# How is life at Mu Sigma

## Normal distribution

The normal distribution is extremely important in statistics because it is an approximation (= approximation) to other distributions:

### SCHEME NORMAL DISTRIBUTION

1) what is the normal distribution? What is the expectation µ what is the scatter σ?

2) What probability is asked for?

a) P (X ≤ a) the probability that the random variable X*at most* assumes the value a

b) P (X ≥ a) the probability that the random variable X has at least the value a

c) P (b ≤ X ≤ c) The probability that the random variable lies between the values b and c.

3) Standardize, i.e. subtract the expected value µ and divide by the scatter σ.

a) P (X ≤ a) = P ($ \ frac {X \; - \; \ mu} {\ sigma} $ ≤ $ \ frac {a \; - \; \ mu} {\ sigma} $)

b) P (X ≥ a) = 1 - P (X

c) P (b ≤ X ≤ c) = P (X ≤ c) - P (X ≤ b) = P ($ \ frac {X \; - \; \ mu} {\ sigma} $ ≤ $ \ frac { c \; - \; \ mu} {\ sigma} $) - P ($ \ frac {X \; - \; \ mu} {\ sigma} $ ≤ $ \ frac {d \; - \; \ mu} {\ sigma} $)

4) since the new random variable X_{ST}: = standard normal distribution is X_{St.} ~ N (0,1) holds and the distribution function of the N (0,1) - distribution is written with Φ, use this w -notation:

a) P (X ≤ a) = ... = Φ ($ \ frac {a \; - \; \ mu} {\ sigma} $)

b) P (X ≥ a) = ... = 1 - Φ ($ \ frac {a \; - \; \ mu} {\ sigma} $)

c) P (b ≤ X ≤ c) = ... = Φ ($ \ frac {c \; - \; \ mu} {\ sigma} $) - Φ ($ \ frac {d \; - \; \ mu} {\ sigma} $).

5) If necessary, use a trick: Φ (- g) = 1 - Φ (g), since the tables for the standard normal division N (0.1) usually only start at the abscissa value 0.

6) Look up the table for the N (0.1) distribution.

**Notice**

whether to use P (X ≤ a) or P (X

P (X ≤ a) = P (X

in some textbooks one finds N (µ, σ) as the “symbol” of the normal distribution, in others N (µ, σ

^{2}We write N (µ, σ), i.e. with the scatter σ instead of the variance σ^{2}.The distance between the expected value µ and the turning points is exactly the same as the scatter σ.

Now for a better understanding

**example**

The height of men in Germany is normally distributed with an expected value of 1.7 m with a dispersion of 0.1 m. What is the probability

a) to be less than 1.65,

b) to be greater than 1.72 as well

c) to have a height between 1.55 m and 1.75 m?

We apply the above scheme to solve this:

Step 1) It is evidently the normal distribution N (1.7; 0.1).

Step 2) Then we write down the task formally

a) P (X ≤ 1.65)

b) P (X ≥ 1.72)

c) P (1.55 ≤ X ≤ 1.75)

We turn the direction, if necessary, to "less than or equal to"

a) you don't have to change anything

b) P (X ≥ 1.72) = 1 - P (X ≤ 1.72)

c) P (1.55 ≤ X ≤ 1.75) = P (X ≤ 1.75) - P (X ≤ 1.55)

Step 3) Then we standardize

a) P (X ≤ 1.65) = P (X_{ST} ≤ $ {{1.65-1.7} \ over {0.1}} $) = P (X_{ST} ≤ - 0,5)

b) P (X ≥ 1.72) = 1 - P (X_{ST} ≤ $ {{1.72-1.7} \ over {0.1}} $) = 1 - P (X_{ST} ≤ 0,2)

c) P (1.55 ≤ X ≤ 1.75) = P (X_{ST} ≤ 0.5) - P (X_{ST} ≤ - 1,5)

Step 4) Φ - Apply notation supplies

a) P (X_{ST} ≤ - 0.5) = Φ (- 0.5)

b) 1 - P (X_{ST} ≤ 0.2) = 1 - Φ (0.2)

c) P (X_{ST} ≤ 0.5) - P (X_{ST} ≤ - 1.5) = Φ (0.5) - Φ (- 1.5)

Step 5) use trick with a) and c), namely Φ (- g) = 1 - Φ (g)

a) Φ (- 0.5) = 1 - Φ (0.5)

b) not necessary because Φ is looked up at a positive point

c) Φ (0.5) - Φ (- 1.5) = Φ (0.5) - [1 - Φ (1.5)] = Φ (0.5) - 1 + Φ (1.5)

Look up step 6) in a table for the N (0.1) distribution:

a) 1 - Φ (0.5) = 1 - 0.691462 = 0.308538

b) 1 - Φ (0.2) = 1 - 0.57926 = 0.42074

c) Φ (0.5) - 1 + Φ (1.5) = 0.691462 - 1 + 0.933193 = 0.624655.

