Pollutant particles are found in discharge from a plant. Particle amounts vary from sample to sample. On average there are 7 particles per cubic meter of discharge.
What is the probability there are x particles in a 1 m3 sample of discharge? (x = 5)
Imagine slicing a cubic meter into 10
segments of 0.1m3 volume.
We could treat these segments as ¡°independent trials¡± and use the Binomial distribution to determine probabilities p(x).
0.25 0.2 0.15 0.1 0.05 0
n= 10
p = 0.7
What is the probability there are x particles in a 1 m3 sample of discharge?
Imagine slicing a cubic meter into n = 10 segments of 0.1m3 volume.
We could treat these segments as ¡°independent trials¡± and use the Binomial distribution to determine probabilities p(x).
0 5 10 15 # of particles
b(x; 10, 0.7) is plotted b(5; 10, 0.7) = 0.1029 E(X) = 10(0.7) = 7
V(X) = 10(0.7)(0.3) = 2.1
Prob
0.25 0.2 0.15 0.1 0.05 0
n= 100
p = 0.07
What is the probability there are x particles in a 1 m3 sample of discharge?
Imagine slicing a cubic meter into n = 100 segments of 0.01m3 volume.
We could treat these segments as ¡°independent trials¡± and use the Binomial distribution to determine probabilities p(x).
0 5 10 15 # of particles
b(x; 100, 0.07) is plotted
b(5; 100, 0.07) = 0.1283
E(X) = 100(0.7) = 7
V(X) = 100(0.07)(0.93) = 6.51
Prob
0.25 0.2 0.15 0.1 0.05 0
n= 1000
p = 0.007
What is the probability there are x particles in a 1 m3 sample of discharge?
Imagine slicing a cubic meter into n = 1000 segments of 0.001m3 volume.
We could treat these segments as ¡°independent trials¡± and use the Binomial distribution to determine probabilities p(x).
0 5 10 15 # of particles
b(x; 1000, 0.007) is plotted
b(5; 1000, 0.007) = 0.1278
E(X) = 1000(0.007) = 7
V(X) = 1000(0.007)(0.993) = 6.951
Prob
0.25 0.2 0.15 0.1 0.05 0
n= 10000
p = 7e-04
What is the probability there are x particles in a 1 m3 sample of discharge?
Imagine slicing a cubic meter into n = 10000 segments of 0.0001m3 volume.
We could treat these segments as ¡°independent trials¡± and use the Binomial distribution to determine probabilities p(x).
0 5 10 15 # of particles
b(x; 10000, 0.0007) is plotted
b(5; 10000, 0.0007) = 0.1277
E(X) = 10000(0.0007) = 7
V(X) = 100(0.0007)(0.9993) = 6.9951
Prob
0.25 0.2 0.15 0.1 0.05 0
n= 1e+05
p = 7e-05
What is the probability there are x particles in a 1 m3 sample of discharge?
Imagine slicing a cubic meter into n = 100000 segments of 0.00001m3 volume.
We could treat these segments as ¡°independent trials¡± and use the Binomial distribution to determine probabilities p(x).
0 5 10 15 # of particles
b(x; 100000, 0.00007) is plotted
b(5; 100000, 0.00007) = 0.1277
E(X) = 100000(0.0007) = 7
V(X) = 100000(0.00007)(0.99993) = 6.99951
Prob
0.25 0.2 0.15 0.1 0.05 0
n= 1e+06
p = 7e-06
What is the probability there are x particles in a 1 m3 sample of discharge?
Imagine slicing a cubic meter into n = 1000000 segments of 0.000001m3 volume.
We could treat these segments as ¡°independent trials¡± and use the Binomial distribution to determine probabilities p(x).
0 5 10 15 # of particles
b(x; 1000000, 0.000007) is plotted
b(5; 1000000, 0.000007) = 0.1277
E(X) = 1000000(0.00007) = 7
V(X) = 1000000(0.000007)(0.999993) = 6.999951
Prob
0.25 0.2 0.15 0.1 0.05
00 5 10 15 # of particles
What is the probability there are x particles in a 1 m3 sample of discharge?
The limiting distribution is the
Poisson distribution. Here the mean is 7.
The mean and the variance are the same.
p(x; 7) is plotted p(5; 7) = 0.1277 E(X) = 7
V(X) = 7 ¦Ò
X
= ¡Ì7 = 2.646
Prob