Normal Distribution Estimation

Grade 11 · statistics · 101 practice problems · read aloud

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Normal Distribution Estimation

What Is It & Why Useful?

The normal distribution is a bell-shaped curve describing many natural phenomena. Estimation lets us approximate probabilities without complex calculations or Z-tables. It's a quick, powerful tool for real-world statistics.

Step-by-Step Guide

  1. Identify: Find the mean (μ) and standard deviation (σ).
  2. Visualize: Sketch the bell curve. Mark μ in the center and μ±1σ, μ±2σ, μ±3σ.
  3. Locate: Place your data value(s) on this sketch.
  4. Apply the Empirical Rule:
    • ≈68% of data falls within μ ± 1σ
    • ≈95% of data falls within μ ± 2σ
    • ≈99.7% of data falls within μ ± 3σ
  5. Estimate: Use the rule to find the approximate probability or percentage.

Worked Examples

Example 1: Test scores are normally distributed with μ=80, σ=5. Estimate the percentage of scores between 75 and 85.

Step 1: μ=80, σ=5.
Step 2: 75 is μ-1σ (80-5). 85 is μ+1σ (80+5).
Step 3: The range 75-85 is within μ ± 1σ.
Step 4: The Empirical Rule states ≈68% of data is here.
Answer: ≈68%

Example 2: Using the same test scores (μ=80, σ=5), estimate the percentage scoring above 90.

Step 1: μ=80, σ=5.
Step 2: 90 is μ+2σ (80+10).
Step 3: The Empirical Rule says ≈95% of data is within μ±2σ. This means ≈5% is outside this range (in both tails).
Step 4: Since the curve is symmetric, only ≈2.5% is in the upper tail above 90.
Answer: ≈2.5%

Common Mistakes ⚠️

Misapplying the Rule: The percentages (68%, 95%, 99.7%) apply ONLY to intervals of 1, 2, and 3 standard deviations from the mean. Don't use them for other intervals.

Forgetting Tail Symmetry: When finding an area in one tail, remember to split the "outside" percentage in half.

Tips & Tricks

Memory Aid: Remember the sequence 68, 95, 99.7. The gaps between these numbers are 27 (95-68), and 4.7 (99.7-95), which get smaller.

Sketch It! 🎨 Always draw the curve. Shade the area you're finding. This prevents logic errors with "above" or "below" a value.

How to Practice

  • Start by sketching curves for different μ and σ values.
  • Find real-world datasets (heights, exam scores) and practice estimating proportions.
  • Create your own problems and solve them, then check with a partner.

Practice problems

6 of the 101, worked through step by step — try them before opening the answer.

1 log₂(32) - log₂(2) = ?

Hint: Remember the logarithmic property that relates subtraction of logs to division of their arguments

Show the answer

Answer: 4

  1. Apply the logarithmic subtraction property: log₂(32) - log₂(2) = log₂(32/2)
  2. Simplify the division: 32/2 = 16
  3. Evaluate log₂(16): 2^4 = 16
  4. Therefore, log₂(16) = 4

The answer is 4.

2 Mere's exam scores are normally distributed with μ=78 and σ=12. Estimate the percentage of scores between 66 and 90.

Hint: Convert the given scores to z-scores using the formula z = (x - μ)/σ, then use the empirical rule to estimate the percentage between these z-scores.

Show the answer

Answer: 68

  1. Calculate the z-score for 66: z = (66 - 78)/12 = -12/12 = -1
  2. Calculate the z-score for 90: z = (90 - 78)/12 = 12/12 = 1
  3. According to the empirical rule for normal distributions, approximately 68% of data falls within 1 standard deviation of the mean (between z = -1 and z = 1)
  4. Therefore, approximately 68% of Mere's exam scores fall between 66 and 90.

The answer is 68.

3 Emma's exam scores are normally distributed with μ=85 and σ=5. What percentage of students scored between 80 and 90?

Hint: Convert the scores to z-scores using the formula z = (x - μ)/σ, then use the standard normal distribution table to find the area between these z-scores.

Show the answer

Answer: 68.27

  1. Calculate z-score for 80: z = (80 - 85)/5 = -5/5 = -1
  2. Calculate z-score for 90: z = (90 - 85)/5 = 5/5 = 1
  3. Look up area to the left of z = 1 in standard normal table: 0.8413
  4. Look up area to the left of z = -1 in standard normal table: 0.1587
  5. Subtract to find area between z = -1 and z = 1: 0.8413 - 0.1587 = 0.6826
  6. Convert to percentage: 0.6826 × 100 = 68.26%

The answer is 68.27% (rounded to two decimal places).

4 Emma's test scores are normally distributed with μ = 70 and σ = 5. Estimate the percentage of scores greater than 80.

Hint: Convert the score to a z-score using the formula z = (x - μ)/σ, then use the standard normal distribution table or the empirical rule to find the area to the right of that z-score.

Show the answer

Answer: 2.28

  1. Calculate the z-score for 80: z = (80 - 70) / 5 = 10 / 5 = 2.00.
  2. The area to the left of z = 2.00 in the standard normal distribution is approximately 0.9772 (from the standard normal table).
  3. The area to the right (greater than 80) is 1 - 0.9772 = 0.0228.
  4. Convert to percentage: 0.0228 × 100 = 2.28%.

The answer is 2.28.

5 Olivia's test scores are normally distributed with μ=85 and σ=5. What percentage of students scored between 80 and 90?

Hint: Convert the given scores to z-scores using the formula z = (x - μ)/σ, then use the empirical rule or standard normal distribution table to find the area between these z-scores.

Show the answer

Answer: 68.27

  1. Calculate z-score for 80: z = (80 - 85)/5 = -5/5 = -1
  2. Calculate z-score for 90: z = (90 - 85)/5 = 5/5 = 1
  3. Using the standard normal distribution table, the area to the left of z = 1 is 0.8413
  4. The area to the left of z = -1 is 0.1587
  5. The area between z = -1 and z = 1 is 0.8413 - 0.1587 = 0.6826
  6. Convert to percentage: 0.6826 × 100 = 68.26%
  7. Using the empirical rule, approximately 68.27% of data falls within 1 standard deviation of the mean Final answer: 68.27%

6 Aroha's test scores are normally distributed with μ = 84 and σ = 6. Estimate the percentage of scores greater than 96.

Hint: Convert the given score to a z-score using the formula z = (x - μ)/σ, then use the standard normal distribution table or the empirical rule to find the area to the right of that z-score.

Show the answer

Answer: 2.28

  1. Calculate the z-score for 96: z = (96 - 84) / 6 = 12 / 6 = 2.00
  2. The area to the left of z = 2.00 in the standard normal distribution is approximately 0.9772 (from the standard normal table).
  3. The area to the right (greater than 96) is 1 - 0.9772 = 0.0228.
  4. Convert to percentage: 0.0228 × 100 = 2.28%.

The answer is 2.28.

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