Pk = e- k 0,1,2,... The (a, b, 0) class. pmf Distribution a b. Po Poisson k! Binomial Pk = (")g*(1-q)m-k, k = 0,1,2,..., m -L | (m+ 1)은 | (1-q)m k Negative binomial PL= (t-1) . k-0,1,2. | (r-1)음 | (1+p)-" (r- 1) (1+ B)-" %3D 1+B 1+ 1+ß Geometric Pk = k = 0, 1,2,... (1+ B)-1 1+B, 1+B) 1+B Three consecutive probabilities of a member of the (a, b,0) class is as follows. 108 54 12 256' P2 = 256 P3 = 256 (a) Determines the entire distribution using the table above. Find p7. (b) Find the expected value and the variance.
Pk = e- k 0,1,2,... The (a, b, 0) class. pmf Distribution a b. Po Poisson k! Binomial Pk = (")g*(1-q)m-k, k = 0,1,2,..., m -L | (m+ 1)은 | (1-q)m k Negative binomial PL= (t-1) . k-0,1,2. | (r-1)음 | (1+p)-" (r- 1) (1+ B)-" %3D 1+B 1+ 1+ß Geometric Pk = k = 0, 1,2,... (1+ B)-1 1+B, 1+B) 1+B Three consecutive probabilities of a member of the (a, b,0) class is as follows. 108 54 12 256' P2 = 256 P3 = 256 (a) Determines the entire distribution using the table above. Find p7. (b) Find the expected value and the variance.
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Author:Sheldon Ross
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Transcribed Image Text:Pk = e- k 0,1,2,...
The (a, b, 0) class.
pmf
Distribution
a
b.
Po
Poisson
k!
Binomial
Pk = (")g*(1-q)m-k, k = 0,1,2,..., m
-L | (m+ 1)은 | (1-q)m
k
Negative binomial
PL= (t-1) . k-0,1,2. | (r-1)음 | (1+p)-"
(r- 1) (1+ B)-"
%3D
1+B
1+
1+ß
Geometric
Pk =
k = 0, 1,2,...
(1+ B)-1
1+B,
1+B)
1+B
Three consecutive probabilities of a member of the (a, b,0) class is as follows.
108
54
12
256'
P2 =
256
P3 =
256
(a)
Determines the entire distribution using the table above. Find p7.
(b)
Find the expected value and the variance.
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