0.9525 = 0.0475

Step 4: Interpret result

About 4.75% of adult males are taller than 75 inches

P(X > 75) = 0.0475
Final answer:

The probability that a randomly selected male is taller than 75 inches is approximately 0.0475 or 4.75%

Applied rules:

• Standardization: Convert to standard normal using Z = (X - μ)/σ

• Complement rule: P(Z > z) = 1 - P(Z ≤ z)

• Normal table: Use z-table to find cumulative probabilities

2 Finding Percentiles
Exercise 2
Scores on a standardized test are normally distributed with mean 500 and standard deviation 100. What score corresponds to the 90th percentile?
Definition:

Percentile: Value below which a percentage of observations fall

90th percentile: Value such that 90% of scores are below it

Inverse transformation: X = μ + Z·σ to convert back from standard normal

Given
μ = 500, σ = 100, percentile = 90%
Z-value
Z₀.₉₀ ≈ 1.28
Score
X = μ + Z·σ
Step 1: Identify parameters

Mean (μ) = 500, Standard deviation (σ) = 100

Step 2: Find z-value for 90th percentile

We need Z such that P(Z ≤ z) = 0.90

From standard normal table: Z₀.₉₀ ≈ 1.28

Step 3: Convert z-score back to original scale

Using X = μ + Z·σ

X = 500 + 1.28 × 100 = 500 + 128 = 628

Step 4: Interpret result

A score of 628 corresponds to the 90th percentile

90th percentile score = 628
Final answer:

The score that corresponds to the 90th percentile is 628

Applied rules:

• Percentile definition: kth percentile has k% of data below it

• Inverse transformation: X = μ + Z·σ

• Standard normal table: Find z-value for given probability

3 Between Two Values
Exercise 3
The weights of bags of flour are normally distributed with mean 5 pounds and standard deviation 0.1 pounds. Find the probability that a randomly selected bag weighs between 4.8 and 5.2 pounds.
Definition:

Probability between values: P(a < X < b) = P(X < b) - P(X < a)

Standardization: Convert both bounds to z-scores separately

Area under curve: Probability equals area under normal curve between two points

Given
μ = 5, σ = 0.1, 4.8 < X < 5.2
Z-scores
Z₁ = -2, Z₂ = 2
Probability
P(-2 < Z < 2)
Step 1: Identify parameters

Mean (μ) = 5 pounds, Standard deviation (σ) = 0.1 pounds

Step 2: Convert both bounds to z-scores

Lower bound: Z₁ = (4.8 - 5)/0.1 = -0.2/0.1 = -2

Upper bound: Z₂ = (5.2 - 5)/0.1 = 0.2/0.1 = 2

Step 3: Calculate probability

P(4.8 < X < 5.2) = P(-2 < Z < 2) = P(Z < 2) - P(Z < -2)

= 0.9772 - 0.0228 = 0.9544

Step 4: Interpret result

About 95.44% of bags weigh between 4.8 and 5.2 pounds

P(4.8 < X < 5.2) = 0.9544
Final answer:

The probability that a randomly selected bag weighs between 4.8 and 5.2 pounds is approximately 0.9544 or 95.44%

Applied rules:

• Probability between values: P(a < X < b) = P(X < b) - P(X < a)

• Standardization: Convert each bound separately

• Normal table: Use to find cumulative probabilities

Normal Distribution Properties
\(f(x) = \frac{1}{\sigma\sqrt{2\pi}} e^{-\frac{1}{2}\left(\frac{x-\mu}{\sigma}\right)^2}, Z = \frac{X-\mu}{\sigma}\)
Normal Distribution Formula
Empirical Rule
68-95-99.7%
Within 1,2,3 std deviations
Standardization
Z = (X-μ)/σ
Convert to N(0,1)
Inverse
X = μ + Z·σ
Convert back to original scale
Key definitions:

Normal Distribution: Bell-shaped, symmetric distribution described by mean μ and std dev σ

Standard Normal: Normal distribution with μ=0, σ=1, denoted N(0,1)

Z-score: Number of standard deviations a value is from the mean

Normal distribution methodology:
  1. Identify parameters: Mean (μ) and standard deviation (σ)
  2. Standardize: Convert to z-scores using Z = (X - μ)/σ
  3. Find probability: Use standard normal table or calculator
  4. Interpret results: Convert back if needed
Tip 1: Remember the empirical rule: 68% within 1σ, 95% within 2σ, 99.7% within 3σ.
Tip 2: Z-scores allow comparison across different normal distributions.
Tip 3: Always sketch the normal curve to visualize the problem.
Tip 4: For "greater than" problems, use complement: P(X > a) = 1 - P(X ≤ a).
Common errors: Forgetting to standardize, misreading tables, incorrect inequality direction.
Exam preparation: Practice standardization, memorize empirical rule, work with percentiles.
Essential formulas:

