# Special Random Variables

Special random variables are special types of distributions. They are found to be common in real life distributions.

## Contents |

## Distributions

The following distributions can be used to calculate probabilities in different experimental trials. Most of these formulas can be found on the formula sheet.

### Binomial

Binomial distributions are used for random experiments that involve **discrete** results or either success or failure.

It is represented as:

- X ~ Bin(n, π)

The probability function for binomial distributions is:

where:

- n = number repetitions of the experiment
- X = number of successes
- π = probability of a success

- E(X) = nπ
- Var(X) = nπ(1-π)

Example:

- If a shaft manufacturer produces shafts of a certain tolerance, with a probability of being off of tolerance at 0.5%, what is the probability of having to throw out more than 2 shafts in a batch of 100.
- It can be seen that the sample size if 100 and the probability is 0.005, therefore:
- X ~ Bin(100, 0.005)
- P(X>=2) = 1 - (P(X=0) + P(X=1))
- = 1 - (100C0 * 0.005
^{0}* 0.995^{100}+ 100C1 * 0.005^{1}* 0.995^{99}= 0.0898

### Hypergeometric

Hypergeometric distributions are similar to binomial distributions, with the difference that the past trial affected the probability of the second trial.

It is represented as:

- X ~ Hyp(N, k, n)

The probability function for hypergeometric distributions is:

where:

- X = successes from the sample
- k = successes in a sample space
- n = size of the sample
- N = size of the sample space

- E(X) = nk/N
- Var(X) = (nk/N)*(1 - k/N)*((N-n)/(N-1))

Example:

- A gear box requires that at least 9 of the 10 gears be in tolerance to function adequately. If the fitter chooses the 10 gears from a selection of 100, with 80 of those gears in tolerance, what is the probability of the gear box functioning?
- It can be seen that the size of the sample space, N, is 100, and the functioning gears, k, is 80. We need to select n gears, being 10 for the gear box, giving:
- X ~ Hyp(100, 80, 10)
- P(X>=9) = P(X=9) + P(X=10)
- = ((80C9)*((100-80)C(10-9)))/(100C10) + ((80C10)*((100-80)C(10-10)))/(100C10) = 0.363

### Poisson

The poisson distribution involves using the mean of the distribution to find the probability of observing certain discrete observation.

It is represented as:

- X ~ P(λ)

The probability function for poisson distributions is:

where:

- λ = mean of the distribution
- X = number of occurrences

- E(X) = λ
- Var(X) = λ

Example:

- If the average number of ibises seen during a day at university is 4.6, what is the probability that there would be no ibises seen on the following day of university?
- X ~ P(4.6)
- P(X=0) = exp(-4.6)*(4.6
^{0})/0! - = 0.01

### Uniform

The uniform distribution involves a linear distribution of probability between the intervals α and β.

It is represented as:

- X ~ U
_{[α,β]}

The probability function for uniform distributions is the **integral** of the following function, where the limits of the integral is the time of the occurrence of the observations:

Or between subintervals a and b:

P(a<X<b) = (b-a)/(β-α)

where:

- α = start of interval
- β = end of interval
- a = start of subinterval
- b = end of subinterval

- E(X) = (α + β)/2
- Var(X) = (β - α)
^{2}/12

Example:

- A passenger arrives at a train platform between 5pm and 6pm, where the trains arrive every 10 minutes. What is the probability that the passenger has to wait less than 3 minutes to catch a train?
- X ~ U
_{[0,60]} - P(7<X<10)+ P(17<X<20) + P(27<X<30) + P(37<X<40) + P(47<X<50) + P(57<X<60)
- = (3+3+3+3+3+3)/60 = 0.3

### Exponential

The exponential distribution is one that allows the calculation of the probability of an event occurring over a certain amount of time. It is represented as:

- X ~ Exp(λ)

The probability function for exponential distributions is the **integral** of the following function, where the limits of the integral is the time of the occurrence of the observations:

where:

- λ = mean of sample space

- E(X) = 1/λ
- Var(X) = 1/λ
^{2}

Example:

- On average, 5 stray cats wonder into a backyard a day. What is the probability that there would be a cat in the backyard less than 10 minutes after scaring another cat off?
- X ~ Exp(5)
- P(X<10) = (from 0 to 10/1440) 5*exp(-5x) dx
- = 1 - exp(-5/144) = 0.034

## End

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