FIND MODE OF BINOMIAL DISTRIBUTION

Problem 1 :

What is the mode the distribution for which mean and standard deviation are 10 and √5 respectively.

Solution :

Mean = np = 10 ---(1)

Standard deviation = √npq = √5

npq = 5----(2)

Applying the value of np in (2)

10q = 5

q = 5/10

q = 1/2

p = 1/2

Applying the value of p in np = 10

n(1/2) = 10

n = 20

Value of (n + 1)p :

= (20 + 1) (1/2)

= 21/2

= 10.5 (not integer)

= 10 + 0.5

The largest integer = 10. So, the required mode is 10.

Problem 2 :

If x is a binomial variate with parameter 15 and 1/3, what is the value of mode of the distribution ?

Solution :

n = 15 and p = 1/3

Value of (n + 1)p :

= (15 + 1) (1/3)

= 16/3

= 5.3 (not a integer)

= 5 + 0.3

= 5 in the largest integer.

So. the mode is 5.

Problem 3 :

For a binomial distribution mean and mode

a) are never equal    b)  are always equal

c)  are equal when q = 0.50      d) do not always exist

Solution :

For a binomial distribution mean and mode are equal when q = 0.50

Problem 4 :

The mean of binomial distribution is 

a) Always more than its variance

b)  always equal to its variance

c)  always less than its variance 

d)  always equal to its standard deviation

Solution :

In binomial distribution there is a comparison between mean and variance.

np > npq

n= mean and npq = variance

Mean > variance

Mean is always more than its variance.

Problem 5 :

For a binomial distribution there may be,

a) one mode     b) two modes

Solution :

There may be two modes.