The Normal Distribution
The Normal Distribution – Introduction
Suppose you were to take a sample of newborn babies and record their weights, what shape do you think that the distribution would be?
Imagine that the average weight of the newborn babies is between 3.4kg and 3.6kg then it would be sensible to get a large number of babies that are close to the interval and the frequencies will then start to decrease as you get further away from this interval.
It would even be possible to predict it would fall to zero but then again only a very small number of babies would be very small or very large.
From the diagram above you can see that the histogram is symmetrical and if a curve is drawn through the midpoint of each bar then the Normal Distribution curve is obtained.
Properties of the Normal Distribution curve
A number of important properties exist for a Normal curve:
- It is bell shaped
- It is symmetrical
- The mean, median and mode all lie on the line of symmetry
- The width of the curve shows the spread of the distribution
How The Normal Distribution Connects With The Standard Deviation
Standard deviation is used when thinking about the spread of a normal curve. For a normal distribution, 68% of the values are within one standard deviation of the mean. And 96% of all values are within two standard deviations of the mean. Finally 99.7% of all values are within three standard deviations of the mean.
Should you find that a number of results were outside of these bounds then it is possible that the sample size is too small or there wasn’t a normal distribution.
In general if the variable X has mean and standard deviation then x which is a particular value of X is transformed into z by the formula
z=\frac{x-\mu}{\sigma}The z score shows how many standard deviations above or below the mean a particular score is.
For instance, a student with a score of 80 on a test with a mean of 60 and a standard deviation of 10. Converting the test score to a z score will give z=\frac{60-80}{10}=2
This means that the original score was 2 standard deviations above the mean.
You can then find the value of \varphi(z)=2 using a Normal distribution table of \varphi(z) and then calculating 1-\varphi(z). φ is pronounced phi
Normal Distribution – Examples
Example
Assuming the distribution of the heights of adult men is normal with the mean being 174cm and standard deviation 7cm find the following probabilities:
- Under 185cm
- Over 185cm
- Over 180cm
- Between 180cm and 185cm
- Under 170cm
Give all answers to 2 significant figures.
Solution
The mean height is μ=174 and the standard deviation is σ=7
- x=185
So z=\frac{185-174}{7}=1.571
\varphi(1.571)=0.9419=0.94So the probability is 0.94
The answer for under 185cm was obtained in part a)
This means that the probability for a man to be over 185cm is:
1-\varphi(1.571)=1-0.9419=0.0581=0.058c) x = 180cm
z=\frac{180-174}{7}=0.857The probability = 1-\varphi(0.857)=1-0.8042=0.1958=0.20
d) The probability is calculated as \varphi(1.571)-\varphi(0.857)=0.9419-0.8042=0.1377=0.14
e) x = 170cm
So z=\frac{170-174}{7}=-0.571
However the value of \varphi(-0.571) as only positive values so the symmetry of the Normal curve needs to be used.
So \varphi(-0.571)=1-\varphi(-0.571)=1-0.716=0.284=0.28
Using the normal distribution curve
All normal curves have essentially the same basic shape so it is possible to scale the two axes and to have one curve fit on top of the other.
A curve for the normal distribution with a mean and standard deviation is given by the function \varphi(x) where:
\varphi(x)=\frac{1}{\sigma \sqrt{2 \pi}} e^{-\frac{1}{2}\left(\frac{x-\mu}{\sigma}\right)}The notation that is used to describe this distribution is N\left(\mu, \sigma^2\right).
Once the parameters of and are known it is possible to know everything there is to know about the distribution.
Quite often the variable X has been transformed to Z using z=\frac{x-\mu}{\sigma} and the distribution then becomes:
\varphi(x)=\frac{1}{\sqrt{2 \pi}} e^{-\frac{1}{2} z^2}Normal Distribution Example
In order to make a particular widget, the company responsible for its manufacturing believes that it is modelled by a normal distribution with a mean of 95 minutes and a standard deviation of 4 minutes.
Based on the above being true find the probability that the time taken to make a widget was:
- Over 97 minutes
- Under 90 minutes
- Between 90 and 97 minutes
Give all answers to 2 d.p
Solution
- μ=95, σ=4
So the probability that a widget took over 97 minutes is:
Z=\frac{97-95}{4}=0.5Therefore the probability is given by: 1-\varphi(5)-1-0.6915-0.3085-0.31
The probability that a widget took under 90 minutes is: z=\frac{90-95}{4}=-1.25
Therefore the probability is given by: 1-\varphi(1.25)=1-0.8944=0.1056=0.11
The probability that a widget takes between 90 and 97 minutes is given by: 1-0.1056-0.3085=0.5859=0.59