Statistical Distributions
A random variable is a variable whose value is dependent on the outcome of a random event. This means the value is not known until the experiment is carried out, however the probabilities of all the outcomes can be modelled with a statistical distribution.
There are multiple ways of writing out a distribution:
The examples above are all the distribution of a fair, six-sided die. The probability of each outcome is the same, so it is called a discrete uniform distribution.
The sum of all probabilities in a distribution must equal 1
The Binomial Distribution
If you repeat an experiment multiples times (each time is known as a 'trial'), you can model the number of successful trials with the random variable, X.
A binomial distribution is used when:
There are a fixed number of trials, n
There are only two possible outcomes (success or fail)
The probability, p, of success is constant
All trials are independent
If the random variable X is distributed binomially with n number of trials and fixed probability of success p, it is noted as:
X∼B(n, p)
Generally, you use a calculator to calculate binomial problems, but you can also use the probability mass function:
There are two forms of the binomial distribution: exact and cumulative:
Exact Binomial Problems
This is when a question asks you to find the probability of there being a specific number of successes out of the number of trials, n.
For example, to find the probability of there being exactly 4 successes for the random variable X∼B(12, 0.4) use an exact binomial distribution.
To find this on a CASIO ClassWix fx-991EX:
Click MENU
Click 7: Distribution
Click 4: Binomial PD
Click 2: Variable
Input your values - in this example, x=4, N=12, p=0.4
Click =
You should find the answer for this example is 0.213. This means there is a 21.3% chance of getting exactly four successful trials, out of a total 12 trials.
Using the table function allows you to see the probabilities for multiple values of x.
Cumulative Binomial Problems
Generally, this is used more, and is used to find the sum of all the probabilities up to and including a certain value of x: P(X ≤ x)
For example, if you want to find the probability of there being up to and including 4 successes (so there could be 0, 1, 2, 3 or 4 successes) for the random variable X∼B(12, 0.4), use the cumulative binomial distribution. This can be done in two ways:
Tables
There are tables with values for this, typically found at the back of formula books. These will have the most commons values for n, and some standard probabilities.
Calculators
To find this on a CASIO ClassWix fx-991EX:
Click MENU
Click 7: Distribution
Click DOWN
Click 1: Binomial CD
Click 2: Variable
Input your values - in this example, x=4, N=12, p=0.4
Click =
Tables and calculators only ever give the probability for 'up to and including x', P(X ≤ x)
Therefore, if you want other forms, such as P(X > a) you need to use the following functions:
For P(X > a), use 1 - P(X ≤ a)
For P(X < a), use P(X ≤ (a-1))
For P(X ≥ a), use 1 - P(X ≤ (a-1))
For P(X ≤ a), use P(X ≤ a)
These rules work because the sum of all the probabilities equals 1.
The Normal Distribution
The normal distribution is used to model continuous random variables. These are variables that can take absolutely any value.
The probability that the continuous random variable takes a particular specific value is always zero, but we can calculate the probability that it takes a value within a certain range. This is because continuous random variables have a continuous probability distribution:
It is modelled as a curved graph
The probability is the area under the curve
The area under the curve can only be defined for ranges, as the area of an infinitely narrow line is zero
Because the area under the graph is the probability, and the sum of all probabilities is 1, the area under the whole graph = 1
You can think of a continuous probability distribution as a histogram with an infinite number of infinitely narrow categories:
The Normal Distribution
A normal distribution is a continuous probability distribution that is bell-shaped and symmetrical about the mean.
μ is the population mean, and is in the middle of the distribution
σ is the standard deviation of the population
σ² is the population variance
The graph is symmetrical about the mean
The graph has a total area of 1
There are points on inflection at μ + σ and μ - σ
If the continuous random variable X is distributed normally with population mean μ and standard deviation σ, it is noted as:
X∼N(μ, σ²)
All things in nature tend to be modelled with a normal distribution (hence the name), especially heights and lengths of members of a population.
It is good to know how the data is spread across the graph:
Around 68% of all data is within one standard deviation of the mean (between the two points of inflection)
Around 95% of all data is within two standard deviations of the mean
Around 99.7% of all data is within three standard deviations of the mean
Example
In the example above, the median is 180 cm, and the standard deviation is 16. Therefore, continuous random variable, X, is modelled: X∼N(180, 16²). Find P(170 < X < 190)
To find this on a CASIO ClassWix fx-991EX:
Click MENU
Click 7: Distribution
Click 2: Normal CD
Input your values - in this example, lower=170, upper = 190, σ=16, μ=180
Click =
You should find the answer for this example is 0.468. This means there is a 46.8% chance of someone's height being between 170 cm and 190 cm.
If only one boundary is specified, e.g. P(X < 190) or P(X > 170), make the other boundary a ridiculously big negative or positive number.
For P(X < 190), the upper boundary is 190, and make the lower one -9999999999999999 or something similar
For P(X > 170), the lower boundary is 170 and make the upper one 9999999999999999 or something similar
Inverse Normal
You can also use the normal distribution backwards, to find limits from probabilities. This is done using the inverse normal distribution function.
Continuing with the example above, where X∼N(180, 16²): find the value of a for which P(X < a) = 0.35.
It is sometimes useful to represent this visually:
To find this on a CASIO ClassWix fx-991EX:
Click MENU
Click 7: Distribution
Click 3: Inverse Normal
Input your values - in this example, area=0.35 σ=16, μ=180
Click =
You should find the answer for this example is 174. This means there is a 35% chance of someone's height being between less than 174 cm
Calculators only ever calculate the area to the left
This means that if you want to find the value of a for which P(X > a) = 0.35, you need to input 0.65 (1-0.35) into your calculator.
Standard Normal
Normally distributed variables can be standardised using coding:
The standard normal distribution has mean 0 and standard deviation 1
If X∼N(μ, σ²), it can be coded into Z∼N(0, 1²) using the equation Z = (X-μ) / σ
Sometimes, the probability P(Z < a) is written as Φ(a)
You use your calculator normally, just enter μ = 0, σ = 1.
Finding μ and σ
Often, you will not know either the mean or the standard deviation of a normal distribution and will have to find it. You will, however, be given a probability, so you can code it into the standard normal distribution and solve.
For example, the random variable X ∼ N(μ, 3²). Given that P(X< 10) = 0.3, find the mean.
Use the inverse normal to find the value for Z when p = 0.3:
Z = -0.524, so rearrange to find μ (you know X=10 and σ = 3)
(3)(-0.524) = 10 - μ
μ = 13.572 = 13.6
If you know neither the mean nor standard deviation, but have two probabilities, set up simultaneous equations and solve these.
Approximating a Normal Distribution
Binomial distributions become difficult to work with when n is large. In these instances, if p is close to 0.5, the model can be approximated with a normal distribution:
If n is large and p ≈ 0.5, then X∼B(n, p) can be approximated as Y∼N(μ, σ²) where μ = np and σ² =np(1-p)
Remember to square root the variance
The binomial distribution is discrete, whereas the normal distribution is continuous. This means you need to apply the continuity correction whenever you approximate a binomial normally. This means that you need to add or subtract 0.5 to account for rounding:
P(X < a) ≈ P(Y < a+0.5)
P(X ≤ a) ≈ P(Y < a+0.5)
P(X = a) ≈ P(a-0.5 < Y < a+0.5)
P(X ≥ a) ≈ P(Y > a+0.5)
P(X > a) ≈ P(Y > a+0.5)
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