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IB Mathematics: Applications and Interpretation · Statistics and Probability

Binomial distribution; mean and variance practice questions.

36 exam-style questions on Binomial distribution; mean and variance — topic 4.8 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
4.8
Subject
IB Mathematics: Applications and Interpretation
Topic
4 · Statistics and Probability
Questions
36 with worked answers
Question types
15 fill-in-the-gap questions · 12 multiple-choice questions · 5 true/false questions · 3 flashcards · 1 matching task
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Sample questions on Binomial distribution; mean and variance

A fair six-sided die is rolled 12 times. Let X be the number of times a 5 is obtained. State the distribution of X.

  • B(12, 5)
  • B(12, 1/6)
  • B(6, 1/12)
  • B(1/6, 12)

Answer: B(12, 1/6)

For a binomial distribution B(n, p), n is the number of trials (12) and p is the probability of success on each trial (1/6 for rolling a 5).

X follows a binomial distribution B(50, 0.4). Calculate the standard deviation of X, to 3 s.f.

  • 3.46
  • 12.0
  • 3.16
  • 20.0

Answer: 3.46

Variance = np(1−p) = 50 × 0.4 × 0.6 = 12, so standard deviation = √12 = 3.46 (3 s.f.).

In a multiple-choice quiz, a student guesses randomly on 10 questions, each with 4 options. Let X be the number of correct answers. Calculate P(X ≥ 1), to 3 s.f.

  • 0.250
  • 0.944
  • 0.0563
  • 0.756

Answer: 0.944

P(X ≥ 1) = 1 − P(X = 0) = 1 − (0.75)¹⁰ = 1 − 0.0563 ≈ 0.944.

A quality-control test finds that 5% of manufactured bolts are defective. In a random sample of 30 bolts, let X be the number of defective bolts. Calculate P(X = 2), to 3 s.f.

  • 0.259
  • 0.0500
  • 0.339
  • 0.812

Answer: 0.259

P(X = 2) = C(30,2)(0.05)²(0.95)²⁸ = 435 × 0.0025 × 0.23783 ≈ 0.259.

A basketball player has a 0.8 probability of scoring each free throw. She takes 6 free throws. Calculate P(exactly 5 scored), to 3 s.f.

  • 0.262
  • 0.393
  • 0.800
  • 0.246

Answer: 0.393

P(X=5) = C(6,5)(0.8)⁵(0.2)¹ = 6 × 0.32768 × 0.2 ≈ 0.393.

For a binomial distribution B(n, p), which expression gives the variance?

  • np
  • √(np)
  • np(1−p)
  • n²p

Answer: np(1−p)

The variance of a binomial distribution is np(1−p); np is the mean, not the variance.

That was 6 of 36 Binomial distribution; mean and variance questions.

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