A discrete random variable x is defined to follow binomial distribution with parameters n and p to be denoted by
x ~ B(n, p)
If the probability of mass function of x is given by
The mode of the binomial distribution
mode (μ_{0}) = the largest integer contained in (n + 1)p
mode (μ_{0}) = (n+1) p and (n+1)p - 1
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.
May 21, 24 08:51 PM
May 21, 24 08:51 AM
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