Page 277 - DCOM203_DMGT204_QUANTITATIVE_TECHNIQUES_I

Quantitative Techniques – I

Notes 2 2 2 2 2 2 2 2
E r 2npE r n p E r 2n p n p
2
E r 2 n p 2 .... (1)
2
Thus, to find we first determine E(r ).
n
r
n
. C p q
Now, E r 2 r 2 n r r n r r r 1 r C p q n r
r
r 1
n n n n r r 1 !
n
n
r
r
r
r r 1 C p q n r . r C p q n r . p q n r np
r r
r 2 r 1 r 2 !r n r !
n n
! n r n r n n 1 . n 2 ! r n r
. p q np . p q np
r 2 r 2 ! n r ! r 2 r 2 ! n r !
n n 2 !
n n 1 p 2 . p r 2 q n r np

r 2 r 2 ! n r !
2
n n 1 p q p n 2 np n n 1 p 2 np
Substituting this value in equation (1), we get
2 2 2 2
n n 1 p np n p np 1 p npq
Or the standard deviation = npq
Remarks: 2 npq mean q , which shows that since 0 < q <1.
3. The values of , , and
3 4 1 2
Proceeding as above, we can obtain
3
E r np npq q p
3
4 2 2 2
E r np 3n p q npq 1 6pq
4
2 2 2
2 n p q q p 2 q p 2
3
Also 1 3 3 3 3
2 n p q npq
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
3n p q npq 1 6pq 1 6pq
4
2 2 2 2 2 3 npq
2 n p q
The above result shows that the distribution is leptokurtic if 6pq < 1, platykurtic if
6pq > 1 and mesokurtic if 6pq = 1.
4. Mode: Mode is that value of the random variable for which probability is maximum.
If r is mode of a binomial distribution, we have

P(r – 1) P(r) P(r + 1)
Consider the inequality P(r) P(r + 1)

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