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Normal Distribution and Lognormal Distribution, Lecture notes of Mathematics

American University of Beirut Mathematics

how to do a lognormal distribution

Typology: Lecture notes

2019/2020

Uploaded on 04/01/2020

farah-jaafar 🇱🇧

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Lecture'10'–'Normal'Distribution'and'Lognormal'Distribution'

Normal'Distribution'

Most observations in real life follow the normal distribution. The reason can be

attributed to the central limit theorem which states, “the sum of sufficient large number

(usually 30 or more) of identical random variables of any probability distribution will

follow the normal distribution”.

Most objects in real life are sums of multiple units. Example – thickness of a book.

The normal distribution is given by,

𝑓𝑥=

1

𝜎2𝜋

𝑒!!.!!!!

!

!𝑓𝑜𝑟 −∞<𝑋<∞

The mean and variance of a normal variable are given by,

𝐸𝑋=𝜇

𝑉𝑋=𝜎!

If a variable follows the normal distribution with parameters  and , then it is

represented as, !

𝑋~𝑁(𝜇,𝜎!)

There is no analytical solution for the integral of normal probability density function. So

we use tables. But we cannot build separate tables of probability for each  and 2. So

we standardize the normal variable by subtracting the mean and dividing by the

standard deviation. The resultant standardized normal variable has a mean zero and a

standard deviation of 1.

𝑍=

𝑋−𝜇

𝜎

Now you can use the tables to compute probabilities.

𝑍~𝑁(0,1)

𝑓𝑧=

1

2𝜋

𝑒!!.!!!!𝑓𝑜𝑟 −∞<𝑍<∞

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Lecture 10 – Normal Distribution and Lognormal Distribution

Normal Distribution

Most observations in real life follow the normal distribution. The reason can be attributed to the central limit theorem which states, “the sum of sufficient large number (usually 30 or more) of identical random variables of any probability distribution will follow the normal distribution”. Most objects in real life are sums of multiple units. Example – thickness of a book. The normal distribution is given by, 𝑓 𝑥 =

!!.! !! !! ! 𝑓𝑜𝑟 − ∞ < 𝑋 < ∞ The mean and variance of a normal variable are given by, 𝐸 𝑋 = 𝜇 𝑉 𝑋 = 𝜎! If a variable follows the normal distribution with parameters and , then it is represented as, 𝑋~𝑁(𝜇, 𝜎!) There is no analytical solution for the integral of normal probability density function. So we use tables. But we cannot build separate tables of probability for each and ^2. So we standardize the normal variable by subtracting the mean and dividing by the standard deviation. The resultant standardized normal variable has a mean zero and a standard deviation of 1. 𝑍 =

Now you can use the tables to compute probabilities. 𝑍~𝑁( 0 , 1 ) 𝑓 𝑧 =

! 𝑓𝑜𝑟 − ∞ < 𝑍 < ∞

Example The population mean and standard deviation of males is 6 ft and 1 ft respectively. A random sample of 40 students is drawn. What is the probability that the average of these 40 students will be less than 5.5 ft? Solution From central limit theorem, 𝑋 ~ 𝑁 6 ,

Lognormal Distribution

If ln X follows the normal distribution, then X is said to follow the lognormal distribution. Pollutant concentrations in rivers, for example, are observed to follow the lognormal distribution. We work with the lognormal variable the same way we work with the normal distribution after making the transformation of X into the log domain. 𝑙𝑛 𝑋~𝑁(𝜃, 𝜔!) 𝐸 𝑋 = 𝑒 !!! ! ! 𝑉 𝑋 = 𝑒!!!! ! 𝑒! ! − 1 Example Suppose that the reaction time in seconds of a person can be modeled by a lognormal distribution with parameter values, = - 0.35 and = 0.2. a) Find the probability that the reaction time is less than 0.6 seconds b) Find the reaction time that is exceeded by 95% of the population. Solution: Data: 𝑙𝑛 𝑋~𝑁(− 0. 35 , 0. 2 !) Part a)

The number of failures is n-x. The probability of a failure is 1 - p. So, the probability of n-x failures is pn-x.
Now the x successes can be arranged in (^)!^!^ 𝐶ways. Now using the multiplicative rule, the binomial formula is obtained. The expected value and variance of a binomial random variable are given by, 𝐸 𝑋 = 𝑛𝑝 𝑉 𝑋 = 𝑛𝑝( 1 − 𝑝) Now, let us assume you are not interested in the probability of x successes in n trials; but interested in the probability of needing x number of trials for r successes. This situation is, in a way, the reverse of the binomial distribution. We compute the probability for any number of successes in n trials using the binomial distribution. Here, we compute the probability for any number of trials required for r successes. The distribution is called negative binomial distribution. The negative binomial distribution is given by, 𝑃 𝑋!" = 𝑥 = (^)!^ !!!!!𝐶𝑝!( 1 − 𝑝)!!! The expected value and variance of XNB are given by, 𝐸 𝑋!" =

It can easily be seen that the first x- 1 successes in the first r- 1 trials are a binomial situation and the binomial distribution result can be used. This result must then be multiplied by p as that is the probability of success in the r th^ trial.