Have you ever wondered, if we repeatedly take samples from a normal distribution, what value appears most frequently? Or in other words, what is the central value that we expect from that distribution?
This is what we call expected value. In the context of normal distribution, expected value has a very interesting and simple property.
For a normal distribution n(x;μ,σ), its expected value is always equal to the mean parameter μ.
Mathematically, we can write:
E(X)=μ
where X is a random variable that follows normal distribution and μ is the mean parameter of that distribution.
The expected value of a continuous random variable is defined as:
E(X)=∫−∞∞x⋅f(x)dx
For normal distribution, the probability density function is:
f(x)=σ2π1e−21(σx−μ)2
Substitute this function into the expected value formula:
E(X)=∫−∞∞x⋅σ2π1e−21(σx−μ)2dx
Now, let's perform the substitution t=σx−μ. Then x=σt+μ and dx=σdt.
E(X)=∫−∞∞(σt+μ)⋅σ2π1e−21t2⋅σdt
E(X)=2π1∫−∞∞(σt+μ)e−21t2dt
E(X)=2πσ∫−∞∞te−21t2dt+2πμ∫−∞∞e−21t2dt
Note that:
First integral∫−∞∞te−21t2dt=0 because the function te−21t2 is an odd function. This means f(−t)=−f(t), so when integrated over the symmetric interval [−∞,∞], the results cancel each other out and equal zero.
Second integral∫−∞∞e−21t2dt=2π because this is the fundamental Gaussian integral. This integral represents the total area under the standard normal distribution curve, which always equals .
Therefore:
E(X)=2πσ⋅0+2πμ⋅2π
E(X)=0+μ=μ
It is proven that the expected value of normal distribution equals its mean parameter.
For every normal distribution X∼N(μ,σ2), we have E(X)=μ. This means we don't need to perform integration every time we calculate the expected value of a normal distribution; we simply use the parameter value μ.
The proof result above provides a very important understanding:
If we have a normal distribution with mean μ and standard deviation σ, then the expected value of that random variable is exactly μ. This means if we take samples in very large numbers and calculate their average, the result will approach the value μ.
Simple example:
If the height of students in your school is normally distributed with mean 165 cm, then the expected value of the height of a randomly selected student is 165 cm.
Suppose the math exam scores in a class are normally distributed with mean μ=75 and standard deviation σ=10. What is the expected value of a student's exam score?
Solution:
Since normal distribution has the property E(X)=μ, the expected value of the exam score is:
E(X)=75
Therefore, the expected value of a student's exam score is 75.
Example2
The weight of newborn babies at a hospital is normally distributed with mean 3.200 grams and standard deviation 400 grams. Determine the expected value of the weight of a baby to be born!
Solution:
Given μ=3.200 grams and σ=400 grams.
The expected value of the weight of a baby to be born is:
Travel time from home to school is normally distributed with mean 25 minutes and standard deviation 5 minutes. Determine the expected value of travel time!
Daily air temperature in Jakarta during June is normally distributed with mean 28∘C and standard deviation 3∘C. What is the expected value of air temperature on a randomly selected day?
Physics exam scores of grade 12 students are normally distributed with mean 78 and standard deviation 12. If a student takes the exam, what is the expected value they will obtain?