Introduction to exact distribution:
The exact distribution refers the probability distribution the probability theory. The probability distribution is used to determine the number of possibility for the occurrence of an event. The most commonly used probability distributions are the binomial distribution, geometric distribution, normal distribution and the gamma distribution. These above mentioned distributions are included in the discrete and continuous probability distribution. The major type of the probability distribution is the discrete probability distribution and the continuous probability distribution. This article has the study about exact distribution.
Types of Exact Distribution:
The major types of the probability distribution are
Discrete probability distribution
Continuous probability distribution
Discrete probability distribution:
The probability for a countable number of occurrences for the event is calculated in the discrete probability distribution.
Continuous probability distribution:
The probability values in this are the continuous ranged value it is calculated in the continuous probability distribution.
Examples for Exact Distribution:
Example 1 to exact distribution:
If X is normally distributed the mean value is 1 and its standard deviation is 2. Determine the value of P (0 ≤ X ≤ 7).
Solution:
The given mu value is 1 and the standard deviation is 2.
Z = `(X- mu)/ sigma`
When X = 0, Z = `(0- 1)/ 2`
= -`1/2`
= -0.5
When X = 7, Z = `(7- 1)/ 2`
= `6/2`
= 3
Therefore,
P (0 ≤ X ≤ 7) = P (-0.5 < Z < 3)
P (0 ≤ X ≤ 7) = P (0 < Z < 0.5) + P (0 < Z < 3) (due to symmetry property)
P (0 ≤ X ≤ 7) = (0. 6915- 0.5) + (0.9987 - 0.5)
P (0 ≤ X ≤ 7) = 0.1915 + 0.4987
P (0 ≤ X ≤ 7) = 0.4987
The value for P (0 ≤ X ≤ 7) is 0.4987.
Between, if you have problem on these topics, please browse expert math related websites for more help on cbse 10th sample papers.
Example 2 to exact distribution:
The probability for destroying the target in only one time is 0.37. Compute the probability that it would be destroyed on the third attempt itself.
Solution:
The probability of destroying the target in one trial is p = 0.37.
The value of the q is calculated by q = 1-p
q = 1- 0.37
q = 0.63
By the geometric distribution, the probability for the success is calculated by using the formula
P(X =x) = q x p, the value of x is 0, 1, 2. . .
The target is destroyed at the third attempt, so x = 3.
P(X = 3) = (0. 63) 3 (0.37)
P(X = 3) = (0.25) (0.37)
P(X = 3) = 0.0925
The probability for destroying the target at the third trial is 0.0925.
The exact distribution refers the probability distribution the probability theory. The probability distribution is used to determine the number of possibility for the occurrence of an event. The most commonly used probability distributions are the binomial distribution, geometric distribution, normal distribution and the gamma distribution. These above mentioned distributions are included in the discrete and continuous probability distribution. The major type of the probability distribution is the discrete probability distribution and the continuous probability distribution. This article has the study about exact distribution.
Types of Exact Distribution:
The major types of the probability distribution are
Discrete probability distribution
Continuous probability distribution
Discrete probability distribution:
The probability for a countable number of occurrences for the event is calculated in the discrete probability distribution.
Continuous probability distribution:
The probability values in this are the continuous ranged value it is calculated in the continuous probability distribution.
Examples for Exact Distribution:
Example 1 to exact distribution:
If X is normally distributed the mean value is 1 and its standard deviation is 2. Determine the value of P (0 ≤ X ≤ 7).
Solution:
The given mu value is 1 and the standard deviation is 2.
Z = `(X- mu)/ sigma`
When X = 0, Z = `(0- 1)/ 2`
= -`1/2`
= -0.5
When X = 7, Z = `(7- 1)/ 2`
= `6/2`
= 3
Therefore,
P (0 ≤ X ≤ 7) = P (-0.5 < Z < 3)
P (0 ≤ X ≤ 7) = P (0 < Z < 0.5) + P (0 < Z < 3) (due to symmetry property)
P (0 ≤ X ≤ 7) = (0. 6915- 0.5) + (0.9987 - 0.5)
P (0 ≤ X ≤ 7) = 0.1915 + 0.4987
P (0 ≤ X ≤ 7) = 0.4987
The value for P (0 ≤ X ≤ 7) is 0.4987.
Between, if you have problem on these topics, please browse expert math related websites for more help on cbse 10th sample papers.
Example 2 to exact distribution:
The probability for destroying the target in only one time is 0.37. Compute the probability that it would be destroyed on the third attempt itself.
Solution:
The probability of destroying the target in one trial is p = 0.37.
The value of the q is calculated by q = 1-p
q = 1- 0.37
q = 0.63
By the geometric distribution, the probability for the success is calculated by using the formula
P(X =x) = q x p, the value of x is 0, 1, 2. . .
The target is destroyed at the third attempt, so x = 3.
P(X = 3) = (0. 63) 3 (0.37)
P(X = 3) = (0.25) (0.37)
P(X = 3) = 0.0925
The probability for destroying the target at the third trial is 0.0925.
