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Normal distribution; normal and inverse normal calculations practice questions.

36 exam-style questions on Normal distribution; normal and inverse normal calculations — topic 4.9 of Statistics and Probability in the IB Mathematics: Applications and Interpretation syllabus — with answers and explanations. Samples below; the full set is free with an account.

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About this topic

Syllabus code
4.9
Subject
IB Mathematics: Applications and Interpretation
Topic
4 · Statistics and Probability
Questions
36 with worked answers
Question types
17 fill-in-the-gap questions · 10 multiple-choice questions · 3 flashcards · 3 true/false questions · 2 diagram tasks · 1 matching task
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Sample questions on Normal distribution; normal and inverse normal calculations

In a normal distribution, approximately what percentage of values lie within one standard deviation of the mean?

  • 50%
  • 68%
  • 95%
  • 99.7%

Answer: 68%

The empirical rule: about 68% within 1 SD, 95% within 2 SD, 99.7% within 3 SD of the mean.

The lifetimes of light bulbs are normally distributed with mean 1200 hours and standard deviation 150 hours. Calculate the probability that a bulb lasts more than 1400 hours, to 3 s.f.

  • 0.0912
  • 0.909
  • 0.500
  • 0.259

Answer: 0.0912

z = (1400 − 1200)/150 = 1.333; P(Z > 1.333) ≈ 0.0912.

The times taken to complete a task are normally distributed with mean 40 minutes and standard deviation 6 minutes. Determine the time exceeded by only 10% of people, to 3 s.f.

  • 47.7 minutes
  • 32.3 minutes
  • 52.0 minutes
  • 43.8 minutes

Answer: 47.7 minutes

The 90th percentile has z = 1.282, so x = 40 + 1.282 × 6 ≈ 47.7 minutes.

A normal distribution has mean 500 and standard deviation 100. Approximately what percentage of values lie between 300 and 700?

  • 68%
  • 95%
  • 99.7%
  • 50%

Answer: 95%

300 and 700 are two standard deviations from the mean, and about 95% of values lie within ±2σ.

The diameters of ball bearings are normally distributed with mean 12.0 mm and standard deviation 0.05 mm. Bearings outside 11.9 mm to 12.1 mm are rejected. Calculate the percentage rejected, to 3 s.f.

  • 4.55%
  • 95.4%
  • 2.28%
  • 0.270%

Answer: 4.55%

11.9 and 12.1 are ±2σ from the mean; P(within) ≈ 0.9545, so rejected ≈ 1 − 0.9545 = 0.0455 = 4.55%.

The daily sales of a shop are normally distributed with mean $2000 and standard deviation $300. Calculate the probability that daily sales are between $1700 and $2300, to 3 s.f.

  • 0.954
  • 0.683
  • 0.500
  • 0.997

Answer: 0.683

$1700 and $2300 are ±1σ from the mean, so P(−1 < Z < 1) ≈ 0.683.

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