Question

Is it possible to have a Binomial Distribution in the following cases:

i. Mean = 5 and Variance = 2.5

ii. P(X = – 2) = 0.1467

iii. P(X less than or not equal to 1) = 0.18

iv. Three parameters i.e. n, p and q

v. P(X = 1) = 0.35

vi. P(X Greater than or equal to 1) = 1.27

vii. .P(X = 3.5) = 0.19

viii. P(1   less than or not equal to   X     less than or not equal to    4) = 0.79.

Answer 1

i.

$$\mu=np=5, Var(X)=npq=2.5$$

$$q=0.5, 0<0.5<1$$

It is possible.

ii. $X$ must be nonnegative.

It is impossible.

iii.

$$P(X<1\ or X\not=1)=1-P(X=1)=0.18$$

It is possible.

iv. Three parameters $n, p, q$ can define a Binomial Distribution.

It is possible.

v. $P(X=1)=0.35$

It is possible.

vi. $P(X\ge1)=1.27$

$$0\le p\le 1$$

It is impossible.

vii. $P(X=3.5)=0.19$

$X$ must be integer.

 It is impossible.

viii.

$$P((X<1\ or X\not=1)\cap(X<4\ or X\not=4))$$

$$=1-P(X=1)-P(X=4)=0.79$$

It is possible.