# How to find the mean/variance of the binomial distribution?

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The binomial distribution describes the probability when a certain trial with only two possible outcomes is repeated ‘n’ number of times.

For example, suppose that a coin is tossed 10 times and we define X=number of times we get heads. Then X follows the binomial probability distribution.

To find the mean of the binomial distribution we use the formula,

• Mean µ = n*p where,

n = total number of trials and,

p = probability of success of a single trial.

Example to find mean of binomial variable: Suppose an archer tries to hit a mark 20 times. The probability of success in a single hit is 0.8. Find the average number of times the archer is able to hit the mark.

Solution: We are given n=20 and,

p= probability of a successful hit=0.9

Then Mean = n*p = 20*0.9 = 18 and so we conclude that the archer hits the target an average of 18 out of 20 times.

To find the variance and standard deviation of the binomial distribution we use the formula,

• Variance σ2= n*p*(1-p)
• Standard Deviation σ = √{n*p*(1-p)}  where,

n = total number of trials and,

p = probability of success of a single trial.

Example to find variance and standard deviation of binomial variable: Suppose a factory manufactures 1000 bulbs. The probability of a bulb being defective in a single hit is 10%. Find the mean and variance of the number of defective bulbs.

Solution: We are given n = 1000 and p=10% = 0.1

Then Mean = n*p = 1000*0.1 = 100 so on average we have a hundred defective bulbs.

Variance σ2= n*p*(1-p) = 1000*0.1*0.9 = 90 and,

Standard Deviation σ = √{n*p*(1-p)}  = √90 = 9.49

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