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AI HL · 13.2

Probability Distributions

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13.2.13

Mixed practice: Poisson distribution

13.2.13

Mixed practice: Poisson distribution

Target: I can find Poisson probabilities for exactly, at most, fewer than, at least, more than and between values, using the calculator. (+5 more)

By the end of this section

Key points

Choosing the calculation

  • Exactly k: Poisson pdf. At most k: Poisson cdf with lower 0 and upper k. Fewer than k is \(P(X \le k-1)\).
  • At least k is \(1 - P(X \le k-1)\). More than k is \(1 - P(X \le k)\).
  • Between two values: one cdf with the lower and upper value in the settings. Count a strict bound one place in: \(P(4 < X \le 8) = P(5 \le X \le 8)\).
  • \(E(X) = Var(X) = \lambda\), so the standard deviation is \(\sqrt{\lambda}\).

Change the period first, then add

  • Rescale \(\lambda\) to the period in the question before you find any probability: 4.5 per hour is 1.5 in 20 minutes.
  • Independent \(Po(a)\) and \(Po(b)\) variables add to give \(Po(a+b)\). Add the means: never multiply them.
  • A probability over several independent days is the daily probability multiplied by itself: three days is \(p^3\).

Working backwards and combining

  • Given a probability and an unknown \(\lambda\), write the equation with the unknown in the Poisson box and use the calculator's solver. Keep the unrounded \(\lambda\).
  • A Poisson probability can be the p of a binomial: p = P(a page has a mistake), then count pages with \(B(n, p)\).
  • Given that: \(P(A \mid B) = P(A \text{ and } B) \div P(B)\).
  • The most likely number is the k with the largest \(P(X = k)\): compare it with its neighbours. For a mean that is not a whole number it is the whole number part. For a whole-number mean, two values tie.
  • For a stock level, find the smallest n with \(P(X \le n) \ge 0.95\) by trying values of n.

Is a Poisson model suitable?

  • The events must happen singly, at a constant average rate, and independently of each other.
  • To criticise the model, say how one event could change the chance of another (a storm causing several power cuts), or why the rate is not constant (rush hour, seasons).

On the calculator

Poisson pdf and cdf

  • Enter λ and x for an exact value. Enter λ, lower and upper for a range, with lower 0 for "at most".
  • Write the settings you used as part of your working.

Screenshot and video to come

Solving for λ

  • Write the equation with λ unknown inside the Poisson pdf or cdf, then solve for λ.
  • Keep the unrounded value of λ for the next part.

Screenshot and video to come

Finding a stock level from cdf values

  • Try values of n in the Poisson cdf with lower 0 until the probability first reaches 0.95.
  • Write the two values either side of 0.95 in your working.

Screenshot and video to come

Worked examples

Problems like the ones in class. Pick one, try it, then open the working.
Exactly and at most

The number of pigeons that land on a statue in a 10 minute period has a Poisson distribution with mean 5.6. Find the probability that exactly 4 pigeons land in a 10 minute period. Then find the probability that at most 4 land.

Working
  1. Define X

    X = number of pigeons in 10 minutes. X ~ Po(5.6)

    The mean is already given for this period.

  2. Exactly 4

    P(X = 4) → From the calculator: Poisson pdf with λ = 5.6, x = 4

    Exactly: use the pdf.

  3. Answer

    P(X = 4) = 0.152

  4. At most 4

    P(X ≤ 4) → From the calculator: Poisson cdf with λ = 5.6, lower = 0, upper = 4

    At most: use the cdf from 0 up to 4.

  5. Answer

    P(X ≤ 4) = 0.342

Video walkthrough coming
More than, and between

The number of ships arriving at a port in one day has a Poisson distribution with mean 7.1. Find the probability that more than 9 ships arrive in a day. Then find the probability that at least 4 and fewer than 8 arrive.

Working
  1. Define X

    X ~ Po(7.1)

    X = number of ships in a day.

  2. Whole numbers

    P(X > 9) = P(X ≥ 10) = 1 − P(X ≤ 9)

    More than 9 means 10 or more.

  3. Calculator

    From the calculator: Poisson cdf with λ = 7.1, lower = 0, upper = 9

    Lower 0, upper 9.

  4. Answer

    P(X > 9) = 1 − 0.8202 = 0.180

  5. Between

    P(4 ≤ X < 8) = P(4 ≤ X ≤ 7)

    Fewer than 8 stops at 7.

  6. Answer

    From the calculator: Poisson cdf with λ = 7.1, lower = 4, upper = 7 → P(4 ≤ X < 8) = 0.507

Video walkthrough coming
A longer stretch of road

The number of potholes on a country road has a Poisson distribution with a mean of 0.7 per kilometre. Find the probability that there are fewer than 3 potholes along a 6 km stretch.

