Standard Deviation Of Probability Distribution
Standard deviation of a probability distribution measures the scattering of the probability distribution with respect to its mean. It is a measure obtained by taking the square root of the variance. Standard deviation can be computed for probability distributions such as binomial distribution, normal distribution, and poison distribution.
The standard deviation of a binomial distribution is σ= √(npq). Here n is the number of trials, p is the probability of success, and q is the probability of failure. Let us check the standard deviations of other probability distributions also, with the help of examples, FAQs.
What Is Standard Deviation Of Probability Distribution?
Standard deviation of probability distribution is the degree of dispersion or the scatter of the probability distribution relative to its mean. It is the measure of the variation in the probability distribution from the mean. The standard deviation of a probability distribution is the square root of its variance.
For n number of observations and the observations are \(x_1, x_2, .....x_n\), then the mean deviation of the value from the mean is determined as \(\sum_{i=1}^{n}\left(x_{i}-\bar{x}\right)^{2}\). However, the sum of squares of deviations from the mean doesn't seem to be a proper measure of dispersion. If the average of the squared differences from the mean is small, it indicates that the observations \(x_i\) are close to the mean \(\bar x\). This is a lower degree of dispersion. If this sum is large, it indicates that there is a higher degree of dispersion of the observations from the mean \(\bar x\). Thus we conclude that \(\sum_{i=1}^{n}\left(x_{i}-\bar{x}\right)^{2}\) is not a reasonable indicator of the degree of dispersion or scatter. Hence we take \(\dfrac{1}{n}\sum_{i=1}^{n}\left(x_{i}-\bar{x}\right)^{2}\) as a proper measure of dispersion and this is called the variance(σ2). The square root of the variance is the standard deviation.
Steps to Calculate Standard Deviation Of A Probability Distribution
The following sequential steps help in easily computing the standard deviation of a probability distribution.
- Find the mean, which is the arithmetic mean of the observations.
- Find the squared differences from the mean. (The data value - mean)2
- Find the average of the squared differences. (Variance = The sum of squared differences ÷ the number of observations)
- Find the square root of variance. (Standard deviation = √Variance)
Standard Deviation Of Different Probability Distributions
The standard deviation of a probability distribution can be taken for normal distribution, binomial distribution, poison distribution.
Normal Distribution
normal distribution or bell curve or the gaussian distribution is the most significant continuous probability distribution in probability. The normal distribution is defined by the probability density function f(x) for the continuous random variable X considered in the system. It is a function whose integral across an interval (say x to x + dx) gives the probability of the random variable X, by considering the values between x and x + dx. Since there will be infinite values between x and x + dx, thus, a range of x is considered,
In a normal distribution, the mean is zero and the standard deviation of the normal probability distribution is 1.
Binomial Distribution
Binomial distribution represents the probability for 'x' successes of an experiment in 'n' trials, given a success probability 'p' and a non-success probability q, such that p + q = 1, for each trial of the experiment. Two parameters n and p are used here in the binomial distribution. The variable ‘n’ represents the number of trials and the variable ‘p’ states the probability of any one(each) outcome. A test that has a single outcome such as success/failure is also called a Bernoulli trial and here we consider the Bernoulli trials for a binomial distribution.The binomial distribution formula is for any random variable X, given by; P(x:n,p) = nC\(_{x}\)px (1-p)n-x = nC\(_{x}\) px (q)n-x
The binomial distribution formula is also written in the form of n-Bernoulli trials where nC\(_{x}\)= n!/x!(n-x)!. Hence, P(x:n,p) = n!/[x!(n-x)!].px.(q)n-x
For a binomial distribution, the mean, variance and standard deviation for the given number of success are represented using the below formulas.
- Mean, μ = np
- Variance, σ2 = npq
- Standard Deviation σ= √(npq)
Poisson Distribution
Poisson distribution is used to represent the probability of happening of a countable number of events in a limited time span. The probability of receiving the number of calls per minute by a call center can be represented as a poisson distribution. Poisson distribution definition is used to model a discrete probability of an event and has a known constant mean rate. In other words, Poisson distribution is used to estimate how many times an event is likely to occur within the given period of time. λ is the Poisson rate parameter that indicates the expected value of the average number of events in the fixed time interval.
Poisson distribution has wide use in the fields of business as well as in biology. In a Poisson distribution, the standard deviation is given by 𝜎= √λt, where λ is the average number of successes in an interval of time t.
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Examples on Standard Deviation Of Probability Distribution
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Example 1: Find the standard deviation of the binomial distribution having a probability of success as 0.8, and has 9 trials.
Solution:
For the probability distribution, we have the probability of success as p = 0.8
Probability of failure. q = 1 - p = 1 - 0.8 = 0.2
Number of trials = n = 9.
Standard deviation of a binomial distribution = \(\sqrt {npq}\) = \(\sqrt {9 x 0.8 x 0.2}\) = \(\sqrt {1.44}\) = 1.2.
Therefore the standard deviation of the binomial probability distribution is 1.2.
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Example 2: Find the mean, variance, and standard deviation of a probability distribution having a probability of success of 0.6, for about 20 trials.
Solution:
The given probability distribution has a probability of success = P = 0.6
The probability of failure = q = 1 - 0.6 = 0.4
Tota number of trials = n = 20
Mean, μ = np = 20 x 0.6 = 1.2
Variance, σ2 = npq = 20 x 1.2 x 0.4 = 0.96
Standard Deviation σ= √(npq) = \(\sqrt {20 x 1.2 x 0.4}\) = \(\sqrt {0.96}\) = 0.97
Therefore the mean is 1,2, variance is 0.96, and the standard deviation of the probability distribution is 0.97.
Practice Questions on Standard Deviation Of Probability Distribution
FAQs on Standard Deviation Of Probability Distribution
What Is Standard Deviation Of Probability Distribution?
Standard deviation of a probability distribution measures the scattering of the probability distribution with respect to its mean. It is a measure obtained by taking the square root of the variance. Standard deviation can be computed for probability distributions such as binomial distribution, normal distribution, and poison distribution. The standard deviation of a binomial distribution is σ= √(npq), and here n is the number of trials, p is the probability of success, and q is the probability of failure.
What Is The Formula Of Standard Deviation Of Probability Distribution?
The formula of the standard deviation of a binomial distribution is σ= √(npq). Here n is the number of trials, p is the probability of success, and q is the probability of failure. The standard deviation for a normal distribution is 1, and for a poison ratio is 𝜎= √λt, where λ is the average number of successes in an interval of time t.
How Do You Calculate The Standard Deviation Of Probability Distribution?
The standard deviation for a probability distribution can be easily computed from the given data, metrics. For a binomial distribution, we need the number of trials n, the probability of success p, and the probability of failure q. For a poison ratio we need λ the average number of successes in an interval of time t. And for a normal distribution, the standard deviation is always equal to 1.
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