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When can the binomial or Poisson be approximated by the normal or Poisson, and how is the continuity correction applied?

Approximate the binomial by the Poisson or the normal, and the Poisson by the normal, under stated conditions, applying a continuity correction where appropriate

A focused answer to the H2 Mathematics outcome on distribution approximations. The conditions for the Poisson and normal approximations to the binomial and the normal approximation to the Poisson, and the continuity correction.

Generated by Claude Opus 4.810 min answer

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  1. What this dot point is asking
  2. The answer
  3. Examples in context
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What this dot point is asking

SEAB wants you to approximate one distribution by another under the stated conditions: the binomial by the Poisson (large nn, small pp), the binomial by the normal (large nn, pp not too extreme), and the Poisson by the normal (large λ\lambda), applying a continuity correction when approximating a discrete variable by a continuous one.

The answer

Poisson approximation to the binomial

When nn is large and pp is small (a common guide is n>50n > 50 and p<0.1p < 0.1), with λ=np\lambda = np moderate,

B(n,p)Po(np).\mathrm{B}(n, p) \approx \mathrm{Po}(np).

No continuity correction is needed because both are discrete.

Normal approximation to the binomial

When nn is large with np>5np > 5 and n(1p)>5n(1 - p) > 5,

B(n,p)N(np, np(1p)).\mathrm{B}(n, p) \approx \mathrm{N}\big(np,\ np(1 - p)\big).

A continuity correction is needed because a discrete variable is being approximated by a continuous one.

Normal approximation to the Poisson

When λ\lambda is large (a common guide is λ>10\lambda > 10),

Po(λ)N(λ,λ).\mathrm{Po}(\lambda) \approx \mathrm{N}(\lambda, \lambda).

Again a continuity correction applies.

The continuity correction

To approximate a discrete count by a continuous normal, widen each integer by half a unit:

  • P(Xk)P(X>k0.5)\mathrm{P}(X \geq k) \approx \mathrm{P}(X > k - 0.5)
  • P(Xk)P(X<k+0.5)\mathrm{P}(X \leq k) \approx \mathrm{P}(X < k + 0.5)
  • P(X=k)P(k0.5<X<k+0.5)\mathrm{P}(X = k) \approx \mathrm{P}(k - 0.5 < X < k + 0.5)

Forgetting the half-unit shift is the commonest error.

Examples in context

Example 1. Polling. A national poll of thousands of voters models the count supporting a party with B(n,p)\mathrm{B}(n, p) but computes it via the normal approximation, since nn is huge and pp moderate, with a continuity correction for exact counts.

Example 2. Call centre load. Calls at λ=50\lambda = 50 per hour are well approximated by N(50,50)\mathrm{N}(50, 50), letting a manager estimate the probability of exceeding capacity using normal tables rather than summing many Poisson terms.

Try this

Q1. State the conditions for approximating the binomial by the Poisson. [2 marks]

  • Cue. nn large and pp small (so npnp is moderate), with λ=np\lambda = np.

Q2. Apply the continuity correction to P(X20)\mathrm{P}(X \geq 20) for a normal approximation. [1 mark]

  • Cue. P(X>19.5)\mathrm{P}(X > 19.5).

Q3. For B(200,0.5)\mathrm{B}(200, 0.5), state the normal approximation's mean and variance. [2 marks]

  • Cue. Mean 100100, variance 200×0.5×0.5=50200 \times 0.5 \times 0.5 = 50.

Exam-style practice questions

Practice questions written in the style of SEAB exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

Original5 marksA biased coin lands heads with probability 0.40.4. It is tossed 100100 times. Using a suitable approximation, find the probability of at least 5050 heads.
Show worked answer →

XB(100,0.4)X \sim \mathrm{B}(100, 0.4). Since nn is large and np=40>5np = 40 > 5, n(1p)=60>5n(1-p) = 60 > 5, use the normal approximation: XN(40,24)X \approx \mathrm{N}(40, 24) (mean np=40np = 40, variance np(1p)=24np(1-p) = 24).

"At least 5050" with continuity correction: P(X50)P(X>49.5)\mathrm{P}(X \geq 50) \approx \mathrm{P}(X > 49.5).

Standardise: z=49.54024=9.54.8991.939z = \dfrac{49.5 - 40}{\sqrt{24}} = \dfrac{9.5}{4.899} \approx 1.939.

P(Z>1.939)10.9737=0.0263\mathrm{P}(Z > 1.939) \approx 1 - 0.9737 = 0.0263.

Markers reward the normal approximation with correct parameters, the continuity correction to 49.549.5, standardising, and the tail probability.

Original4 marksA rare disease affects 0.5%0.5\% of a population. In a sample of 400400, use a suitable approximation to find the probability of exactly 11 case.
Show worked answer →

XB(400,0.005)X \sim \mathrm{B}(400, 0.005). Since nn is large and pp is small with np=2np = 2, approximate by the Poisson: XPo(2)X \approx \mathrm{Po}(2).

P(X=1)=e2211!=2e22×0.1353=0.2707\mathrm{P}(X = 1) = \dfrac{e^{-2} 2^1}{1!} = 2e^{-2} \approx 2 \times 0.1353 = 0.2707.

Markers reward the Poisson approximation with λ=np=2\lambda = np = 2, the Poisson formula, and the probability.

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