Definition 1: The probability density function of the normal distribution is defined as:
Here is the constant e = 2.7183…, and is the constant π = 3.1415… which are described in Built-in Excel Functions.
The normal distribution is completely determined by the parameters µ and σ. It turns out that µ is the mean of the normal distribution and σ is the standard deviation. We use the abbreviation N(µ, σ) to refer to a normal distribution with mean µ and standard deviation σ.
As we shall see, the normal distribution occurs frequently and is very useful in statistics.
Excel Functions: Excel provides the following functions regarding the normal distribution:
NORMDIST(x, μ, σ, cum) where cum takes the values TRUE and FALSE
NORMDIST(x, μ, σ, FALSE) = probability density function value f(x) for the normal distribution
NORMDIST(x, μ, σ, TRUE) = cumulative probability distribution value F(x) for the normal distribution
NORMINV(p, μ, σ) is the inverse of NORMDIST(x, μ, σ, TRUE)
NORMINV(p, μ, σ) = the value x such that NORMDIST(x, μ, σ, TRUE) = p
Excel 2010/2013 provide the following additional functions: NORM.DIST, which is equivalent to NORMDIST, and NORM.INV, which is equivalent to NORMINV.
Example 1: Create a graph of the distribution of IQ scores using the Stanford-Binet scale.
This distribution is known to be the normal distribution N(100, 16). To create the graph, we first create a table with the values of the probability density function f(x) for for values of x = 50, 51, …, 150. This table begins as Figure 1.
Figure 1 – Probability density function for IQ
The value of f(x) for each x is calculated using the NORMDIST function with cum = FALSE. The probability density curve is now created as a line chart using the techniques described in Line Charts. From Figure 2, you can see that the curve in this chart has the characteristic bell shape of the normal distribution.
Figure 2 – IQ as a normal curve
Observation: The basic parameters of the normal distribution are as follows:
- Mean = median = mode = µ
- Standard deviation = σ
- Skewness = kurtosis = 0
The function is symmetric about the mean with inflection points (i.e. the points where there curve changes from concave up to concave down or from concave down to concave up) at x = μ ± σ.
As can be seen from Figure 3, the area under the curve in the interval μ – σ < x < μ + σ is approximately 68.26% of the total area under the curve. The area under the curve in the interval μ – 2σ < x < μ + 2σ is approximately 95.44% of the total area under the curve and the area under the curve in the interval μ – 3σ < x < μ + 3σ is approximately 99.74% of the area under the curve.
Figure 3 – Areas under the normal curve
Given the symmetry of the curve, this means that the area under the curve where x > μ + σ is 16.13%, i.e. (100% – 68.26%) / 2. The area under the curve where x > μ + 2σ is 2.28% and the area under the curve where x > μ + 3σ is 0.13%.
It also turns out that 95% of the area under the curve is in the interval -1.96 < x < 1.96. This will be important when considering the critical value for α = .05.
Property 1: If x has normal distribution N(μ, σ) then the linear transform y = ax + b, where a and b are constants, has normal distribution N(aμ+b, aσ).
Property 2: If x1 and x2 are independent random variables, and x1 has normal distribution N(μ1, σ1) and x2 has normal distribution N(μ2, σ2) then x1 + x2 has normal distribution N(μ1+μ2, σ) where
Observation: Click here for addition characteristics of the normal distribution function (using calculus), as well as a proof of Property 1 and 2.
Example 2: A charity group prepares sandwiches for the poor. The weights of the sandwiches are distributed normally with mean 150 grams and standard deviation of 25 grams. One sandwich is chosen at random (this is a random sample of size one). What is the probability that this sandwich will weigh between 145 and 155 grams?
NORMDIST(145, 150, 25, TRUE) = .42074 = probability that weight is less than 145 grams
NORMDIST(155, 150, 25, TRUE) = .57926 = probability that weight is less than 155 grams
The answer therefore = .57926 – . 42074 = .15852 = 15.85%.