# Binomial Mean and Standard Deviation – Probability | Class 12 Maths

• Last Updated : 27 Jan, 2022

Binomial distribution is the probability distribution of no. of Bernoulli trials i.e. if a Bernoulli trial is performed n times the probability of its success is given by binomial distribution. Keep in mind that each trial is independent of another trial with only two possible outcomes satisfying the same conditions of Bernoulli trials.

Consider the case of tossing a coin n times, the probability of getting exactly x no. of heads /tails can be calculated using the binomial distribution. If in the same case tossing of a coin is performed only once it is the same as Bernoulli distribution.

A random variable X which takes values 1,2,…..n is said to follow binomial distribution if its probability distribution function is given by

### P(X = r) = nCr pr qn-r

Where,

r = 0, 1,2……, n, where p, q>0 such that p+q=1

p = probability of success of an event

q = probability of failure of an event

## The mean or Expected value of the binomial distribution

The mean of the binomial distribution is the same as the average of anything else which is equal to the submission of the product of no. of success and probability at each success.

Mean = ∑r r. P(r)

= ∑r r  nCr pr qn-r

= ∑r r n/r  n-1Cr-1 p.pr-1 qn-r    [as nCr= n/r  n-1Cr-1]

= np ∑r n-1Cr-1 pr-1 q(n-1)-(r-1)

= np(q+p)n-1       [by binomial theorem i.e. (a+b)n = ∑k=0 nCk an bn-k ]

=np [as p+q=1]

Therefore, Mean=np

## The variance of binomial distribution

We know, variance is the measurement of how spread the numbers are from the mean of the data set. Similarly, the variance of the binomial distribution is the measurement of how to spread the probability at each no. of success from the mean probability which is the average of the squared differences from the mean.

Variance = (∑r  r2. P(r)) – Mean2

= ∑r  [r(r-1)+r] nCr pr qn-r – (np)2

= ∑r  r(r-1) nCr  pr qn-r + ∑r  r nCr pr qn-r – (np)2

= ∑r r(r-1) n/r (n-1)/(r-1)  n-2Cr-2 p2 pr-2 qn-r +np – (np)2

= n(n-1)p2 {∑r  n-2Cr-2  pr-2 qn-r } +np – (np)2

= n(n-1) p2 (q+p)n-2 + np – n2p2         [by binomial theorem i.e. (a+b)n = ∑k=0 nCk an bn-k ]

= n2p2 -np2 +np-n2p2                        [as p+q=1]

= np-np2

= np(1-p)

= npq

Therefore, Variance=npq

## The standard deviation of binomial distribution

Standard deviation is also a standard measure to find out how to spread out are the no. from the mean value.

Standard Deviation = (Variance)1/2

= (npq)1/2

Example 1. A coin is tossed five times. What is the probability of getting exactly 3  times head? Also find the mean, variance, and standard deviation.

Solution:

n = 5 (no. of trials)

p = probability of getting head at each trial

=1/2

q = 1-1/2 = 1/2

r = 3 ( no. of successes i.e. getting a head)

P(X=r) = nCr pr qn-r

= 5C3 (1/2)3 (1/2)5-3

= 5!/(3!*2!) 1/8 * 1/4

= 10 *(1/8)*(1/4)

= 5/16

Mean = np

= 5 * 1/2 = 5/2

Variance = npq = 5 * 1/2 * 1/2

= 5/4

Standard deviation = (5/4)1/

= 51/2/2

Example 2. A die is tossed thrice. What is the probability of getting an even number? What are the mean, variance, and standard deviation of the binomial distribution?

Solution:

Here, n = 3(no. of trials)

p = probability of getting an even number during each trial

p = 3/6=1/2 [ 2,4,6 are even no. in dice]

q = 1-1/2 =1/2

r= 1( no. of successes i.e. getting a even no. )

P(X=r)= nCr * pr * qn-r

=3C (1/2)1  (1/2)3-1

= 3!/(1!*2!) 1/2 * 1/4

= 3 * (1/2) * (1/4)

= 3/8

Mean = np

= 3 * 1/2 = 3/2

Variance = npq = 3 * 1/2 * 1/2

= 3/4

Standard deviation= (3/4)1/2

=31/2/2

Example 3. If the probability of defective bolts is 0.1, find the mean, variance, and standard deviation for the distribution of defective bolts in a total of 500 bolts.

Solution:

Considering as a case of binomial distribution ,

n = 500( no. of trials which we can are no. of bolts here)

p = probability of one defective bolt during each trial

p = 0.1

Q=1-0.1 =0.9

Mean = np

= 500 * 0.1 =50

Variance = npq

= 500 * 0.1 * 0.9 = 45

Standard Deviation = (variance)1/2

= (45)1/2 = 6.71

Example 4. Two cards are drawn successively from a pack of 52 cards with replacement. Find the probability distribution for no. of aces. Also find the mean, variance, and standard deviation.

Solution:

n = 2(no. of trials)

p = probability of getting an ace in each trial

= 4/52 =1/13

q = 1-1/13 =12/13

r = no. of successes i.e no. of aces (0,1,2)

P(X=r) = 2Cr (1/13)r (12/13)2-r

For r = 0

P(0) = 2C0 (1/13) (12/13) 2-0

= 144/169

For r=1

P(1) = 2C1 (1/13)1 (12/13)2-1

= 24/169

For r=2

P(2) = 2C2 (1/13) 2 (12/13)2-2

=1/169

Therefore, probability distribution can be given as :

Mean =np

= 2 * 1/13 =2/13

Variance = npq

= 2 * (1/13) * (12/13)

= 24/16

Standard Deviation= (variance)1/2

= (24/169)1/2 = 0.376

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