[Math] the ‘y’ axis in normal distribution

Question

I have this graph for questions on normal distribution:

Question 1. What is the probability that a participant selected at random will require more than 500 hours to complete the program?

Answer is $0.5$ by looking at the graph.

Question 2. What is the "probability" that a candidate selected at random will take between 500 and 650 hours to complete the training program?

The answer is $0.4332$.

Now, my question is where is the probably density function i.e. Y axis? How do I draw the Y-axis? Is "probably density function" same as just probability that I am finding?

I would like if someone can draw that Y-axis and answer it using Y-axis. So, I can understand its use.

Answer 1

Areas in graphs in the Question are not to scale and may be confusing. (Red Z with arrows is inappropriate.) Vertical scales are missing.

(2) $$P(500 \le X \le 650) = P\left(\frac{500-500}{100}\le \frac{X-\mu}{\sigma} \le \frac{650 - 500}{100}\right)\\ =P(0 \le Z \le 1.5) = 0.4332.$$

On original scale of $\mathsf{Norm}(\mu = 500, \sigma = 100),$ the desired area is between the vertical dashed lines.

On standard normal scale:

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Addenda:(1) The area under every density function must be $1.$ Except for a linear transformation, all normal density functions are the same shape. Under any normal density curve, the area between $\mu \pm \sigma$ is about 68% of the entire area. (As the horizontal scale, indicated by $\sigma,$ increases, the height of the curve decreases.)

• For the standard normal distribution the interval $\mu \pm \sigma$ has length $2$ and the distribution reaches a maximum height of about 0.4.

• In $\mathsf{Norm}(\mu = 500, \sigma = 100)$ the interval $\mu \pm \sigma$ amounts to $(400, 600$ of length 200 (100 times as long as for standard normal), and the maximum height is about 0.004 ($\frac{1}{100}$th as tall).

(2) Finding the probability that a standard normal random variable $Z$ lies in the the interval $[0, 1]:$ From printed tables, we know that $P(0 \le Z \le 1) = 0.3413.$ Here is a rough idea how that probability can be computed.

In the figure below the interval $[0, 1]$ is divided into ten sub-intervals, each of width $0.1$ by points at $0.0, 0.1, 0.2, \dots, 0.9, 1.0.$ The respective centers of these intervals are at points $0.05, 0.15, \dots, 0.95,$ respectively. A rectangle is erected above each interval; its height is such that the center of the top crosses the standard normal density curve.

For example, the first rectangle has height $\phi(0.05) = \frac{1}{\sqrt{2\pi}}e^{-.5(0.05^2} = 0.3984439$ and width $0.1.$ So the area of the first rectangle is about $0.03984.$ The second rectangle has width $0.1$ and is a little shorter, and so on.

One can show that the total area of all ten of these tall, thin rectangles is $0.3414$ to four decimal places. Of course, the combined area of the rectangles is not exactly the same as $P(0 \le Z \le 1) = 0.3413$ because the straight-line tops of the rectangles do not exactly match the standard normal density curve. But the approximation is very good. [Computation in R.]

w = .1;  z = seq(.05, .95, by=.1);  h = dnorm(z)
w*h
[1] 0.03984439 0.03944793 0.03866681 0.03752403 0.03605270
[6] 0.03429439 0.03229724 0.03011374 0.02779849 0.02540591
sum(w*h)
[1] 0.3414456

If we used 100 rectangles each with width 0.01, then their sum would be 0.3413, matching the probability given in printed tables, correct to more than four decimal places.

There is no way to use individual points to find $P(0 \le Z \le 1)$ because there are infinitely many points. But we can find this probability to any desired degree of accuracy by using sufficiently many sufficiently thin rectangles. Nowadays, we have somewhat better ways of computing standard normal probabilities, but the original printed tables were made using almost exactly the same rectangle method described above. (Imagine rooms full of people slaving away at mechanical calculators for many hours.)