Poisson Distribution worksheet

1. P(X = 2), exactly 2: 0.15

1. Start with the Poisson formula. It gives the chance of exactly k events when they happen on their own at an average of λ per stretch

   $P\left(\text{X = k}\right)$$=\frac{e^{\text{−}\lambda }\times {\lambda }^k}{k\text{!}}$

2. Put in your numbers: the average λ = 4 and the count k = 2

   $P\left(\text{X = 2}\right)$$=\frac{e^{\mathrm{-4}}\times 4^2}{2\text{!}}$$=\frac{0.0183156389\times 16}{2}$$=0.14652511111$

3. Add the chances of 0 through 2 events to get the chance of at most 2

   $P\left(\text{X ≤ 2}\right)$$=0.0183156389+0.0732625556+0.1465251111$$=0.238103305554$

4. The other answers follow from that total. Fewer than k drops the exactly-k chance; more than k and at least k are what's left of 100%

   $P\left(\text{X < 2}\right)=0.091578194444\text{,}P\left(\text{X > 2}\right)=0.761896694446\text{,}P\left(\text{X ≥ 2}\right)=0.908421805556$

5. For a Poisson count, the average and the variance (spread) both equal λ, so the standard deviation σ is the square root of λ

   $\sigma$$=\sqrt{4}$$=2$

2. P(X = 2), exactly 2: 0.08

1. Start with the Poisson formula. It gives the chance of exactly k events when they happen on their own at an average of λ per stretch

   $P\left(\text{X = k}\right)$$=\frac{e^{\text{−}\lambda }\times {\lambda }^k}{k\text{!}}$

2. Put in your numbers: the average λ = 5 and the count k = 2

   $P\left(\text{X = 2}\right)$$=\frac{e^{\mathrm{-5}}\times 5^2}{2\text{!}}$$=\frac{0.006737947\times 25}{2}$$=0.084224337489$

3. Add the chances of 0 through 2 events to get the chance of at most 2

   $P\left(\text{X ≤ 2}\right)$$=0.006737947+0.033689735+0.0842243375$$=0.124652019483$

4. The other answers follow from that total. Fewer than k drops the exactly-k chance; more than k and at least k are what's left of 100%

   $P\left(\text{X < 2}\right)=0.040427681995\text{,}P\left(\text{X > 2}\right)=0.875347980517\text{,}P\left(\text{X ≥ 2}\right)=0.959572318005$

5. For a Poisson count, the average and the variance (spread) both equal λ, so the standard deviation σ is the square root of λ

   $\sigma$$=\sqrt{5}$$=2.2360679775$

3. P(X = 1), exactly 1: 0.15

1. Start with the Poisson formula. It gives the chance of exactly k events when they happen on their own at an average of λ per stretch

   $P\left(\text{X = k}\right)$$=\frac{e^{\text{−}\lambda }\times {\lambda }^k}{k\text{!}}$

2. Put in your numbers: the average λ = 3 and the count k = 1

   $P\left(\text{X = 1}\right)$$=\frac{e^{\mathrm{-3}}\times 3^1}{1\text{!}}$$=\frac{0.0497870684\times 3}{1}$$=0.149361205104$

3. Add the chances of 0 through 1 events to get the chance of at most 1

   $P\left(\text{X ≤ 1}\right)$$=0.0497870684+0.1493612051$$=0.199148273471$

4. The other answers follow from that total. Fewer than k drops the exactly-k chance; more than k and at least k are what's left of 100%

   $P\left(\text{X < 1}\right)=0.049787068368\text{,}P\left(\text{X > 1}\right)=0.800851726529\text{,}P\left(\text{X ≥ 1}\right)=0.950212931632$

5. For a Poisson count, the average and the variance (spread) both equal λ, so the standard deviation σ is the square root of λ

   $\sigma$$=\sqrt{3}$$=1.7320508076$

4. P(X = 3), exactly 3: 0.2

1. Start with the Poisson formula. It gives the chance of exactly k events when they happen on their own at an average of λ per stretch

   $P\left(\text{X = k}\right)$$=\frac{e^{\text{−}\lambda }\times {\lambda }^k}{k\text{!}}$

