Reduced normal distribution: how to transform

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Let's now teach how statisticians memorized an entire distribution to facilitate probability calculations

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Introduction

When we have a random variable with a normal distribution, our main interest is to obtain the probability that this random variable takes on a value within a given interval.

Standard normal distribution

Let us see how to proceed by means of a concrete example. Let X be the random variable that represents the diameters of the screws produced by a certain machine. Suppose that this variable has a normal distribution with mean 2 cm and standard deviation 0.04 cm.

There may be interest in knowing the probability that a screw has a diameter with a value between 2 and 2.05 cm. It is easy to note that this probability, indicated by:

\[ P(2 < X < 2.05) \]

Belongs to the shaded area in the figure below:

Normal example 01
Normal example 01

The direct calculation of this probability requires more advanced mathematical knowledge. To avoid this need, we use the transformation to equate this probability to some already known ones.

Every normal distribution has characteristics similar to a standard normal distribution, that is, one that has a normal distribution with mean 0 and standard deviation 1. The probabilities of a normal distribution with mean 0 and standard deviation 1 are already known; therefore, transformations are made from any normal distribution so that it is equivalent to the standard normal distribution.

Therefore:

“To calculate the probabilities of a normal distribution, it must first be transformed into a distribution with mean ZERO and standard deviation 1”

- Professor Leon

To transform the mean of a variable into zero, each datum of this variable must be subtracted by its mean. To transform the standard deviation of a variable into 1, each datum of this variable must be divided by the standard deviation.

Thus, we have the following formula to transform any normal data set into a standard normal distribution data set:

\[ z = \frac{x - \bar{x}}{\sigma} \]

\[ \text{z : value of the variable in the standard normal distribution} \]

\[ \bar{x}\text{ : mean of the data set} \]

\[ \sigma\text{ : standard deviation of the data set} \]

We can then write:

\[ P(\bar{x} < X < x) = P(0 < Z < z) \]

In the case of the problem described earlier:

\[ P(2 < X < 2.05) = P(0 < Z < z) \]

What would z be in this case? By the formula:

\[ z = \frac{2.05 - 2}{0.04} = 1.25 \]

The advantage of knowing z, as was said earlier, is that it is part of the standard normal distribution, which is already studied using the table below:

Normal table
Normal table

In this case, the value of z must be looked up in this table, where it is possible to verify that the z of 1.25 has value 39435. This means that a screw having a diameter between 2 and 2.05 has a probability of 39.435%.

\[ P(2 < X < 2.05) = P(0 < Z < z) = 39.43\% \]

Let us calculate a few more probabilities:

(a) P(–1.73 < Z < 0) = P(0 < Z < 1.73) = 0.4582, due to the symmetry of the curve.

(b) P(Z > 1.73) = 0.5 – P(0 > Z > 1.73) = 0.5 – 0.4582 = 0.0418, since P(Z > 0) = 0.5 = P(Z < 0).

(c) P(Z < –1.73) = P(Z > 1.73) = 0.0418.

(d) P(0.47 < Z < 1.73) = P(0 < Z < 1.73) – P(0 < Z < 0.47) = 0.4582 – 0.1808 = 0.2774.

Now suppose that X is a r.v. N( μ , σ² ), with μ = 3 and σ 2 = 16, and that we want to calculate P(2 < X < 5). We have:

\[ P(2 < X < 5) = P(\frac{2 - \bar{x}}{\sigma} < \frac{X - \bar{x}}{\sigma} < \frac{5 - \bar{x}}{\sigma}) = 39.43\% \]

\[ P(\frac{2 - 3}{4} < Z < \frac{5 - 3}{4}) = P(\frac{-1}{4} < Z < \frac{2}{4}) \]

\[ P(-0.25 < Z < 0.5) = P(-0.25 < Z < 0) + P(0 < Z < 0.5) = 0.0987 + 0.1915 = 0.2902 \]

Examples

The deposits made at Banco da Ribeira during the month of January are normally distributed, with a mean of \$10,000.00 and a standard deviation of \$1,500.00. A deposit is selected at random from among all those referring to the month in question. Find the probability that the deposit is:

a. R$10,000.00 or less;

b. at least $10,000.00;

c. an amount between $12,000.00 and $15,000.00;

d. greater than $20,000.00.

Solution:

a.

\[ P(X < 10,000) = P(Z < \frac{10,000 - 10,000}{1,500}) = P(Z < 0) = 0.5 = 50\% \]

b.

\[ P(X >= 10,000) = P(Z >= \frac{10,000 - 10,000}{1,500}) = P(Z >= 0) = 0.5 = 50\% \]

c.

\[ P(12,000 < X < 15,000) = P(\frac{12,000 - 10,000}{1,500} < Z < \frac{15,000 - 10,000}{1,500}) = P(1.333 < Z < 3.333) = 0.09133 = 9.133\% \]

d.

\[ P(X > 20,000) = P(Z > \frac{20,000 - 10,000}{1,500}) = P(Z > 6.667) = 0,.....1 = 0\% \]

Exercises

1) Given that Z is a variable with a standard normal distribution, calculate:

a. P(O < Z < 1 .44)

b. P(- 0.85 < Z < O)

2) Given that Z is a variable with a standard normal distribution, calculate:

c. P(-1.48 < Z < 2.05)

d. P(0.72 < Z < 1 .89)

3) Given that Z is a variable with a standard normal distribution, calculate:

e. P(Z > - 2.03)

f. P(Z > 1 .08)

4) Given that Z is a variable with a standard normal distribution, calculate:

g. P(Z < - 0.66)

h. P(Z < 0.60)

5) A standardized schooling test has a normal distribution with mean 100 and standard deviation 1O. Determine the probability that an individual taking the test has a score:

a. greater than 120;

b. greater than 80;

c. between 85 and 115;

d. greater than 100.

6) The weights of 600 students are normally distributed with mean 65.3 kg and standard deviation 5.5 kg. Determine the number of students who weigh:

a. between 60 and 70 kg;

b. more than 63.2 kg;

c. less than 68 kg.

7) The lifespan of a certain electronic component has a mean of 850 days and a standard deviation of 40 days. Knowing that the lifespan is normally distributed, calculate the probability that this component lasts:

a. between 700 and 1,000 days;

b. more than 800 days;

c. less than 750 days.

8) The weekly wages of industrial workers are normally distributed around the mean of R$ 500, with a standard deviation of R$ 40. Calculate the probability that a worker has a weekly wage between R$ 490 and R$ 520.

9) Given Z as a variable with a standard normal distribution, calculate:

a. P(-1.25 < Z < O)

b. P(-0.5 < Z < 1.48)

10) Given Z as a variable with a standard normal distribution, calculate:

a. P(0.8 < Z < 1.23)

b. P(Z > 0.6)

Answer Key

1) a. 42.51%; b. 30.23%;

2) c. 91.04%; d. 20.64%;

3) e. 97.88%; f. 14.01%;

4) g. 25.46%; h. 72.58%;

5) a. 2.28%; b. 97.72%; c. 86.64%; d. 50%;

6) a. 63.38%; b. 64.80%; c. 68.79%;

7) a. 99.98%; b. 89.44%; c. 0.63%;

8) 29.02%

9) a. 39.44%; b. 62.20%

10) a. 10.26%; b. 27.42%

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