Normal Distribution

5.3

Normal Distribution

When the sample \(n\) becomes large, we can estimate the answer by using a normal distribution.
the conditions needed to determine whether the size \(n\) is large enough or not:

1.	\(np \geqslant 10\), where \(p\) is the probability of ‘success’

2.	\(n(1-p) \geqslant 10\), where \((1-p)\) is the probability of 'failure'

Normal Distribution, \(X \sim \text{N}(\mu, \sigma^2)\)

1.	A probability function of a continuous random variable

2.	The distribution is symmetrical with the most of the data clustered around the centre close to the mean

3.	The probabilities for the data further the mean taper off equally in both directions

\(\text{Mean }=\text{ Median }=\text{ Mode}\)
The graph is symmetrical about an axis at the centre of the normal distribution
\(50\%\) of the data values is less than the mean and the other \(50\%\) is greater than the mean

Features of Normal Distribution function graph

(a)	The curve is bell-shaped and is symmetrical about a vertical line that passes through the mean, \(\mu\)

(b)	The curve has a maximum value at the axis of symmetry, \(X=\mu\)

(c)	The mean, \(\mu\) divides the region under the graph into two equal parts.

(d)	Both ends of the curve extend indefinitely without touching the \(x\)-axis

(e)	The total area under the graph is equal to the total probability of all outcomes, that is \(1 \text{ unit}^2\)

Their positions and the width of the graphs depend on their respective mean, \(\mu\) and standard deviation, \(\sigma\)
Based on the graph below, The area under the graph for \(X\) from \(a\) to \(b\) represents the probability of \(X\) occurring for the value of \(X\) from \(a\) to \(b\) and is written as:


	\(P(a \text{ < }X \text{ < }b) = P(a \leqslant X \leqslant b)\)

Example:

Example

The diagram above shows a normal distribution function graph which is symmetrical at \(X=35\).

(a)	State the mean value, \(\mu\).

(b)	Express the shaded region in probability notation.

(c)	If the probability of the shaded region is \(0.64\), find \(P(X\text{ < }28)\).

Solution:

(a)	\(\mu=35\)

(b)	\(P(28\text{ < }X\text{ < }42)\)

(c)	Since the graph is symmetrical at \(X=35 \text{ and }X=28\) and \(X=42\) are both \(7\) units respectively to the left and right of the mean, then \(\begin{aligned} P(X\text{ < }28)&=P(X\text{ > }42)\\\\ &=\dfrac{1-0.64}{2}\\\\ &=0.18 \end{aligned}\)

Law of Large Numbers


	The larger the size of a sample, the value of the experimental mean gets closer to the theoretical mean value of the population.

Standard Normal Distribution


	A normal distribution whose mean and standard deviation are \(0 \text{ and }1\) respectively.

A continuous random variable \(X \sim \text{N}(\mu, \sigma^2)\) with mean, \(\mu\) and standard deviation, \(\sigma\) can be standardised by changing it to another continuous random variable \(Z\) whose mean is 0 and standard deviation is 1 by using the following formula:


	\(Z = \dfrac{X- \mu}{\sigma}, \text{ where }Z \sim \text{N}(0,1)\)

Remark


	\(\begin{aligned} \text{Mean, E}(Z) &= \text{E}\begin{pmatrix}{\dfrac{X-\mu}{\sigma}}\end{pmatrix}\\\\ &=\dfrac{1}{\sigma}[\text{E}(X)-\mu]\\\\ &=\dfrac{1}{\sigma}[\mu-\mu]\\\\ &=0 \end{aligned}\)		\(\begin{aligned} \text{Var}(Z) &= \text{Var}\begin{pmatrix}{\dfrac{X-\mu}{\sigma}}\end{pmatrix}\\\\ &=\dfrac{1}{\sigma^2}[\text{Var}(X)-0]\\\\ &=\dfrac{1}{\sigma^2}[\sigma^2]\\\\ &=1 \end{aligned}\)

Example

(a)	A continuous random variable \(X\) is normally distributed with mean \(30\) nd a standard deviation of \(8\). Find the \(z\)-score if \(X = 42\).

(b)	The heights of buildings in Kampung Pekan are normally distributed with a mean of \(23 \text{ m}\) and a variance of \(25\text{ m}^2\), find the height of the building if the standard score is \(0.213\).


Solution:

(a)	Given \(X = 42, \mu = 30 \text{ and } \sigma = 8.\)

	\(\begin{aligned} Z &= \dfrac{X-\mu}{\sigma}\\\\ &=\dfrac{42-30}{8}\\\\ &=1.5 \end{aligned}\)

(b)	Given \(\mu=23, \sigma^2=25 \text{ and }z\text{-score } = 0.213.\)

	\(\begin{aligned} \text{Thwn, }\sigma&=\sqrt{25}\\ &=5\\ \end{aligned}\) Therefore, \(\begin{aligned} Z &= \dfrac{X-\mu}{\sigma}\\\\ 0.213&=\dfrac{X-23}{5}\\\\ 1.065&=X-23\\\\ X&=24.065\text{ m} \end{aligned}\)

the way to convert this probability of the event to a standard normal distribution with a continuous random variable \(Z\) is as follows:


	\(\begin{aligned} P(a \text{ < }X \text{ < }b) &= P\begin{pmatrix}\dfrac{a-\mu}{\sigma} \text{ < } \dfrac{X-\mu}{\sigma} \text{ < } \dfrac{b-\mu}{\sigma}\end{pmatrix}\\\\ &=P\begin{pmatrix}\dfrac{a-\mu}{\sigma} \text{ < } Z \text{ < } \dfrac{b-\mu}{\sigma}\end{pmatrix} \end{aligned}\)

The diagram below shows the relation between the normal distribution graph and the standard normal distribution graph:

Example

	The lengths of a type of screw produced by a factory can be considered as normally distributed with a mean of \(10.6\text{ cm}\) and a standard deviation of \(3.2\text{ cm}\). Represent the probability that a screw randomly chosen from the factory has a length between \(8.4\text{ cm}\) and \(13.2\text{ cm}\) where \(Z\) is a standard continuous random variable.


	Solution:

	Let \(X\) represent the length of the screw produced by the factory.

	Given \(\mu=10.6 \text{ and }\sigma =3.2\) \(\begin{aligned} &P(\text{Length of screw is between }8.4\text{ cm and }13.2\text{ cm})\\\\ &= P(8.4 \text{ < } X \text{ < } 13.2)\\\\ &=P\begin{pmatrix}\dfrac{8.4-10.6}{3.2} \text{ < } \dfrac{X-\mu}{\sigma} \text{ < } \dfrac{13.2-10.6}{3.2}\end{pmatrix}\\\\ &=P(-0.6875 \text{ < } Z\text{ < } 0.8125) \end{aligned}\)