Working
  1. One kilometre

    E(X) = Var(X) = 0.7

    For a Poisson variable the mean and the variance are the same.

  2. Rescale

    Y = number of potholes in 6 km. Y ~ Po(0.7 × 6) = Po(4.2)

    Six times the length means six times the mean.

  3. Calculator

    P(Y < 3) = P(Y ≤ 2) → From the calculator: Poisson cdf with λ = 4.2, lower = 0, upper = 2

    Fewer than 3 means 0, 1 or 2.

  4. Answer

    P(Y < 3) = 0.210

Video walkthrough coming
Two kinds of post

A small office receives first-class letters at a mean rate of 3.2 per day and second-class letters at a mean rate of 4.9 per day. Each number of letters has a Poisson distribution and the two are independent. Find the probability that more than 9 letters arrive in one day altogether.

Working
  1. Add the means

    T = first-class + second-class. T ~ Po(3.2 + 4.9) = Po(8.1)

    Independent Poisson variables add: so do their means.

  2. Upper tail

    P(T > 9) = 1 − P(T ≤ 9)

    More than 9: take P(T ≤ 9) from 1.

  3. Calculator

    From the calculator: Poisson cdf with λ = 8.1, lower = 0, upper = 9

  4. Answer

    P(T ≤ 9) = 0.7041, so P(T > 9) = 1 − 0.7041 = 0.296

Video walkthrough coming
Finding the mean

The number of power cuts in a town in one year has a Poisson distribution with mean λ. The probability of at least one power cut in a year is 0.9. Find λ.

Working
  1. Write the equation

    P(X ≥ 1) = 1 − P(X = 0) = 0.9

    At least one is the opposite of none.

  2. Equation for the solver

    1 − Poisson pdf(λ, x = 0) = 0.9

    Put the unknown λ in the Poisson box.

  3. Solve

    Solve for λ: λ = 2.303…

    Use the calculator's solver, not a formula.

  4. Next part

    Then P(X = 2) = 0.265 with λ = 2.303…

    Keep the unrounded λ for later parts.

Video walkthrough coming
Tiles in a box

The number of cracks in a ceramic tile has a Poisson distribution with mean 0.45. A box holds 20 tiles. Find the probability that exactly 5 tiles in a box have at least one crack.

Working
  1. One tile

    X = number of cracks in a tile. X ~ Po(0.45)

  2. Probability for one tile

    p = P(X ≥ 1) = 1 − P(X = 0) = 1 − 0.6376 = 0.3624…

    Check: the chance a tile has at least one crack.

  3. Binomial for the box

    Y = number of tiles with a crack. Y ~ B(20, 0.3624…)

    Each tile is a trial; use the unrounded p.

  4. Calculator

    From the calculator: binomial pdf with n = 20, p = 0.3624…, x = 5

  5. Answer

    P(Y = 5) = 0.113

Video walkthrough coming
Stock for a week

A hardware shop sells ladders at a mean rate of 2.2 per day, with a Poisson distribution. The shop is restocked every 7 days. Find the smallest number of ladders it should hold at the start of the week so that it is at least 95% certain that demand for the week does not exceed the stock.

Working
  1. Rescale

    W = demand in 7 days. W ~ Po(2.2 × 7) = Po(15.4)

    The stock must last a whole week.

  2. Condition

    Need P(W ≤ n) ≥ 0.95

    Demand must not be more than the stock n.

  3. Search

    From the calculator: Poisson cdf with λ = 15.4, lower = 0, upper = 21 → 0.934

    Try values of n. This is below 0.95.

  4. Next value

    upper = 22 → 0.958

    This is at least 0.95.

  5. Answer

    n = 22

    The smallest n that works.

Video walkthrough coming
Given at least one

The number of buses arriving at a stop in a 15 minute period has a Poisson distribution with mean 2.3. Given that at least one bus arrives, find the probability that exactly 3 buses arrive.

Working
  1. Write it

    P(X = 3 | X ≥ 1) = P(X = 3 and X ≥ 1) ÷ P(X ≥ 1)

    If X = 3 then X ≥ 1 is already true.

  2. Top line

    P(X = 3) = 0.2033 From the calculator: Poisson pdf with λ = 2.3, x = 3

    X = 3 is inside X ≥ 1, so the top is just P(X = 3).

  3. Bottom line

    P(X ≥ 1) = 1 − P(X = 0) = 1 − 0.1003 = 0.8997

  4. Answer

    0.2033 ÷ 0.8997 = 0.226

Video walkthrough coming