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

Binomial distribution is the probability distribution of no. of Bernoulli trials i.e. if a Bernoulli trail is performed n times the probability of its success is given by binomial distribution. Keeping in mind that each trial is independent of other trial with only two possible outcomes satisfying 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 binomial distribution. If in the same case tossing of a coin is performed only once it is 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) = **^{n}C_{r} p^{r} q^{n-r}

^{n}C

_{r}p

^{r}q

^{n-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

## Mean or Expected value of binomial distribution

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

Mean= ∑_{r}r. P(r)

= ∑_{r }r^{n}C_{r}p^{r}q^{n-r}

= ∑_{r}r n/r^{n-1}C_{r-1}p.p^{r-1}q^{n-r }[as^{n}C_{r}= n/r^{ n-1}C_{r-1}]

= np ∑_{r}^{n-1}C_{r-1}p^{r-1}q^{(n-1)-(r-1)}

= np(q+p)^{n-1 }[by binomial theorem i.e. (a+b)^{n}= ∑_{k=0}^{n}C_{k}a^{n}b^{n-k }]

=np [as p+q=1]

Therefore,Mean=np

## 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 binomial distribution is the measurement of how 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}r^{2}. P(r)) – Mean^{2}

= ∑_{r}[r(r-1)+r]^{n}C_{r}p^{r}q^{n-r}– (np)^{2}

= ∑_{r}r(r-1)^{n}C_{r}p^{r}q^{n-r }+ ∑_{r}r^{n}C_{r }p^{r}q^{n-r}– (np)^{2}

= ∑_{r}r(r-1) n/r (n-1)/(r-1)^{n-2}C_{r-2}p^{2}p^{r-2}q^{n-r }+np – (np)^{2}

= n(n-1)p^{2}{∑_{r}^{n-2}C_{r-2}p^{r-2}q^{n-r }} +np – (np)^{2}

= n(n-1) p^{2}(q+p)^{n-2}+ np – n^{2}p^{2 }[by binomial theorem i.e. (a+b)^{n}= ∑_{k=0}^{n}C_{k}a^{n }b^{n-k}]

= n^{2}p^{2}-np^{2}+np-n^{2}p^{2 }[as p+q=1]

= np-np^{2}

= np(1-p)

= npq

Therefore,Variance=npq

## Standard deviation of binomial distribution

Standard deviation is also a standard measure to find out how 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 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/

= 5

^{1/2}/2

**Example 2. A die is tossed thrice. What is the probability of getting an even number. What is the mean, variance and standard deviation of 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)=

^{n}C_{r }* pr * q^{n-r}=

^{3}C_{1 }(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 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 :

X 0 1 2 P(X) 144/169 24/169 1/169 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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