Normal Distribution And Standardising

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Normal distribution and standardising
X ~ N(μ, σ2)
The only values of the normal distribution that are tabulated are from Z ~ N(0, 1). Not many distributions will have a mean of 0 and a variance of 1 however, so we need to convert any normal distribution of X into the normal distribution of Z!
This is done usingthe formula:

Where μ is the mean and σ is the standard deviation.
Example:
If X ~ N(100, 64) find P(X > 108)
P(X > 108) so our first step is to use the formula for Z on both sides of the inequality.
So,

This becomes:

We can now fill in the values for the mean μ, and the standad deviation σ.
In this case, the mean μ = 100 and the variance σ2 = 64.
Therefore the standarddeviation σ = √64 = 8

By drawing our bell and reading from the standard normal table we obtain...
P(Z > 1) | = 1 − Φ (1) |
| = 1 − 0.8413 = 0.1587 |

In the next example we will use the same distribution of X and the same steps, but this time we will only write down the necessary workings. Try and follow, using the above example if you get stuck! Each step will only appear when youclick that you are ready (if you ever are!)
Example:
If X ~ N(100, 64) calculate P(109 < X < 118)
De-standardising
De-standardising is very much like standardising until the final few steps where we need to work backwards with the standard normal tables. These problems occur when we are given the probability of an event but where one of z, μ or σ is unknown.
Example:
If a random variable Xis normally distributed with mean 56 and standard deviation 5, calculate the value of a, if:
1. P(X > a) = 0.9744
2. P(X > a) = 0.2358
First let's look at our distribution...
X ~ N(56, 52)
and

the standardising formula.
Now if:
1. P(X > a) = 0.9744
So standardising:

Then looking at the bell we see...

...and looking backwards from the tables, 'from the mainbelly'
P(Z > -1.95) = 0.9744
Hence the 2 parts highlighted in red must be equal.

Rearranging, a = -1.95 × 5 + 56 = 46.25
2. P(X > a) = 0.2358
So, standardising:

By looking at the bell we see can see why in this example we subtract 0.2358 from 1.

Handy Hint: |
If probability is less than 0.5 as this is not given in main 'belly' of the tables we will need to subtract from 1. |Then looking backwards from the tables, 'from the main belly' we obtain...
P(Z > 0.72) = 0.7642
Hence the 2 parts highlighted in red will be equal giving

And rearranging gives: a = 5 × 0.72 + 56 = 59.6
Finding mu or sigma
μ = mu
σ = sigma
We will use the same method of de-standardising for problems involving finding μ, or σ, or both. The latter may require the solution of linearsimultaneous equations.
Example:

The Normal Distribution

A tennis ball firing machine fires balls with a distance that is normally distributed. The mean distance, μ, is unknown and the standard deviation is 0.8m.
If 5% of balls go further than 20m find the mean distance, μ.
First let's look at our distribution and the associated bell...
X ~ N(μ, 0.82)

If, P(X > 20) = 0.05, thestandardising we obtain:

From looking backwards at the tables (in the main 'belly') we also see that:
P(Z > 1.64) = 0.95
Note: This isn't exactly 1.64 but is close enough for our calculation.
Hence the 2 parts highlighted in red are equal, giving...

rearranging this gives
μ = 20 − 1.64 × 0.8 = 18.7m (1 d.p.)
Solutions to Standard Normal Distribution Exercises
These are the solutions tothe standard normal distribution exercises. You are strongly advised to work out your own solutions before you look at these.
Exercise 1
Suppose the heights of adult females is normally distributed with a mean of 66 inches and a standard deviation of 1.75 inches. Suppose that the heights of adult males is normally distributed with a mean of 70 inches and a standard deviation of 2.2 inches. Jill...