# Why and where do we use normal distribution

content
"Preliminary remarks
“The normal distribution, a continuous random variable
"Remarks
»The normal distribution as an approximation of the binomial distribution
“The approach area
»The transformation into the standard normal distribution
“Examples
"Remarks

##### Preliminary remarks

The normal distribution is often introduced differently. It describes a continuous random variable, so it can be introduced as a counterpart to our discrete distribution functions. On the other hand, it also approximates the binomial distribution and is often used as a computational aid.

We will therefore keep the two following sections completely independent, i.e. introduce the normal distribution twice.

##### The normal distribution, a continuous random variable

If we remember, we also counted the bars, i.e. actually the area of ​​these, for the binomial distribution and other discrete random variables. With continuous random variables we have a density function \ (f \) and the area below it represents the probability. Many continuous random variables \ (X \) are normally distributed. The normal distribution is a so-called bell curve that is symmetrical about the expected value \ (\ mu \). It is marked with \ (N (\ mu, \ sigma) \). In the following we see the graph of \ (N (2000, 50) \). This normal distribution very closely approximates the burning time of lightbulbs with an expected life of 2000 hours and a fluctuation \ (\ sigma \) of 50 hours. The area from \ (- \ infty \) to 1975 is hatched. This area now corresponds to the probability that the light bulb shines for less than 1975 hours, i.e. \ (P (X \ leq 1975) \). How do we get our probability now? Let \ (f \) be the density function and \ (F \) its cumulative distribution function, then according to the above consideration we have
\ begin {align *}
P (X \ leq 1975) = \ int _ {- \ infty} ^ {1975} f (x) dx.
\ end {align *}
What is our \ (f \) so that we can solve the problem? This "short" formula, which depends on \ (\ mu \) and \ (\ sigma \), is hidden behind the normal distribution
\ begin {align *}
f (x) = \ frac {1} {\ sigma \ sqrt {2 \ pi}} e ^ {- \ frac {1} {2} (\ frac {x- \ mu} {\ sigma}) ^ 2}
\ end {align *}
and the area we get over
\ begin {align *}
F (x) = \ frac {1} {\ sigma \ sqrt {2 \ pi}} \ int _ {- \ infty} ^ xe ^ {- \ frac {1} {2} (\ frac {t- \ mu} {\ sigma}) ^ 2} dt.
\ end {align *}
Sadly, this integral cannot be solved manually. So you either need a technical aid in the applications (Geogebra, graphic calculator) or you can use a trick that we will now look at.

A special normal distribution is the \ (N (0,1) \) - standard normal distribution. It is often abbreviated with \ (\ phi \) and its distribution function then appropriately with \ (\ Phi \). The values ​​of the cumulative distribution function are simply tabulated and we can convert any other normal distribution into this.

##### Remarks

We have already stated several times that for a continuous random variable \ (P (X = k) = 0 \) applies. Now we know why, because if \ (f \) is the corresponding density function and \ (F \) is its distribution function, then it holds
\ begin {align *}
P (X = k) = \ int_k ^ k f (x) dx = F (k) -F (k) = 0.
\ end {align *}

For the same reason, with continuous random variables it does not matter whether we want to calculate \ (P (X

The idea is now to find a similar looking function that is similar to the histogram and thus approximates the area. This exists and is the so-called normal distribution \ (f \) as a function of \ (\ mu \) and \ (\ sigma \)
\ begin {align *}
f (x) = \ frac {1} {\ sigma \ sqrt {2 \ pi}} e ^ {- \ frac {1} {2} (\ frac {x- \ mu} {\ sigma}) ^ 2} .
\ end {align *}

##### The approximation surface we get over
\ begin {align *}
\ frac {1} {\ sigma \ sqrt {2 \ pi}} \ int_ {20} ^ {3} e ^ {- \ frac {1} {2} (\ frac {t- \ mu} {\ sigma} ) ^ 2} dt = F (30) -F (20) = 0.7267.
\ end {align *}

