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Normal distribution

The normal distribution

The normal distribution

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Tutor: Bilal

Summary

The normal distribution

In a nutshell

The normal distribution is a distribution used to model continuous random variables, and it appears in many real-life situations.


Definition

A continuous random variable is a random variable that can take infinitely many values.



Shape of the normal distribution

The graph of a normal distribution curve is shown below.


Maths; Normal distribution; KS5 Year 13; The normal distribution


Due to its shape, it is also known as a ​bell curve.


Parameters of the normal distribution

The normal distribution depends on two parameters: the mean (μ\mu​) and variance (σ2\sigma^2) of the data set.

  • The mean affects the horizontal displacement of the graph.
  • The variance affects the "steepness" of the graph.

Maths; Normal distribution; KS5 Year 13; The normal distribution
Maths; Normal distribution; KS5 Year 13; The normal distribution

Low value of σ2\sigma^2​​

High value of σ2\sigma^2​​


This makes sense; variance represents the spread of data. A higher variance should mean that the data is more spread out, as the second graph above shows.



Finding probabilities from a normal distribution

If a continuous random variable XX​ follows a normal distribution with mean μ\mu and variance σ2\sigma^2, then write:


XN(μ,σ2)\boxed{X \sim N(\mu, \sigma^2)}​​


The probability that XX​ lies within an interval is given by the area under the corresponding curve within the interval.


Maths; Normal distribution; KS5 Year 13; The normal distribution


A=P(aXb)A = P(a \le X \le b)​​


Note that because continuous random variables have an infinite number of possibilities, the probability of any particular event occuring is zero. Therefore, the normal distribution is only useful when trying to find a cumulative probability.



Properties of the normal distribution

Here are some properties that are worth learning about the normal distribution.


  • The normal distribution curve is symmetrical about the mean.
  • The mean, median and mode of the normal distribution are all the same.
  • Approximately 68%68\%​ of the data in a normal distribution lies within one standard deviation of the mean.
  • Approximately 95%95\%​ of the data in a normal distribution lies within two standard deviations of the mean.
  • Approximately 99.7%99.7\%​ of the data in a normal distribution lies within three standard deviations of the mean.
  • The total area under a normal distribution curve is equal to 11​.


Example 1

The random variable XX follows a normal distribution with mean 100100 and variance 1616. Find the following probabilities.

i) P(X100)P(X \leq 100)

ii) P(92X108)P(92 \leq X \leq 108)

iii) P(100X<104)P(100 \leq X \lt 104)


Part i):

Use the fact that the normal distribution is symmetrical about the mean:

P(X100)=0.5P(X \leq 100) = \underline{0.5}​​


Part ii):

Use the fact that 95%95\% of data lies within two standard deviations of the mean:

σ2=16σ=4P(92X108)=P(1002(4)X100+2(4))=P(μ2σXμ+2σ)=0.95\sigma^2=16 \rightarrow \sigma=4\\ \begin{aligned}P(92 \leq X \leq 108) &= P(100-2(4) \leq X \leq 100+2(4))\\&=P(\mu - 2\sigma \leq X \leq \mu + 2\sigma)\\&=0.95 \end{aligned}​​


P(92X108)=0.95P(92 \leq X \leq 108) = \underline{0.95}​​


Part iii):

Use the fact that 68%68\% of the data lies within one standard deviation of the mean combined with the fact that the normal distribution is symmetrical about the mean:

P(100X104)=P(μXμ+σ)=12×P(μσμμ+σ)=12×0.68=0.34\begin{aligned} P(100 \leq X \leq 104) &= P( \mu \leq X \leq \mu + \sigma)\\&= \dfrac12 \times P(\mu - \sigma \leq \mu \leq \mu +\sigma)\\&= \dfrac12 \times 0.68\\&=0.34 \end{aligned}​​


P(100X104)=0.34P(100 \leq X \leq 104) = \underline{0.34}​​



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