# Help to understand normal distribution?

1. Jul 25, 2013

### LDC1972

1. The problem statement, all variables and given/known data
Metal strips are produced mean length 150cm, standard deviation 10cm.
Find probability that length of random strip is:
a/ shorter than 165cm
b/ longer than 170cm
c/ between 145 and 155cm

2. Relevant equations

Z = x - μ / σ

θ (Z) - θ (z) = average probability (question c)

3. The attempt at a solution

a/ Z = 165 - 150 / 10 = 1.5
1.5 from standardised table = 0.9332 = 93.32% probability length shorter than 165cm

b/ Z = 170 - 150 / 10 = 2
2 from standardised table = 0.9772

Now here I did 1 - 0.9772 = 0.0228 = 2.28% probability length longer than 170cm.

I did this not because I understand the curve graph, but because without subtracting 1 the probability was obviously wrong.

c/ as per a/ to get Z = -0.5 for 145cm and Z = 0.5 for 155cm
used tables to get 0.6915 for 0.5.
Then did:
θ = 0.6915 - (1 - 0.6915) = 0.383 = 38.3% between 145 and 155 cm

Again, I had to -1 to make equation work anyway, BUT also to give a sensible probability.

I'm assuming answers are right? But how do we say when to -1 or not. I know it' to left or right of mean etc, but simply can't get head round it?

Any help welcome as always!

Thanks guys.

Oh I know θ should be O with vertical line, but couldn't find it on the selection box, so made do..

2. Jul 25, 2013

### Ray Vickson

The equation you wrote is incorrect. It reads as $$Z = x - \frac{\mu}{\sigma}$$ but should be
$$Z = \frac{X-\mu}{\sigma}$$
You need to learn to use parentheses, like this: z = (x-μ)/σ .

Similarly, it is false to say that 165 - 150 / 10 = 1.5, because 165 - 150 / 10 = 165 - 15 = 150; perhaps you mean (165 - 150)/10? If so, that is what you should write.

Anyway, to answer your questions: it depends on what sorts of tables/calculator buttons you are using. One type of standardized table is that of $\Phi(z)$, which is the area under the standard normal density curve to the left of z; that is:
$$\Phi(z) = \int_{-\infty}^z \frac{1}{\sqrt{2 \pi}} e^{-t^2/2} \, dt.$$
Then, the area to the right of z is $1-\Phi(z).$ The area between $z_1$ and $z_2 \;(z_2 > z_1)$ is just $\Phi(z_2) - \Phi(z_1),$ because it is the area to the left of z_2 minus the area to the left of z_1 (leaving only the stuff between z_1 and z_2). That's all there is to it!

3. Jul 25, 2013

### CompuChip

If you're using the standardized table, you need to keep in mind that it gives cumulative probabilities for $Z \le z$.
I can put it another way if you know that the probability is given by the area under the normal curve: the standardized table gives you left-tailed probabilities such as the following:

In the first example you gave, this is exactly what you want: the probability that Z < 1.5 can be found by reading off z = 1.5 from the table.
In the second example, you want the probability that Z > 2. This is not like the colored area above, but it is actually like the white area. The table does not provide this. However, you know that the total area under the curve must be 1, which corresponds to "the probability of finding some value is 100%". So to find the white area you can find the colored area and subtract that from 1. This is called the complement rule, in algebraic language it reads $P(Z \ge z) = 1 - P(Z \le z)$ and it follows directly from the fact that $P(Z \ge z) + P(Z \le z) = 1$ (orange area + white area = total area) must hold.

Now suppose that you want to calculate the shaded area in this graph:

Assuming that the boundaries are -1.8 and 0.8, the shaded area would be $-1.8 \le z \le 0.8$. Can you express this entirely in terms of $z \le \cdots$? Hint: you can start with $z \le 0.8$ but then you are taking too much, so you need to subtract that bit again.