What does the cumulative distribution function \ (F (x) = P (X \ leq x) \) look like with which we can calculate \ (F (30) -F (20) \)? Sadly it is
\ begin {align *}
P (X \ leq x) =: F (x) = \ frac {1} {\ sigma \ sqrt {2 \ pi}} \ int _ {- \ infty} ^ xe ^ {- \ frac {1} {2} (\ frac {t- \ mu} {\ sigma}) ^ 2} dt.
\ end {align *}
cannot be solved manually. Fortunately, there is a trick, we can use a secondary calculation to convert every normal distribution into a certain normal distribution, the \ (N (0,1) \) - standard normal distribution. It is often abbreviated with \ (\ phi \) and its distribution function then appropriately with \ (\ Phi \).

[A picture will be added here soon.]

The approximation by the normal distribution works well from \ (\ sigma> 3 \), this is a proven rule of thumb, otherwise the error of the area is often too large ##### The transformation into the standard normal distribution

If you don't have a graphical calculator at hand, you need the standard normal distribution. Colloquially, we shift and compress any normal distribution \ (N (\ mu \ sigma) \) to \ (N (0; 1) \). In the following we did it for \ (N (2; 0.5) \). As written, one cannot give an explicit formula for the cumulative distribution function \ (\ Phi \) of \ (\ phi \), since the integral can only be solved numerically. The values ​​of the standard normal distribution are then simply tabulated and readable.

[A table will be added here soon.]

So that we can always uniquely identify the transformed random variable, we do not denote it with \ (X \) but with \ (z \). The formula then applies
\ begin {align *}
z = \ frac {X- \ mu} {\ sigma}.
\ end {align *}

What does this mean exactly? If we have a normally distributed random variable with \ (\ mu = 1.80 \) and standard deviation \ (\ sigma = 0.09 \), this could represent the height of men in Germany, for example, and would like to know what the probability is We are interested in the fact that a randomly selected man is shorter than 1.7 meters
\ begin {align *}
P (X \ leq 1.7)
\ end {align *}
where \ (X \) represents our height in cm. Now we can use the cumulative distribution function
\ begin {align *}
P (X \ leq 1.7) = \ frac {1} {0.09 \ sqrt {2 \ pi}} \ int _ {- \ infty} ^ xe ^ {- \ frac {1} {2} (\ frac {t-1.7} {0.09}) ^ 2} dt
\ end {align *}
do not solve manually as described several times. So we shift the normal distribution \ (N (1.80; 0.09) \) to \ (N (0; 1) \) and get our new random variable
\ begin {align *}
z & = \ frac {X- \ mu} {\ sigma}
& = \ frac {1.70-1.80} {0.09} = - 1.11.
\ end {align *}
Now we are no longer looking for \ (P (X \ leq 1.7) \) but \ (P (z \ leq -1.1) = \ Phi (-1.1) \) and read our table
\ begin {align *}
P (z \ leq -1.1) = \ Phi (-1.1) = 0.13 = P (X \ leq 1.7)
\ end {align *}
as a result. If you carry out an approximation of the binomial distribution, you sometimes still use the so-called continuity correction, you can find more about this in the examples.

##### Examples

Pilots: In 2016, Lufthansa defined the permitted size for prospective pilots. In future, it will only show pilots with the following body size, they must not be smaller than 1.65 or larger than 1.98. On the basis of \ (\ mu = 1.80 \) and \ (\ sigma = 0.09 \), calculate what percentage of the male population cannot become pilots due to their height.

### Solution with the help of a program:

We solve the example first (very quickly) with a graphic program and then manually with the help of the standard normal distribution. We have a constant random size \ (X \) that represents the height of a man. We want to know what percentage is less than 1.65 and how much greater than 1.98. So we are looking for \ (P (X \ leq 1.65) \) and \ (P (X \ geq 1.98) \).  and our graphical calculator gives us the results \ (P (X \ leq 1.65) = 0.0478 \) and \ (P (X \ geq 1.98) = 0.0228 \). Alternatively, one could also work with the counter-event because it applies
\ begin {align *}
P (X \ leq 1.65 \ text {or} X \ geq 1.98) & = 1-P (1.65 & =1-0,9295=0,0705.
\ end {align *} ### Alternative, manual, solution:

