binomial distribution practice problems

In mathematics, binomial distribution is one of the interesting topics in probability theory and statistics. Binomial distribution is one of the main type’s theoretical frequency distributions. Binomial distribution is also called as a Bernoulli experiment.  Binomial distribution is the process of the number of success in a sequence for the n independent trails. Each trails gains the success of probability p. The following are the example and practice problems in binomial distribution.

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Binomial distribution practice problems – Mean Variance and standard deviation:

Mean:

μ = E[x] = np

Standard deviation:

σ = `sqrt(npq)`

Variance:

E[x^2] = σ^2 = npq

Binomial distribution practice problems – Example problems:

Here we solve some example problems based on the binomial distribution

Example 1:

A coin is tossed twelve times. Calculate the expected number of heads, variance and the standard deviation by using the binomial distribution.

Solution:

Given

Let coin tossed for ten times, so n = 12

If we toss a coin means, probability of getting head is p = `1/ 2`

Probability of getting tails is denoted by q

q = 1 – p                        [p + q = 1]

q = 1 – `1/ 2`

q = `1/2`

Mean:

μ = E[x] = np

= 12 (`1/2` )

= (`12/2` )

μ = E[x] = 6

Variance:

E[x^2] = σ^2 = npq

= 12(`1/2` )(`1/2` )

= 12(`1/4` )

= (`12/4` )

= 3

E[x^2] = σ^2 = 3

Standard deviation:

σ = sqrt(3)

= 1.732

σ = 1.732

Answer:

Mean = μ = E[x] = 6

Variance = σ^2 = E[x^2] = 3

Standard deviation = σ = 1.732

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Example 2:

A die is rolled for 50 times. Calculate the expected number , variance and the standard deviation by using the binomial distribution.

Solution:

Given

Let die is rolled for three times, so n = 50

If we roll a die means, probability of getting is p = `1/ 6`

Probability of not getting is denoted by q

q = 1 – p                        [p + q = 1]

q = 1 – `1/ 6`

q =` 5/6`

Mean:

μ = E[x] = np

= 50 (`1/6` )

= `50/6`

= `25/ 3`

μ = E[x] = 8.33 (or) `25/ 3`

Variance:

E[x^2] = σ^2 = npq

= 50(`1/6` )(`5/6` )

= 50(`5/36` )

= `125/18`

E[x^2] = σ^2 = `125/18` = 6.94

Standard deviation:

σ = `sqrt(6.94)`

= 2.64

σ = 2.64

Answer:

Mean = μ = E[x] = 8.33 (or) =`25/3`

Variance = σ^2 = E[x^2] = 6.94 (or) = `125/18`

Standard deviation = σ = 2.64

Binomial distribution practice problems – practice problems:

Problem 1:

A coin is tossed for twenty times. Calculate the expected number of heads, variance and the standard deviation by using the binomial distribution.

Solution:

Mean = μ = E[x] = 10

Variance = σ^2 = E[x^2] = 5

Standard deviation = σ = 2.23

Problem 2:

A coin is tossed for fifteen times. Calculate the expected number of heads, variance and the standard deviation by using the binomial distribution.

Solution:

Mean = μ = E[x] = 7.5

Variance = σ^2 = E[x^2] = 3.75

Standard deviation = σ = 1.94