Binomial Distribution Examples
Binomial distribution is a very important topic of statistics. To learn it better, it is very important to workout binomial distribution examples. This page is based on binomial distribution examples. Workout these examples here and gain your quality statistics help.
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 trail gains the success of probability p. Tutor will help the students with step by step solutions for the given problem. The following are the example problems in binomial distribution examples tutor.
Looking out for more help on Binomial Distribution Table in math by visiting listed websites.
Binomial distribution Formula
Formulas:
P(X = x) = nCx px (1 – p)(n-x)
Where
n = number of events
x = number of successful events
p = in a single trail probability of success
nCx = `(n!)/((n - x)!(x!))`
1 – p = probability of failure
Mean (or) Expected value:
μ = E[x] = np
Standard deviation:
σ = `sqrt(npq)`
Variance:
E[x2] = σ2 = npq
Binomial distribution Problems
My forthcoming post is on Binomial Calculator and simple math problems for kids will give you more understanding about Math.
Here we solving some examples based on the binomial distribution
Example 1:
A coin is tossed for eighty-eight times. Determine the expected value of heads, variance and the standard deviation by using the binomial distribution.
Solution:
Given
Let coin tossed for eighty-eight times, so n = 88
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
= 88 (`1/2` )
= (`88/2` )
μ = E[x] = 44
Variance:
E[x2] = σ2 = npq
= 88(`1/2` )(`1/2` )
= 88(`1/4` )
= (`88/4` )
= 22
E[x2] = σ2 = 22
Standard deviation:
σ = `sqrt (npq)`
σ = `sqrt(22)`
= 4.69
σ = 4.69
Answer:
Mean = μ = E[x] = 44
Variance = E[x2] = σ2] = 22
Standard deviation = σ = 4.69
Example 2:
A coin is tossed for fifteen times. Determine the probability of getting exactly 10 heads.
Solution:
Given
Let coin tossed for fifteen times, so n = 15
If we toss a coin means, probability of getting trail is p = `1/ 2`
Probability of getting head is denoted by q
q = 1 – p [p + q = 1]
q = 1 – `1/ 2`
q = `1/ 2`
P(X = x) = nCx px (1 – p)(n-x)
P(X = 10) = 15C10 `(1/2)^(10)` `(1/2)^(15-10)`
= 15C10 (0.5)10 (0.5)5
= 15C10 (0.0009) (0.0312)
Calculate 15C10:
Using the formula
nCx = `(n!)/((n - x)!(x!))`
15C10 = `(15!)/((15 - 10)!(10!))`
= `(15!)/((5!)(10!))`
= `(1307674368000)/((120)(3628800))`
= `(1307674368000)/(435456000)`
= 3003
P(X = 10) = 3003 (0.0009) (0.0312)
= 3003 (0.00002)
= 0.0843
The probability of getting exactly 10 head is 0.0843
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 trail gains the success of probability p. Tutor will help the students with step by step solutions for the given problem. The following are the example problems in binomial distribution examples tutor.
Looking out for more help on Binomial Distribution Table in math by visiting listed websites.
Binomial distribution Formula
Formulas:
P(X = x) = nCx px (1 – p)(n-x)
Where
n = number of events
x = number of successful events
p = in a single trail probability of success
nCx = `(n!)/((n - x)!(x!))`
1 – p = probability of failure
Mean (or) Expected value:
μ = E[x] = np
Standard deviation:
σ = `sqrt(npq)`
Variance:
E[x2] = σ2 = npq
Binomial distribution Problems
My forthcoming post is on Binomial Calculator and simple math problems for kids will give you more understanding about Math.
Here we solving some examples based on the binomial distribution
Example 1:
A coin is tossed for eighty-eight times. Determine the expected value of heads, variance and the standard deviation by using the binomial distribution.
Solution:
Given
Let coin tossed for eighty-eight times, so n = 88
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
= 88 (`1/2` )
= (`88/2` )
μ = E[x] = 44
Variance:
E[x2] = σ2 = npq
= 88(`1/2` )(`1/2` )
= 88(`1/4` )
= (`88/4` )
= 22
E[x2] = σ2 = 22
Standard deviation:
σ = `sqrt (npq)`
σ = `sqrt(22)`
= 4.69
σ = 4.69
Answer:
Mean = μ = E[x] = 44
Variance = E[x2] = σ2] = 22
Standard deviation = σ = 4.69
Example 2:
A coin is tossed for fifteen times. Determine the probability of getting exactly 10 heads.
Solution:
Given
Let coin tossed for fifteen times, so n = 15
If we toss a coin means, probability of getting trail is p = `1/ 2`
Probability of getting head is denoted by q
q = 1 – p [p + q = 1]
q = 1 – `1/ 2`
q = `1/ 2`
P(X = x) = nCx px (1 – p)(n-x)
P(X = 10) = 15C10 `(1/2)^(10)` `(1/2)^(15-10)`
= 15C10 (0.5)10 (0.5)5
= 15C10 (0.0009) (0.0312)
Calculate 15C10:
Using the formula
nCx = `(n!)/((n - x)!(x!))`
15C10 = `(15!)/((15 - 10)!(10!))`
= `(15!)/((5!)(10!))`
= `(1307674368000)/((120)(3628800))`
= `(1307674368000)/(435456000)`
= 3003
P(X = 10) = 3003 (0.0009) (0.0312)
= 3003 (0.00002)
= 0.0843
The probability of getting exactly 10 head is 0.0843