**Notice**

The results can be checked by looking at the density function and the size of the hatched area. The total area below the density function is 1, the total area in question then has the calculated proportion.

We deal with another type of task here separately:

**example**

What height do 30% of German men exceed?

This modification of the task leads to “looking from the inside out” and is therefore treated separately. The result of the probability, namely 30%, is already known, the task now is to find the height that is exceeded with a probability of 30%: P (X ≥ a) = 0.3.

Exactly the opposite was true: the body size was known, but the probability was sought.

One solves the problem according to the o.e. **Scheme**:

Step 1) X ~ N (1.7; 0.1)

Step 2) P (X ≥ a) = 0.3, i.e. case b), because there is a minimum probability. Here you can see what is being asked for.

We turn the symbol around so that one can apply the distribution function:

P (X ≥ a) = 0.3 $ \ Leftrightarrow $ 1 - P (X ≤ a) = 0.3 $ \ Leftrightarrow $ - P (X ≤ a) = - 0.7 $ \ Leftrightarrow $ P (X ≤ a) = 0.7.

Step 3) Standardize P (X ≤ a) = 0.7 $ \ Leftrightarrow $ P (X_{ST} ≤ ) = 0,7.

Step 4) Use the Φ notation:

P (X_{ST} ≤ $ \ frac {a \; - \; 1.7} {0.1} $) = 0.7 $ \ Leftrightarrow $ Φ ($ \ frac {a \; - \; 1.7} {0.1 } $) = 0.7.

Step 5) was not necessary here

Step 6) Look from the inside out: you look up the inside of the table to see where 0.7 is reached or just exceeded and then you look for which numbers this is happening outside. Here the number 0.7 is exceeded for the first time at 0.53. So:

$ \ frac {a \; - \; 1.7} {0.1} $ = 0.53 $ \ Leftrightarrow $ a = 0.53 · 0.1 + 1.7 = 1.753.

This means that 30% of German men are taller than 1,753 m.

**Notice**

You can see that the tables for the N (0.1) distribution only start at 0.5. What happens if you ask what height 40% of German men exceed?

Here you need 0.4 as a value within the table and you won't find a number. The following trick, which applies specifically to the standard normal distribution (since it is symmetrical around zero), helps:

x_{α} = - x_{1 - }_{α}.

So x_{0,4} = - x_{1 - 0,4} = - x_{0,6}. So you look up the 0.6 - fractile and write a minus sign in front of it. 60% of German men are more than 1.675m tall.

You can see in the example calculated above that we still calculated quite imprecisely. We took the value in the table that just exceeded the fractile we were looking for, but did not find it. We can't completely erase this mistake, but we can**linear interpolation** which we will not go into further here.

So the 0.7 fractile of the N (0.1) distribution is roughly 0.5244. This estimate is*more accurate* than 0.53.

**Notice**

The density function of N (0,1) is

The distribution function of the standard normal distribution is

**Videos**

**Exercise (right-wrong questions about normal distribution)**

The following statements are true or false. Decide.

a) With normal distribution, the mode, expected value and median always match.

b) If any normally distributed random variable is standardized, one always obtains a standard normally distributed random variable.

c) Approx. 95% of all values lie in the 1 - σ - range of the normal distribution.

d) With the normal distribution it is important to standardize. The process of centering is an important part of this.

e) In the case of the standard normal distribution, intermediate values that are not specified in the distribution function table of the N (0.1) distribution can be precisely calculated using linear interpolation.

##### Solution:

a) With normal distribution, the mode, expected value and median always match.

**Correct**.

b) If any normally distributed random variable is standardized, one always obtains a standard normally distributed random variable.

**Correct**.

c) Approx. 95% of all values lie in the 1 - σ - range of the normal distribution.

**Not correct**, approx. 68% of all values are in the 1 · σ range.

d) With the normal distribution it is important to standardize. The process of centering is an important part of this.

**Correct**. Centering means subtracting the expected value μ from the sought value x. Standardizing means then dividing by the standard deviation σ: $ \ frac {x \; - \; \ mu} {\ sigma} $.

e) In the case of the standard normal distribution, intermediate values that are not specified in the distribution function table of the N (0.1) distribution can be precisely calculated using linear interpolation.

**Not correct**, they can only be interpolated (= approximated).

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