Normal Distribution

A specific type of a continuous distribution is the normal distribution. The normal distribution is in the shape of a bell curve and is a common occurring probability distribution in nature, or is used as a close approximation for the measurement of variables.

Bell Curve

The normal probability density function.

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

The Bell curve has the following key properties:

It is symmetrical around the y axis
The mean of the function is zero
The standard deviation is one.

General Normal Probability Distribution

It is possible to translate the graph by shifting to a new mean and to change the spread of the the normal distribution by dividing our new x term by the standard deviation and multiplying the y term by the standard deviation.

$\frac{1}{\sqrt{2\pi}}e^{\frac{-x^{2}}{2}} \rightarrow \frac{1}{\sigma\sqrt{2\pi}}e{-\frac{1}{2}(\frac{x-\mu}{\sigma})^{2}}$

This gives us a new equation of on over sigma times the square root of two pi times e to the power of negative a half of x minus mew over sigma all squared. Where sigma is our standard deviation and mew is the the mean value.

Spread of the Normal Distribution

The main strength of the normal distribution is that we know what percentage of the distribution lies within each standard deviation of the mean.

$\mu-\sigma \leq X \leq \mu +\sigma$ includes about 68% of the values

$\mu-2\sigma \leq X \leq \mu +2\sigma$ includes about 95% of the values

$\mu-3\sigma \leq X \leq \mu +3\sigma$ includes about 99.7% of the values

Standardised Values

To help us solve the general normal distribution, we can apply the general normal distribution to our original normal distribution. This we can do by linking our, z value, the standardise value to our x value.

$z=\frac{x-\mu}{\sigma}$

When n is positive the value lies above the mean and when n is negative the value lies below the mean. The z value can be thought of as the point on the original normal curve that incorporates the same amount of the distribution as the x point on the general normal curve.

Practice Exam Question

Question 5 from the 2010 exam 1

Part a asks to find the probability that X is greater than five for a normally distributed random variable with a mean of five. Given that normal distributions are symmetrical we know that 50% of the distributions lie either side of the mean. Therefore the probability that we get a response greater than 5 is 0.5

Part B, asks us to find the equivalent standardise value for the probability x is greater than 7, but at the other end the distribution for when the standardised value is less than b.

Step 1 Find the value of z above which the probability is the same as x is greater than 7.

$z=\frac{x-\mu}{\sigma} \newline = \frac{7-5}{3}=\frac{2}{3}$

Given the probability of the normal distribution is symmetrical around zero, the probability that z is less than b is the same as the probability that z is greater than negative b. Therefore if we times our z value of two thirds by negative one we will get our b value.

$b=-\frac{2}{3}$

The symmetrical properties of the normal distribution are key to answering many of the normal distribution questions.

Feedback

Want to suggest an edit? Have some questions? General comments? Let us know how we can make this resource more useful to you.

While we endeavour to provide you with great study material - we’re not qualified teachers, as such Engage cannot guarantee the validity of the information here.

The Victorian Curriculum and Assessment Authority (VCAA) does not endorse this website and makes no warranties regarding the correctness or accuracy of its content. VCE is a registered trademark of the VCAA. Current and past VCE exams and related content can be accessed directly at www.vcaa.vic.edu.au

Contents