Question

Find the mean for the values of n and p when the conditions for the binomial distribution are met.
   n = 100, p = 0.4

24

40

4.9

60

and

Which is a probability distribution?

X	1	2	3	4	5
P(X)	1/20	2/20	3/20	4/20	5/20

X	2	4	6	8	15
P(X)	1/19	3/19	4/19	4/19	7/19

X	1	2	3	4	5
P(X)	1/10	1/10	1/10	1/10	1/10

None of the above

Answer 1

1)here as we know that mean =np=100*0.4 =40

2) as for probability distribution sum of probabiliteis must be 1 as well each probability should be between 0 and 1.

i)

here sum of probabilities =(1/20)+(2/20)+(3/20)+(4/20)+(5/20)=15/20 which is not equal to 1 ; hence it is not valid,

ii)

here sum of probabilities =(1/19)+(3/19)+(4/19)+(4/19)+(7/19)=19/19=1 ; is equal to 1 as well all point probabilities are between 0 and 1 ; hence it is vaild probability distribution

iii)

sum of probabiltiies=(1/10)+(1/10)+(1/10)+(1/10)+(1/10) =5/10=1/2 which is not equal to 1 ; hence it is not valid probability distribution