Subject: Business Statistics
A binomial random variable is one that counts the number of successful trials across numerous independent, identical Bernoulli trials, and a binomial distribution is the probability distribution of a binomial random variable. The probability distribution for discrete random variables is called the Poisson distribution. Simeon D. Poisson (1781–1840), a French mathematician, is honored by the name of this formula, which is used to estimate the likelihood that a given number of events will take place during a given period of time or space. When f or is more than 10, it is preferable to use the normal distribution. A specific type of these symmetric distributions that may be defined by two parameters is the Normal distribution, (i) µ = The mean of the distribution (ii) ƒ = The standard deviation of the distribution.
A Bernoulli trial is one in which there are just two possible outcomes: true or false or success or failure. The Bernoulli trial sample space is given by S={success,failure}. The random variable that counts the number of successes in a single Bernoulli trial is called a Bernoulli random variable. For instance, if X is a random variable, it provides, based on given values, the number of successful attempts.
If X is a Bernoulli random variable, we may write,
X ~ BER (p)
Where ~ is read as “is distributed as” and BER stands for Bernoulli.
The experiment must be a Bernoulli Process in order to satisfy the requirement for a binomial random variable.
An experiment needs to possess the following three characteristics in order to be a Bernoulli process:
A binomial random variable is one that counts the number of successful trials across numerous independent, identical Bernoulli trials, and a binomial distribution is the probability distribution of a binomial random variable.
Binomial Probability Function:
Using the two parameters n and p, we can characterize the distribution of a binomial random variable as follows:
X ~ B (n, p)
to indicate that X is Binomially distributed with n number of independent trials and p probability of success is each trial. The letter B stands for binomial.
Let us analyze the probability that the number of successes X in the n trials is exactly x (obviously number of failures are n-x) i.e. X = x and x = 0,1,2,…………. n; as n trials are made, at the best all n can be successes. Now we know that there are nCx ways of getting x successes out of n trials. We also observe that each of these nCx possibilities has px(1-p)n-x probability of occurrence corresponding to x successes and (n-x) failures.
Therefore, P(X = x) = nCx p x (1-p)n-x for x = 0,1,2,………, n
This equation is the Binomial probability formula. If we denote the probability of failure as q then the Binomial probability formula is
P(X = x) = nCx p x qn-x for x = 0,1,2,………, n
Important formulas:
The probability distribution for discrete random variables is called the Poisson distribution. Simeon D. Poisson (1781–1840), a French mathematician, is remembered in the name of this formula, which is used to estimate the likelihood of a given number of events occurring during a given period of time or area. Events that occur over time or space include:
There are numerous Poisson distributions, just like the binomial distribution. The average rate at which the event happens determines each Poisson distribution. The rate is denoted by the Greek letter "L" (lambda, which means "rate") The Poisson distribution has just one parameter. The likelihood that X specific occurrences occur in the specified space or time is given as follows:
p(X)=\frac{λ^xe^{-λ}}{x!}
Where, x is the number of successes λis the mean of the Poisson distribution and e = 2.71828 (the base of natural logarithms).
The random variable X counts the number of successes in Poisson Process. A Poisson process corresponds to a Bernoulli process under the following conditions: ¾ the number of trials n, is infinitely large i.e. n → ∝¾ the constant probability of success p, for each trial is infinitely small i.e. p → 0 (obviously q → 1) ¾ np = λ is finite.
A Poisson distribution has an identical mean and variance, which are both equal to λ. The square root of λ is the Poisson distribution's standard deviation.
It is preferable to use the normal distribution if or is greater than 10. A specific type of these symmetric distributions that may be defined by two parameters is the Normal distribution.
Changing the values of µ and σ alter the positions and shapes of the distributions.
If X is Normally distributed with mean µ and standard deviation σ, we write X∼N(µ, σ2) µ and σ are the parameters of the distribution.
Assuming the probability of male birth as ½, find the probability distribution of number of boys out of 5 births.
Let the random variable X measures the number of boys out of 5 births. Clearly X is a binomial random variable. So we apply the Binomial probability function to calculate the required probabilities.
X ~ B (5, ½)
P(X = x) = nCx p x qn-x for x = 0, 1, 2, 3, 4, 5
The probability distribution of X is given below:
X=x | 0 | 1 | 2 | 3 | 4 | 5 |
P(X=x) | \(\frac{1}{32}\) | \(\frac{5}{32}\) | \(\frac{10}{32}\) | \(\frac{10}{32}\) | \(\frac{5}{32}\) | \(\frac{1}{32}\) |
(a) The required probabilities are:
(i) P(X ≥1)
= 1- P(X = 0)
= 1- 1/32
= 31/32
(ii) P(X ≤ 3) = P(X = 0)+ P(X = 1)+ P(X = 2)+ P(X = 3)
= 1/32 + 5/32 + 10/32 + 10/32
= 26/32
(b) Out of 960 families with 5 children, the expected number of families with:
(i) at least one boy = 960 * P(X ≥1)
= 960 * 31/32
= 930
(ii) at most 3 boys = 960 * P(X ≤ 3)
= 960 * 26/32
= 720
At a coffee shop the average number of customers-arrivals during a specified period of 15 minutes is 2. If the arrival process is well described by a Poisson process, find the probability that during a given period of 15 minutes:
(a) no customer will arrive.
(b) at least two customers will arrive.
(c) atmost three customers will arrive.
(d) between 1 and 3 customers will arrive.
Let X denote the number of cars arrivals during the specified period of 15 minutes. So X ~ POI ( μ)
We apply the Poisson probability function P(X = x) = p(X)=\frac{λ^xe^{-λ}}{x!} as follows:
a) Probability of no customer will arrive
p(X=0)=\frac{2^0e^{-2}}{0!}
=0.1353
b) Probability of at least two customers will arrive
P(X ≥2)
=1-[ P(X = 0) + P(X = 1)]
=1-[\frac{2^0e^{-2}}{0!}+\frac{2^1e^{-1}}{1!}]
=0.5940
c) at most three customers will arrive
=P(X ≤ 3)
=P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3)
= 0.8571
d) P(between 1 and 3 customers will arrive)
= P(1 ≤ X ≤ 3)
= P(X ≤ 3) - P(X = 0)
= 0.8571 –0.1353
= 0.7218
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