Page 186 - DMTH404_STATISTICS

Statistics

Notes To find this probability, we assume that the first r trials are successes and remaining n - r trials
are failures. Since different trials are assumed to be independent, the probability of this sequence
is
p q q. ....

p p. ....
.
q
.

      r n-r



g
 b

r times n r times i.e. p q .
Since out of n trials any r trials can be success, the number of sequences showing any r trials as
n
success and remaining (n - r) trials as failure is C , where the probability of r successes in each
r
r n r
trial is p q . Hence, the required probability is ( )P r = n C p q  , where r = 0, 1, 2, ...... n.
r n-r
r
Writing this distribution in a tabular form, we have
r 0 1 2  n Total
0 n
2 n
n
r
P ( ) n C p q n C p q n 1 n C p q 2  n C p q 0 1
0 1 2 n
It should be noted here that the probabilities obtained for various values of r are the terms in the
n
binomial expansion of (q + p) and thus, the distribution is termed as Binomial Distribution.
n r n- r
P(r) = C p q is termed as the probability function or probability mass function (p.m.f.) of
r
the distribution.
14.1.2 Summary Measures of Binomial Distribution
(a) Mean
The mean of a binomial variate r, denoted by m , is equal to E(r), i.e.,
n n
r n r
n
r
r
m = E ( ) = å rP ( ) = å . r C p q  (note that the term for r = 0 is 0)
r
r= 0 r= 1
n r . ! n ( . n n  ) 1 !
n
r n r
r n r
= å . p q  = å . p q 
r= 1 ( ! r n r )! r= 1 (r  1 ) ( ! n r )!
n (n  ) 1 ! 1

q
+
np= å . p r 1 n r = np (q p+ ) n = np q p = 1
r= 1 (r  1 ) ( ! n r )!
(b) Variance
2
The variance of r, denoted by s , is given by
2 2
2
E r E
E r npù
+
s = é ë  ( ) r ù 2 = é ë  û 2 = E r é ë 2  2npr n p ù û
û
2 2
2 2
2 2
= E ( ) 2r 2  npE ( ) r + n p = E ( ) 2r 2  n p + n p
( ) n p= E r 2  2 2 .... (1)
2
Thus, to find s , we first determine E(r ).
2
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