Normal Distribution

The normal distribution (Exhibit 34.17) is said to represent an elementary “truth about the general nature of reality”. It is a family of distributions, each defined by two parameters — mean (μ) and standard deviation (s).

The probability distribution function of a random variable X∼N(μ,σ) is:

$$f(x)=\frac{1}{\sqrt{2πσ^2}} e^{-\frac{(x-μ)^2}{2σ^2}}$$

Standard Normal Distribution

The normal distribution X∼N(0,1), μ = 0 and σ = 1, is called the standard normal distribution, and its distribution function is:

$$f(x)=\frac{e^{-x^2/2}}{\sqrt{2π}}$$

A value from any normal distribution can be transformed into its corresponding value on a standard normal distribution using the following formula:

$$Z =\frac{X - μ}{σ}$$

where Z is the value on the standard normal distribution, X is the value on the original distribution, μ is the mean of the original distribution, and σ is the standard deviation of the original distribution.

$$E(Z)= E\biggl(\frac{X - μ}{σ}\biggr)=\frac{E(X) - μ}{σ}=0$$ $$Var(Z)=Var\biggl(\frac{X - μ}{σ}\biggr)=\frac{Var(X)}{σ^2}=1$$

To computing probabilities for normal distributions, convert to standard value and lookup the distribution function tables for the standard normal distribution:

$$P(X≤b)→P\biggl(Z≤\frac{b - μ}{σ}\biggr)$$

Note: The probabilities may also be obtained using the NORM.DIST function in Excel. The Excel NORM.INV function returns the variable value, given the parameters and the probability.

If a random variable follows a normal distribution, we can expect that:

• 68% of the variable’s observations fall within a range of ±1 standard deviation from the mean.
• 90% of the observations fall within a range of ± 1.65 standard deviations from the mean, as depicted in Exhibit 34.18.
• 95% of observations lie within 1.96 standard deviations from the mean.

Approximating Binomial with Normal

According to the Central Limit Theorem, the normal distribution can be used as an approximation to the binomial distribution, if the sample size is large enough.

If X is Binomial (n, p):

$$E(X) = np$$ $$Var(V) = np(1-p)$$

The approximate equivalent normal distribution is N[np,√(np(1-p))], provided np≥ 5 and n(1- p)≥ 5 (thumb rule).