14. Normal Probability Distributions
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The Normal Probability Distribution is very common in the field of statistics.
Whenever you measure things like people's height, weight, salary, opinions or votes, the graph of the results is very often a normal curve.
The Normal Distribution
A random variable X whose distribution has the shape of a normal curve is called a normal random variable.
`f(X)=1/(sigmasqrt(2pi))e^((xmu)^2 //2\ sigma^2``mu`
This random variable X is said to be normally distributed with mean μ and standard deviation σ if its probability distribution is given by
`f(X)=1/(sigmasqrt(2pi))e^((xmu)^2 "/"2\ sigma^2`
Properties of a Normal Distribution

The normal curve is symmetrical about the mean μ;

The mean is at the middle and divides the area into halves;

The total area under the curve is equal to 1;

It is completely determined by its mean and standard deviation σ (or variance σ^{2})
Note:
In a normal distribution, only 2 parameters are needed, namely μ and σ^{2}.
Area Under the Normal Curve using Integration
The probability of a continuous normal variable X found in a particular interval [a, b] is the area under the curve bounded by `x = a` and `x = b` and is given by
`P(a<X<b)=int_a^bf(X)dx`
and the area depends upon the values of μ and σ.
[See Area under a Curve for more information on using integration to find areas under curves. Don't worry  we don't have to perform this integration  we'll use the computer to do it for us.]
The Standard Normal Distribution
It makes life a lot easier for us if we standardize our normal curve, with a mean of zero and a standard deviation of 1 unit.
If we have the standardized situation of μ = 0 and σ = 1, then we have:
`f(X)=1/(sqrt(2pi))e^(x^2 "/"2`
We can transform all the observations of any normal random variable X with mean μ and variance σ to a new set of observations of another normal random variable Z with mean `0` and variance `1` using the following transformation:
`Z=(Xmu)/sigma`
We can see this in the following example.
Example 1
Say `μ = 2` and `sigma = 1/3` in a normal distribution.
The graph of the normal distribution is as follows:
The following graph (that we also saw earlier) represents the same information, but it has been standardized so that μ = 0 and σ = 1 (with the above graph superimposed for comparison):
The two graphs have different μ and σ, but have the same area.
The new distribution of the normal random variable Z with mean `0` and variance `1` (or standard deviation `1`) is called a standard normal distribution. Standardizing the distribution like this makes it much easier to calculate probabilities.
Formula for the Standardized Normal Distribution
If we have mean μ and standard deviation σ, then
`Z=(Xmu)/sigma`
Since all the values of X falling between x_{1} and x_{2} have corresponding Z values between z_{1} and z_{2}, it means:
The area under the X curve between X = x_{1} and X = x_{2}
equals
the area under the Z curve between Z = z_{1} and Z = z_{2}.
Hence, we have the following equivalent probabilities:
P(x_{1} < X < x_{2}) = P(z_{1} < Z < z_{2})
Example 2
Considering our example above where `μ = 2`, `σ = 1/3`, then
Onehalf standard deviation = `σ/2 = 1/6`, and
Two standard deviations = `2σ = 2/3`
So `1/2` s.d. (standard deviation) to 2 s.d. to the right of `μ = 2` will be represented by the area from `x_1=13/6 = 2 1/6 ~~ 2.167` to `x_2=8/3 = 2 2/3~~ 2.667`. This area is graphed as follows:
The area above is exactly the same as the area
z_{1} = 0.5 to z_{2} = 2
in the standard normal curve:
Percentages of the Area Under the Standard Normal Curve
A graph of this standardized (mean `0` and variance `1`) normal curve is shown.
Standard Normal Curve showing percentages μ = 0, σ = 1.
In the above graph, we have indicated the areas between the regions as follows:
−1 ≤ Z ≤ 1 68.27%
−2 ≤ Z ≤ 2 95.45%
−3 ≤ Z ≤ 3 99.73%
This means that `68.27%` of the scores lie within `1` standard deviation of the mean.
This comes from: `int_1^1 1/(sqrt(2pi))e^(z^2 //2)dz=0.68269`
Also, `95.45%` of the scores lie within `2` standard deviations of the mean.
This comes from: `int_2^2 1/(sqrt(2pi))e^(z^2 //2)dz=0.95450`
Finally, `99.73%` of the scores lie within `3` standard deviations of the mean.
This comes from: `int_3^3 1/(sqrt(2pi))e^(z^2 //2)dz=0.9973`
The total area from `∞ < z < ∞` is `1`.
The zTable
The areas under the curve bounded by the ordinates z = 0 and any positive value of z are found in the zTable. From this table the area under the standard normal curve between any two ordinates can be found by using the symmetry of the curve about z = 0. We can also use Scientific Notebook, as we shall see.
Go here for the actual zTable.
Example 3
Find the area under the standard normal curve for the following, using the ztable. Sketch each one.
(a) between z = 0 and z = 0.78
(b) between z = −0.56 and z = 0
(c) between z = −0.43 and z = 0.78
(d) between z = 0.44 and z = 1.50
(e) to the right of z = −1.33.
Example 4
Find the following probabilities:
(a) P(Z > 1.06)
(b) P(Z < 2.15)
(c) P(1.06 < Z < 4.00)
(d) P(1.06 < Z < 4.00)
Example 5
It was found that the mean length of `100` parts produced by a lathe was `20.05\ "mm"` with a standard deviation of `0.02\ "mm"`. Find the probability that a part selected at random would have a length
(a) between `20.03\ "mm"` and `20.08\ "mm"`
(b) between `20.06\ "mm"` and `20.07\ "mm"`
(c) less than `20.01\ "mm"`
(d) greater than `20.09\ "mm"`.
Example 6
A company pays its employees an average wage of `$3.25` an hour with a standard deviation of `60` cents. If the wages are approximately normally distributed, determine
 the proportion of the workers getting wages between `$2.75` and `$3.69` an hour;
 the minimum wage of the highest `5%`.
Example 7
The average life of a certain type of motor is `10` years, with a standard deviation of `2` years. If the manufacturer is willing to replace only `3%` of the motors because of failures, how long a guarantee should she offer? Assume that the lives of the motors follow a normal distribution.
Application  The Stock Market
Sometimes, stock markets follow an uptrend (or downtrend) within `2` standard deviations of the mean. This is called moving within the linear regression channel.
Here is a chart of the Australian index (the All Ordinaries) from 2003 to Sep 2006.
Image source: incrediblecharts.com.
The upper gray line is `2` standard deviations above the mean and the lower gray line is `2` standard deviations below the mean.
Notice in April 2006 that the index went above the upper edge of the channel and a correction followed (the market dropped).
But interestingly, the latter part of the chart shows that the index only went down as far as the bottom of the channel and then recovered to the mean, as you can see in the zoomed view below. Such analysis helps traders make money (or not lose money) when investing.
Image source: incrediblecharts.com.
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