• Standardization: Z = (X - μ)/σ

• Inverse transformation: X = μ + Z·σ

• Probability between values: P(a < X < b) = P(X < b) - P(X < a)

• Complement rule: P(X > a) = 1 - P(X ≤ a)

Solution: Exercises 4 to 5
4 Sampling Distribution
Exercise 4
The amount of coffee dispensed by a machine is normally distributed with mean 8 oz and standard deviation 0.5 oz. If a sample of 25 cups is taken, find the probability that the sample mean is less than 7.9 oz.
Definition:

Sampling Distribution: Distribution of sample statistics (like sample mean)

Central Limit Theorem: Sample mean approaches normal distribution regardless of population distribution

Sample Mean Parameters: Mean = μ, Std Dev = σ/√n

Given
μ = 8, σ = 0.5, n = 25, X̄ < 7.9
Std error
σ/√n = 0.5/5 = 0.1
Z-score
Z = (7.9-8)/0.1 = -1
Step 1: Identify population parameters

Population mean (μ) = 8 oz, Population standard deviation (σ) = 0.5 oz

Step 2: Find sampling distribution parameters

Sample mean distribution: Mean = μ = 8 oz

Standard error = σ/√n = 0.5/√25 = 0.5/5 = 0.1 oz

Step 3: Calculate z-score for sample mean

Z = (X̄ - μ)/(σ/√n) = (7.9 - 8)/0.1 = -0.1/0.1 = -1

Step 4: Find probability

P(X̄ < 7.9) = P(Z < -1) = 0.1587

P(X̄ < 7.9) = 0.1587
Final answer:

The probability that the sample mean is less than 7.9 oz is approximately 0.1587 or 15.87%

Applied rules:

• Sampling distribution: X̄ ~ N(μ, σ²/n)

• Standard error: σ_X̄ = σ/√n

• Standardization: Z = (X̄ - μ)/(σ/√n)

5 Quality Control
Exercise 5
Battery lifetimes are normally distributed with mean 50 hours and standard deviation 5 hours. A battery is considered defective if it lasts less than 40 hours or more than 60 hours. What percentage of batteries are defective?
Definition:

Defective items: Those falling outside acceptable limits

Union of events: P(A or B) = P(A) + P(B) for mutually exclusive events

Outlier detection: Values beyond certain standard deviations

Given
μ = 50, σ = 5, def < 40 or > 60
Z-scores
Z₁ = -2, Z₂ = 2
Defect rate
P(X < 40) + P(X > 60)
Step 1: Identify parameters

Mean (μ) = 50 hours, Standard deviation (σ) = 5 hours

Step 2: Calculate z-scores for defective limits

Lower limit: Z₁ = (40 - 50)/5 = -10/5 = -2

Upper limit: Z₂ = (60 - 50)/5 = 10/5 = 2

Step 3: Find probabilities for defective ranges

P(X < 40) = P(Z < -2) = 0.0228

P(X > 60) = P(Z > 2) = 1 - P(Z ≤ 2) = 1 - 0.9772 = 0.0228

Step 4: Calculate total defect rate

P(defective) = P(X < 40) + P(X > 60) = 0.0228 + 0.0228 = 0.0456

Defect rate = 4.56%
Final answer:

Approximately 4.56% of batteries are defective

Applied rules:

• Union of events: P(A or B) = P(A) + P(B) for mutually exclusive

• Complement rule: P(Z > z) = 1 - P(Z ≤ z)

• Quality control: Defects occur outside specification limits

Detailed Summary: Normal Distributions
\(Z = \frac{X-\mu}{\sigma}, P(a < X < b) = \Phi\left(\frac{b-\mu}{\sigma}\right) - \Phi\left(\frac{a-\mu}{\sigma}\right)\)
Standardization and Probability Calculation
Comprehensive definitions:

Normal Distribution: A continuous probability distribution that is symmetric and bell-shaped, characterized by its mean μ and standard deviation σ

Standard Normal Distribution: A normal distribution with mean 0 and standard deviation 1, denoted as N(0,1)

Z-score: The number of standard deviations a data point is from the mean, calculated as Z = (X - μ)/σ

Empirical Rule (68-95-99.7 Rule): Approximately 68% of data falls within 1σ, 95% within 2σ, and 99.7% within 3σ of the mean

Complete methodology:
  1. Problem identification: Identify if normal distribution applies, note μ and σ
  2. Standardization: Convert raw scores to z-scores using Z = (X - μ)/σ
  3. Probability calculation: Use standard normal table or calculator
  4. Result interpretation: Convert back to original units if needed
Tip 1: Sketch the normal curve and shade the area representing the probability you're looking for.
Tip 2: Remember that P(X = a) = 0 for continuous distributions, so P(X ≤ a) = P(X < a).
Tip 3: For "at least" problems, use: P(X ≥ a) = 1 - P(X < a).
Tip 4: Z-scores allow comparison of values from different normal distributions.
Common applications: Heights, weights, test scores, measurement errors, biological traits.
Key properties: Symmetric, unimodal, mean=median=mode, total area under curve = 1.
Essential formulas and rules:

• Standardization: Z = (X - μ)/σ

• Inverse transformation: X = μ + Z·σ

• Probability between values: P(a < X < b) = P(X < b) - P(X < a)

• Complement rule: P(X > a) = 1 - P(X ≤ a)

• Sampling distribution: X̄ ~ N(μ, σ²/n), SE = σ/√n

• Empirical rule: 68-95-99.7% within 1-2-3 standard deviations

Visualization: Normal Distribution Concepts
Normal Distribution Characteristics
Explore the characteristics of normal distributions:
Standard deviation effects, z-score interpretations, and probability calculations
Standard Normal: N(0,1), General Normal: N(μ,σ²)

Analysis: The chart shows how normal distributions vary with different parameters.

  • Standard deviation affects spread: larger σ means wider curve
  • Mean affects center: shifts curve left/right
  • Area under curve represents probability

Questions & Answers

Question: I don't understand why we need to standardize normal distributions. Why can't we just work with the original values?

Answer: Standardization is crucial for several reasons:

  • Universal reference: Once standardized, we can use the same standard normal table for any normal distribution
  • Comparison: Z-scores allow comparing values from different normal distributions with different means and standard deviations
  • Simplicity: We only need to memorize one table (standard normal) instead of having tables for every possible combination of μ and σ
  • Interpretation: Z-scores tell us exactly how many standard deviations away from the mean a value is

For example, a score of 85 on a test with mean 80 and SD 5 has z-score (85-80)/5 = 1, which is equivalent to a score of 95 on a test with mean 90 and SD 5, since (95-90)/5 = 1. Both are exactly 1 standard deviation above their respective means.

Question: What is the empirical rule (68-95-99.7) and how do I use it?

Answer: The empirical rule (also called the 68-95-99.7 rule) describes the approximate percentage of data within certain standard deviations of the mean in a normal distribution:

  • 68%: Within 1 standard deviation (μ ± σ)
  • 95%: Within 2 standard deviations (μ ± 2σ)
  • 99.7%: Within 3 standard deviations (μ ± 3σ)

Use it for:

  • Quick estimates without calculations
  • Checking reasonableness of exact calculations
  • Understanding the spread of data in normal distributions

For example, if heights are N(70, 3²), then about 95% of people have heights between 64 and 76 inches (70 ± 2×3).

Question: What's the difference between a normal distribution and a standard normal distribution?

Answer: The key differences are:

  • Normal Distribution: Any distribution of the form N(μ, σ²) with any mean μ and standard deviation σ
  • Standard Normal Distribution: Specifically N(0, 1) with mean 0 and standard deviation 1
  • Relationship: Any normal distribution can be converted to standard normal using Z = (X - μ)/σ
  • Usage: We standardize to use the standard normal table for probability calculations

The standard normal distribution serves as a universal reference that allows us to compute probabilities for any normal distribution by converting to z-scores.

The shape remains the same (bell curve), but standardization changes the scale and center of the distribution.

Question: How do I handle "between" probability problems with normal distributions?

Answer: For "between" probability problems, use the formula: P(a < X < b) = P(X < b) - P(X < a)

Steps:

  1. Standardize both bounds: Z_a = (a - μ)/σ and Z_b = (b - μ)/σ
  2. Find P(X < b) = P(Z < Z_b) using standard normal table
  3. Find P(X < a) = P(Z < Z_a) using standard normal table
  4. Subtract: P(a < X < b) = P(Z < Z_b) - P(Z < Z_a)

For example, if X ~ N(100, 15²) and you want P(85 < X < 115):

Z₁ = (85-100)/15 = -1, Z₂ = (115-100)/15 = 1

P(85 < X < 115) = P(Z < 1) - P(Z < -1) = 0.8413 - 0.1587 = 0.6826

Question: Why do we use the complement rule for "greater than" problems?

Answer: We use the complement rule because standard normal tables typically provide cumulative probabilities P(Z ≤ z), not P(Z > z).

The complement rule states: P(Z > z) = 1 - P(Z ≤ z)

This is particularly useful because:

  • Standard normal tables usually only go up to positive z-values
  • It's easier to subtract from 1 than to look up negative z-values
  • It ensures accuracy by using the primary table values

For example, to find P(Z > 1.5):

P(Z > 1.5) = 1 - P(Z ≤ 1.5) = 1 - 0.9332 = 0.0668

This approach works for any probability calculation where the table doesn't directly provide the needed value.