Calculate and interpret probabilities given the discrete uniform and the binomial distribution functions.
Due to the central limit theorem, the Normal distribution is probably the most important random distribution of all. It was found by the German mathematician Carl Friedrich Gauß.
We distinguish between
normal distributions. The univariate normal distribution describes a single random variable. The multivariate normal distribution describes several related random variables. When a group of several random variables is normally distributed, each univariate random variable is also normally distributed. But this is not true the other way around.
A normal distribution is not skewed at all, but it is symmetric. Therefore, the mean, median and mode of a normal distribution are all equal. The parameter of skewness equals 0.
Its kurtosis, as a measure of peakedness, equals 3; its excess kurtosis equals 0.
The density function of a normal distribution is calculated as
f(x) = 1/(σ∙√(2∙π))∙exp(-(x - μ)2/(2∙σ2)).
Construct a binomial tree to describe the movement of stock prices.
Describe the continuous uniform distribution. Calculate and interpret probabilities, given a continuous uniform probability distribution.
The standard normal distribution N(0,1) is the normal distribution with mean μ = 0 and standard deviation σ = 1. We use it to calculate probabilities of any (!) normal distribution N(μ,σ2).
How do we calculate probabilities using the normal distribution?
METHOD FOR CALCULATING NORMAL DISTRIBUTION:
1. Which probability do we want to calculate?
a) P(X< d),
b) P(X > d), or
c) P(d < X < e)?
2. To find the answer, do the following:
a) do not change,
b) change to P(X > d) = 1 – P(X ≤ d), or
c) do not change.
3. Standardize, i.e. subtract the mean and divide by the standard deviation. In this manner, we change from the normal distribution N(μ,σ2) to the standard normal distribution N(0,1):
a) P(X < d) = P((X – μ)/σ < (d - μ)/σ)
= P(XSt < (d - μ)/σ))
= φ(d - μ)/σ))
b) P(X > d) = 1 – P(X ≤ d)
= 1 – P((X – μ)/σ ≤ (d – μ)/σ)
= 1 – P(XSt ≤ (d – μ)/σ)
= 1 - φ((d - μ)/σ)
c) P(d < X < e) = P((d – μ)/σ < (X – μ)/σ < (e – μ)/σ)
= P(d – μ)/σ < XSt < (e – μ)/σ)
= φ((e – μ)/σ) - φ((d - μ)/σ)
4. Look up the values of φ in the table of the standard normal distribution. Sometimes we must use φ(-x) = 1 - φ(x).
Suppose Tom's weight is normally distributed with a mean of 75kg and a standard deviation of 3kg. What is the probability that he weighs, at most, 72 kg today?
Using equation a) from the aforementioned method, we calculate
P(X ≤ 72) = P(XSt≤ (72 – 75)/3) standardize
= P((XSt≤ - 1)
= φ(-1) change the notation to φ
= 1 - φ(1) use φ(-x) = 1 - φ(x)
= 1 - 0.8413
In the former example, calculate the probability that Tom weighs more than 76kg.
Using equation b), we calculate
P(X > 76) = 1 - P(X ≤ 76)
= 1 - P(XSt≤ (76 – 75)/3) standardize
= 1 – φ(-0.333) change the notation to φ
= 1 – [1 – φ(0.333)] use φ(-x) = 1 - φ(x)
= 0.6293 look up the value.
Using the data from the second last example, calculate the probability that Tom weighs between 72 and 77kg.
Using equation c) from the aforementioned method, we calculate
P(72 ≤ X ≤ 77) = P(72 – 75)/3 ≤ XSt≤ (77 – 75)/3) standardize
= φ(2/3) – φ(-1) use the φ-notation
= φ(0.667) – [1- φ(1)] φ(-x) = 1 - φ(x)
= φ(1) + φ(0.667) – 1
= 0.8413 + 0.7486 – 1 look up the values
The probability that any normally distributed random variable ranges one, two, or three standard deviations around the mean is given as 68.26 percent, 95.44 percent, and 99.61 percent respectively.
We calculate the second result here; try the others on you own!
P(μ–2·σ≤ X ≤ μ+2·σ) = P((μ – 2·σ- μ)/σ ≤ XSt≤ μ + 2·σ- μ)/σ)
= P(–2·σ/σ ≤ XSt≤ 2·σ/σ)
= P(–2 ≤ XSt≤ 2)
= φ(2) – φ(-2)
= φ(2) – [1 – φ(2)]
= 2∙0.9772 – 1 = 95.44 percent.