FIND MEAN AND STANDARD DEVIATION OF BINOMIAL DISTRIBUTION

Problem 1 :

6 coins are tossed 512 times. Find the expected frequency of heads. Also compute the mean and standard deviation of the number number of heads.

Solution :

Number of coins tossed (n) = 6

Probability of getting heads = 1/2

not getting heads q = 1/2

Getting 0 heads :

Getting 1 head :

Getting 2 heads :

Getting 3 heads :

Getting 4 heads :

Getting 5 heads :

Getting 6 heads :

Expected Frequency :

Mean :

Mean = np

= 6 (1/2)

= 3

Standard deviation :

Variance = npq

Standard deviation = √npq

= √6x(1/2)x(1/2)

= √1.5

= 1.22

Problem 2 :

What is the standard deviation of the number of recoveries among 48 patients when the probability of recovering is 0.75 ?

a)  36     b)  81     c)  9       d) 3

Solution :

Number of patients = 48

Probability of recovering p = 0.75

q = 0.25

Standard deviation = √npq

= √48(0.75)(0.25)

= √9

= 3

Problem 3 :

X is a binomial variable with n = 20. What is the mean of X if it is known that X is symmetric ?

Solution :

Here n = 20

Since mean is symmetric p = q

p = q = 0.5

Mean = np

= 20(0.5)

= 10

Problem 4 :

What is the number of trails of binomial distribution having mean and standard deviation as 3 and 1.5 respectively ?

a)  2    b)  4    c)  8       d)  12

Solution :

mean = np = 3 ----(1)

Standard deviation = √npq = √1.5

npq = 1.5

Applying (1)

3q = 1.5

q = 1.5/3

q = 0.5

p = 0.5

np = 3

n(0.5) = 3

n = 3/0.5

n = 6

So, the number of trials is 6.

Problem 5 :

The mean of a binomial distribution with parameters n and p is 

a) n(1 - p)   b)  np (1 - p)    c)  np      d) √np(1 - p)

Solution :

Mean = np

So, option c is correct.

Problem 6 :

The variance of the binomial distribution with parameters n and p is 

a) np²(1 - p)     b) √np(1 - p)     c) nq(1 - p)     d) n²p²(1 - p)²

Solution :

Variance = npq

p + q = 1

q = 1 - p or p = 1 - q

Applying the value, we get

= np(1 - p) or nq(1 - q)

So, option c is correct.