Further normal distribution

Further normal distribution

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 using the formula:

Conversion 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,

Example

This becomes:

Example

We can now fill in the values for the mean μ, and the standard deviation σ.

In this case, the mean μ = 100 and the variance σ2 = 64.

Therefore the standard deviation σ = √64 = 8

Example

By drawing our bell and reading from the standard normal table we obtain...

P(Z > 1) = 1 − Φ (1)
= 1 − 0.8413 = 0.1587

Example
Further normal distribution

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 you click that you are ready (if you ever are!)

Example:

If X ~ N(100, 64) calculate P(109 < X < 118)

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 X is 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

Example

the standardising formula.

Now if:

1. P(X > a) = 0.9744

So standardising:

Example

Then looking at the bell we see...

P(X > a) = 0.9744

...and looking backwards from the tables, 'from the main belly'

P(Z > -1.95) = 0.9744

Hence the 2 parts highlighted in red must be equal.

Example

Rearranging, a = -1.95 × 5 + 56 = 46.25

2. P(X > a) = 0.2358

So, standardising:

Example

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

Example

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.

Example

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

Example

And rearranging gives: a = 5 × 0.72 + 56 = 59.6

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 linear simultaneous equations.

Example:

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)

Example

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

Example

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...

Example

rearranging this gives

μ = 20 − 1.64 × 0.8 = 18.7m (1 d.p.)

S-cool exclusive!!