We have to "push" our \ (N (1.80; 0.09) \) normal distribution to \ (\ phi = N (0; 1) \) as before. With the transformation, \ (x_1 = 1.65 \) and \ (x_2 = 1.98 \) become
\ begin {align *}
z_1 = \ frac {1.65-1.80} {0.09}, & \ qquad z_2 = \ frac {1.98-1.80} {0.09} \
z_1 = -1.6667, & \ qquad z_2 = 2.
\ end {align *}
Now applies
\ begin {align *}
P (X \ leq 1.65) = P (z \ leq -1.6667) = \ Phi (-1.6667) = 0.0478
\ end {align *}
as
\ begin {align *}
P (X \ geq 1.98) & = P (z \ geq 2) = 1-P (z <2) \
& = 1- \ Phi (2) = 1-0.9772 = 0.0228.
\ end {align *}
Our solution is then \ (0.0228 + 0.0478 = 0.0706 \).

Dodger: According to statistics from 2015, almost 2 percent of people in an Austrian metropolis drive without a ticket, so-called fare dodgers. Almost 22,000 people are checked every day. What is the probability of controlling between 400 and 500 people?

### Solution with the help of a program:

We solve the example first (very quickly) with a graphic program and then manually with the help of the standard normal distribution. In addition, we consider the continuity correction in the manual case. For the approximation by the normal distribution we need \ (\ mu \) and \ (\ sigma \). We calculate this from \ (n = 22000 \) and \ (p = 0.02 \):
\ begin {align *}
\ mu = n \ cdot p, & \ qquad \ sigma = \ sqrt {= n \ cdot p \ cdot (1-p)} \
\ mu = 440, & \ qquad \ sigma \ approx 20.77.
\ end {align *}
Then we are interested in the event \ (P (400 \ leq X \ leq 500) \) which corresponds to the following area under the \ (N (440; 20.77) \) and our calculation program gives us the result \ (P (400 \ leq X \ leq 500) = 0.971 \).

### Alternative, manual, solution:

As before, we set up the associated normal distribution \ (N (440; 20.77) \) from the values ​​of the binomial distribution. Now we shift this to \ (\ phi = N (0; 1) \). With the transformation, \ (x_1 = 400 \) and \ (x_2 = 500 \) become
\ begin {align *}
z_1 = \ frac {400-440} {20.77}, & \ qquad z_2 = \ frac {500-440} {20.77} \
z_1 = -1.92585, & \ qquad z_2 = 2.88878.
\ end {align *}
Now applies
\ begin {align *}
P (400 \ leq X \ leq 500) & = P (-1.92585 \ leq z \ leq 2.88878) \
& = \ Phi (2.88878) - \ Phi (-1.92585) = \
& = 0,9981-0,0271=0,971.
\ end {align *}

### Alternative, manual, solution, with continuity correction:

The continuity correction aims to improve the approximation. If we zoom in to the point \ (x_1 = 400 \) of our approximation, we see that the calculated area of ​​the normal distribution starts from \ (x_1 = 400 \), but the bars of the histogram of the binomial distribution actually start from 399.5. In the language of analysis, "the normal distribution for the hatched integral \ (\ int_ {400} ^ {500} \) forgets half of a bar (checkered), a better approximation would therefore be intuitive \ (\ int_ {399.5} ^ {500.5} \) This is done by the so-called continuity correction. During the transformation we get different values ​​\ (z_1 \) and \ (z_2 \) due to the correction by 0.5
\ begin {align *}
z_1 = \ frac {(400-0.05) -440} {20.77}, & \ qquad z_2 = \ frac {(500 + 0.05) -440} {20.77} \
z_1 = -1.94993, & \ qquad z_2 = 2.91286.
\ end {align *}

The larger \ (n \) is in the approximation by the normal distribution, the less important the continuity correction becomes. It is therefore often left out in school.

##### Remarks

With normal distribution, you work almost exclusively with rounded values, which is why different approaches lead to rounding differences in the results.