Page 187 - DMTH404_STATISTICS

Unit 14: Theoretical Probability Distributions

Notes
n

2

. C p q
E r
Now, ( ) = å r 2 n r r n r = é ë ( r r  ) 1 + ù û n r r n r
r C p q
r= 1
n n n ( r r  ) 1 !n
n
r n r
n
r n r
r n r
= å ( r r  ) 1 C p q  + å . r C p q  = å . p q  + np
r
r
r= 2 r= 1 r= 2 ( ! r n r )!
n ! n n ( n n  1 ) ( . n  ) 2 !
r n r
r n r
= å . p q  + np = å . p q  + np
r= 2 (r  2 ) ( ! n r )! r= 2 (r  2 ) ( ! n r )!
n (n  ) 2 !

n= (n  ) 1 p 2 å . p r 2 q n r + np
r= 2 (r  2 ) ( ! n r )!
2
n= (n  1 p 2 ) n 2 + np = ( n n  ) 1 p + np
) (q p+
Substituting this value in equation (1), we get
2 2 2 2
s = ( n n  ) 1 p + np n p = np ( 1 p ) = npq

Or the standard deviation = npq

2 2
=
´
Note s = npq mean q , which shows that s < mean , since 0 < q <1.
(c) The values of m , m , b and b
3 4 1 2
Proceeding as above, we can obtain

3
m = E (r np ) = npq (q p )
3
4 2 2 2
m = E (r np ) = 3n p q + npq (1 6pq )
4
2 2 2
m 3 2 n p q (q p ) 2 (q p ) 2
Also b = = =
1 3 3 3 3
m n p q npq
2
The above result shows that the distribution is symmetrical when
1
p = q = , negatively skewed if q < p, and positively skewed if q > p
2
2 2 2
m 3n p q + npq (1 6pq ) (1 6pq )
b = 4 = = 3 +
2 2 2 2 2
m 2 n p q npq
The above result shows that the distribution is leptokurtic if 6pq < 1, platykurtic if 6pq >
1 and mesokurtic if 6pq = 1.
(d) Mode

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