Mean and Variance of the Normal Distribution

Expert reviewed 21 July 2024 7 minute read


  • identify the numerical and graphical properties of data that is normally distributed
  • calculate probabilities and quantiles associated with a given normal distribution using technology and otherwise, and use these to solve practical problems
    • identify contexts that are suitable for modelling by normal random variables, eg the height of a group of students
    • recognise features of the graph of the probability density function of the normal distribution with mean 𝜇 and standard deviation 𝜎, and the use of the standard normal distribution

Note:

Video coming soon!

Review of the Difference Between Continuous and Discrete Distribution

  • Discrete Distributions: Deal with countable outcomes and are described by a PMF. Examples include the roll of a die and the number of students in a class.
  • Continuous Distributions: Deal with uncountable outcomes within a range and are described by a PDF. Examples include heights and weights.

What is the Expected Value or Mean of a Continuous Distribution?

As we have discussed in previous modules, the mean or expected value ****of a distribution is a measure of a central value. It provides an average outcome of a random variable over many trials. For a random variable XX, the mean is denoted by E(X)E(X) or μ\mu.

When dealing with continuous distributions, we must alter our previous understanding of calculating the mean or expected value. As a continuous distribution has a probability density function (PDF), usually denoted as f(x)f(x), the formula for the mean is changed to incorporate it. When working with a continuous distribution over the interval [a,b][a,b], the formula for the mean or expected value is given by:

E(X)=abxf(x)dxE(X)=\int_a^bxf(x)dx

Where,

  • xx is a possible value of the random variable XX
  • f(x)f(x) is the probability density function within the interval [a,b][a,b]

What is the Variance of a Continuous Distribution?

The variance ****of a distribution measures the spread or dispersion of the random variable around the mean. It is denoted by Var(X)Var(X) or σ2\sigma^2. The variance gives an idea of how much the values of the random variable differ from the mean value. For a continuous distribution, the formula to calculate the variance is given by:

Var(X)=E(X2)μ2=abx2f(x)dxμ2Var(X)=E(X^2)−\mu^2=\int_a^bx^2f(x)dx-\mu^2

Where,

  • xx is a possible value of the random variable XX
  • f(x)f(x) is the probability density function within the interval [a,b][a,b]
  • μ\mu and E(X)E(X) is the respective mean and expected value of the distribution

Practice Question 1

Determine the mean and standard deviation of the continuous distribution, which has a PDF y=125xy=\frac{1}{25}x, and lies within the limits 0x100\leq x\leq 10.

First we must determine the mean of the distribution, by employing the formula above:

E(X)=05(x×x25)dx=[x375]05=53750275=53E(X)=\int_0^{5}(x\times \frac{x}{25})dx\\=[\frac{x^3}{75}]_0^{5}\\=\frac{5^3}{75}-\frac{0^2}{75}\\=\frac{5}{3}

Now that we have found the mean, we can employ the formula to determine the variance of the distribution.

σ2=05(x2x25)dx(53)2=[x4100]05259=(2540100)2593.47\sigma^2=\int_0^{5}(x^2\frac{x}{25})dx-(\frac{5}{3})^2\\=[\frac{x^4}{100}]_0^{5}-\frac{25}{9}\\=(\frac{25}{4}-\frac{0}{100})-\frac{25}{9}\\\approx3.47