Introduction to sum of Poisson random variables:
The sum of the Poisson random variables is the mean value for the Poisson distribution. Poisson distribution is the limited form of the binomial distribution. The sum (mean) is calculated by using the number of the possibilities and the probability for the occurrence of the event. The example for the Poisson distribution is the number of cars passing through the certain street in the time period of t. This article has the information about the sum of the Poisson random variables.
Formula for the Sum of the Poisson Random Variables:
The formula used to find the sum (mean) of the Poisson distribution is
`lambda` = n p
Where n is the number of trial for the event and the p is the probability of the success.
Examples for the Sum of the Poisson Random Variables:
Examples 1 for the sum of the Poisson random variables:
Two coins are tossed for 33 times. Using the Poisson distributions calculate the sum (mean) of head for the coin.
Solution:
The number of times the coin tossed is 33. So the value of n is 33.
The probability for getting one head with one coin is `1/2` .
The probability for getting two head with two coins are p = `(1/2) ^2` = `1/4` .
The sum (mean) `lambda ` = n p
`lambda ` = 33 (1/4)
` lambda` = `33 /4`
` lambda ` = 8.25
The sum (mean) for getting the head for the two coins is 8.25.Having problem with practice pre algebra keep reading my upcoming posts, i will try to help you.
Examples 2 for the sum of the Poisson random variables:
Four coins are tossed for 122 times. Using the Poisson distributions calculate the sum (mean) of head for the coin.
Solution:
The number of times the coin tossed is 122. So the value of n is 122.
The probability for getting one head with one coin is `1/2` .
The probability for getting four head with four coins are p = `(1/2) ^4` = `1/16` .
The sum (mean) `lambda` = n p
`lambda` = `122 (1/16)`
`lambda` = `122 /16`
`lambda` = 7.625
The sum (mean) for getting the head for the four coins is 7.625.
Examples 3 for the sum of the Poisson random variables:
Five coins are tossed for 150 times. Using the Poisson distributions calculate the sum (mean) of head for the coin.
Solution:
The number of times the coin tossed is 150. So the value of n is 150.
The probability for getting one head with one coin is `1/2` .
The probability for getting five head with five coins are p = `(1/2) ^5` = `1/32` .
The sum (mean) `lambda` = n p
`lambda` = `150 (1/32)`
`lambda ` = `150 /32`
`lambda` = 4.6875
The sum (mean) for getting the head for the five coins is 4.6875.
The sum of the Poisson random variables is the mean value for the Poisson distribution. Poisson distribution is the limited form of the binomial distribution. The sum (mean) is calculated by using the number of the possibilities and the probability for the occurrence of the event. The example for the Poisson distribution is the number of cars passing through the certain street in the time period of t. This article has the information about the sum of the Poisson random variables.
Formula for the Sum of the Poisson Random Variables:
The formula used to find the sum (mean) of the Poisson distribution is
`lambda` = n p
Where n is the number of trial for the event and the p is the probability of the success.
Examples for the Sum of the Poisson Random Variables:
Examples 1 for the sum of the Poisson random variables:
Two coins are tossed for 33 times. Using the Poisson distributions calculate the sum (mean) of head for the coin.
Solution:
The number of times the coin tossed is 33. So the value of n is 33.
The probability for getting one head with one coin is `1/2` .
The probability for getting two head with two coins are p = `(1/2) ^2` = `1/4` .
The sum (mean) `lambda ` = n p
`lambda ` = 33 (1/4)
` lambda` = `33 /4`
` lambda ` = 8.25
The sum (mean) for getting the head for the two coins is 8.25.Having problem with practice pre algebra keep reading my upcoming posts, i will try to help you.
Examples 2 for the sum of the Poisson random variables:
Four coins are tossed for 122 times. Using the Poisson distributions calculate the sum (mean) of head for the coin.
Solution:
The number of times the coin tossed is 122. So the value of n is 122.
The probability for getting one head with one coin is `1/2` .
The probability for getting four head with four coins are p = `(1/2) ^4` = `1/16` .
The sum (mean) `lambda` = n p
`lambda` = `122 (1/16)`
`lambda` = `122 /16`
`lambda` = 7.625
The sum (mean) for getting the head for the four coins is 7.625.
Examples 3 for the sum of the Poisson random variables:
Five coins are tossed for 150 times. Using the Poisson distributions calculate the sum (mean) of head for the coin.
Solution:
The number of times the coin tossed is 150. So the value of n is 150.
The probability for getting one head with one coin is `1/2` .
The probability for getting five head with five coins are p = `(1/2) ^5` = `1/32` .
The sum (mean) `lambda` = n p
`lambda` = `150 (1/32)`
`lambda ` = `150 /32`
`lambda` = 4.6875
The sum (mean) for getting the head for the five coins is 4.6875.
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