2. Put in your numbers: the average λ = 4 and the count k = 3

   $P\left(\text{X = 3}\right)$$=\frac{e^{\mathrm{-4}}\times 4^3}{3\text{!}}$$=\frac{0.0183156389\times 64}{6}$$=0.195366814813$

3. Add the chances of 0 through 3 events to get the chance of at most 3

   $P\left(\text{X ≤ 3}\right)$$=0.0183156389+0.0732625556+0.1465251111+0.1953668148$$=0.433470120367$

4. The other answers follow from that total. Fewer than k drops the exactly-k chance; more than k and at least k are what's left of 100%

   $P\left(\text{X < 3}\right)=0.238103305554\text{,}P\left(\text{X > 3}\right)=0.566529879633\text{,}P\left(\text{X ≥ 3}\right)=0.761896694446$

5. For a Poisson count, the average and the variance (spread) both equal λ, so the standard deviation σ is the square root of λ

   $\sigma$$=\sqrt{4}$$=2$

5. P(X = 2), exactly 2: 0.22

1. Start with the Poisson formula. It gives the chance of exactly k events when they happen on their own at an average of λ per stretch

   $P\left(\text{X = k}\right)$$=\frac{e^{\text{−}\lambda }\times {\lambda }^k}{k\text{!}}$

2. Put in your numbers: the average λ = 3 and the count k = 2

   $P\left(\text{X = 2}\right)$$=\frac{e^{\mathrm{-3}}\times 3^2}{2\text{!}}$$=\frac{0.0497870684\times 9}{2}$$=0.224041807655$

3. Add the chances of 0 through 2 events to get the chance of at most 2

   $P\left(\text{X ≤ 2}\right)$$=0.0497870684+0.1493612051+0.2240418077$$=0.423190081127$

4. The other answers follow from that total. Fewer than k drops the exactly-k chance; more than k and at least k are what's left of 100%

   $P\left(\text{X < 2}\right)=0.199148273471\text{,}P\left(\text{X > 2}\right)=0.576809918873\text{,}P\left(\text{X ≥ 2}\right)=0.800851726529$

5. For a Poisson count, the average and the variance (spread) both equal λ, so the standard deviation σ is the square root of λ

   $\sigma$$=\sqrt{3}$$=1.7320508076$

6. P(X = 3), exactly 3: 0.09

1. Start with the Poisson formula. It gives the chance of exactly k events when they happen on their own at an average of λ per stretch

   $P\left(\text{X = k}\right)$$=\frac{e^{\text{−}\lambda }\times {\lambda }^k}{k\text{!}}$

2. Put in your numbers: the average λ = 6 and the count k = 3

   $P\left(\text{X = 3}\right)$$=\frac{e^{\mathrm{-6}}\times 6^3}{3\text{!}}$$=\frac{0.0024787522\times 216}{6}$$=0.08923507836$

3. Add the chances of 0 through 3 events to get the chance of at most 3

   $P\left(\text{X ≤ 3}\right)$$=0.0024787522+0.0148725131+0.0446175392+0.0892350784$$=0.151203882777$

4. The other answers follow from that total. Fewer than k drops the exactly-k chance; more than k and at least k are what's left of 100%

   $P\left(\text{X < 3}\right)=0.061968804417\text{,}P\left(\text{X > 3}\right)=0.848796117223\text{,}P\left(\text{X ≥ 3}\right)=0.938031195583$

5. For a Poisson count, the average and the variance (spread) both equal λ, so the standard deviation σ is the square root of λ

   $\sigma$$=\sqrt{6}$$=2.4494897428$

7. P(X = 2), exactly 2: 0.04

1. Start with the Poisson formula. It gives the chance of exactly k events when they happen on their own at an average of λ per stretch

   $P\left(\text{X = k}\right)$$=\frac{e^{\text{−}\lambda }\times {\lambda }^k}{k\text{!}}$

2. Put in your numbers: the average λ = 6 and the count k = 2

   $P\left(\text{X = 2}\right)$$=\frac{e^{\mathrm{-6}}\times 6^2}{2\text{!}}$$=\frac{0.0024787522\times 36}{2}$$=0.04461753918$

3. Add the chances of 0 through 2 events to get the chance of at most 2

   $P\left(\text{X ≤ 2}\right)$$=0.0024787522+0.0148725131+0.0446175392$$=0.061968804417$

4. The other answers follow from that total. Fewer than k drops the exactly-k chance; more than k and at least k are what's left of 100%

   $P\left(\text{X < 2}\right)=0.017351265237\text{,}P\left(\text{X > 2}\right)=0.938031195583\text{,}P\left(\text{X ≥ 2}\right)=0.982648734763$

5. For a Poisson count, the average and the variance (spread) both equal λ, so the standard deviation σ is the square root of λ

   $\sigma$$=\sqrt{6}$$=2.4494897428$

8. P(X = 2), exactly 2: 0.01

1. Start with the Poisson formula. It gives the chance of exactly k events when they happen on their own at an average of λ per stretch

   $P\left(\text{X = k}\right)$$=\frac{e^{\text{−}\lambda }\times {\lambda }^k}{k\text{!}}$

2. Put in your numbers: the average λ = 8 and the count k = 2

   $P\left(\text{X = 2}\right)$$=\frac{e^{\mathrm{-8}}\times 8^2}{2\text{!}}$$=\frac{0.0003354626\times 64}{2}$$=0.010734804093$

3. Add the chances of 0 through 2 events to get the chance of at most 2

   $P\left(\text{X ≤ 2}\right)$$=0.0003354626+0.002683701+0.0107348041$$=0.013753967744$

4. The other answers follow from that total. Fewer than k drops the exactly-k chance; more than k and at least k are what's left of 100%

   $P\left(\text{X < 2}\right)=0.003019163651\text{,}P\left(\text{X > 2}\right)=0.986246032256\text{,}P\left(\text{X ≥ 2}\right)=0.996980836349$

5. For a Poisson count, the average and the variance (spread) both equal λ, so the standard deviation σ is the square root of λ

   $\sigma$$=\sqrt{8}$$=2.8284271247$

9. P(X = 1), exactly 1: 0.01

1. Start with the Poisson formula. It gives the chance of exactly k events when they happen on their own at an average of λ per stretch

   $P\left(\text{X = k}\right)$$=\frac{e^{\text{−}\lambda }\times {\lambda }^k}{k\text{!}}$

2. Put in your numbers: the average λ = 6 and the count k = 1

   $P\left(\text{X = 1}\right)$$=\frac{e^{\mathrm{-6}}\times 6^1}{1\text{!}}$$=\frac{0.0024787522\times 6}{1}$$=0.01487251306$

3. Add the chances of 0 through 1 events to get the chance of at most 1

   $P\left(\text{X ≤ 1}\right)$$=0.0024787522+0.0148725131$$=0.017351265237$

4. The other answers follow from that total. Fewer than k drops the exactly-k chance; more than k and at least k are what's left of 100%

   $P\left(\text{X < 1}\right)=0.002478752177\text{,}P\left(\text{X > 1}\right)=0.982648734763\text{,}P\left(\text{X ≥ 1}\right)=0.997521247823$

5. For a Poisson count, the average and the variance (spread) both equal λ, so the standard deviation σ is the square root of λ

   $\sigma$$=\sqrt{6}$$=2.4494897428$

10. P(X = 4), exactly 4: 0.18

1. Start with the Poisson formula. It gives the chance of exactly k events when they happen on their own at an average of λ per stretch

   $P\left(\text{X = k}\right)$$=\frac{e^{\text{−}\lambda }\times {\lambda }^k}{k\text{!}}$

2. Put in your numbers: the average λ = 5 and the count k = 4

   $P\left(\text{X = 4}\right)$$=\frac{e^{\mathrm{-5}}\times 5^4}{4\text{!}}$$=\frac{0.006737947\times 625}{24}$$=0.175467369768$

3. Add the chances of 0 through 4 events to get the chance of at most 4

   $P\left(\text{X ≤ 4}\right)$$=0.006737947+0.033689735+0.0842243375+0.1403738958+0.1754673698$$=0.440493285065$

4. The other answers follow from that total. Fewer than k drops the exactly-k chance; more than k and at least k are what's left of 100%

   $P\left(\text{X < 4}\right)=0.265025915297\text{,}P\left(\text{X > 4}\right)=0.559506714935\text{,}P\left(\text{X ≥ 4}\right)=0.734974084703$

5. For a Poisson count, the average and the variance (spread) both equal λ, so the standard deviation σ is the square root of λ

   $\sigma$$=\sqrt{5}$$=2.2360679775$