IntroductionWhen we have a random variable with a normal distribution, our main interest is to find the probability that it takes a value within a given interval. In this lesson, we will see how to transform any normal distribution into the standard normal distribution and use its table to calculate these probabilities.Standard normal distributionLet's see how to proceed with a concrete example. Let X be the random variable representing the diameters of the screws produced by a certain machine. Suppose this variable has a normal distribution with mean 2 cm and standard deviation 0.04 cm.We may want to know the probability that a screw has a diameter between 2 and 2.05 cm. This probability, written as\[ P(2 < X < 2.05) \]corresponds to the shaded area in the figure below:Normal example 01Calculating this area directly requires integrals that have no simple solution. To get around this, we transform the variable so that the probability we want equals a probability that has already been tabulated.Every normal distribution can be converted into the standard normal distribution (also called the reduced normal distribution), that is, the normal distribution with mean 0 and standard deviation 1. The probabilities of this distribution have already been calculated and are available in tables; so all we need to do is transform the normal distribution of the problem into one equivalent to the standard normal.In other words:“To calculate the probabilities of a normal distribution, first transform it into a distribution with mean ZERO and standard deviation 1”- Professor LeonTo make the mean of a variable equal to zero, we subtract the mean from each value. To make the standard deviation equal to 1, we divide the result by the standard deviation.This gives us the formula that turns any value of a normal variable into the corresponding value of the standard normal (standardization):\[ z = \frac{x - \mu}{\sigma} \]\[ z\text{ : corresponding value in the standard normal distribution} \]\[ \mu\text{ : mean of the distribution} \]\[ \sigma\text{ : standard deviation of the distribution} \]Since the mean is transformed into zero, we can write:\[ P(\mu < X < x) = P(0 < Z < z) \]In the problem described above:\[ P(2 < X < 2.05) = P(0 < Z < z) \]What is the value of z in this case? Using the formula:\[ z = \frac{2.05 - 2}{0.04} = 1.25 \]The advantage of knowing z is that it belongs to the standard normal distribution, whose probabilities are in the table below. The body of the table gives the area between 0 and z, that is, P(0 < Z < z):Normal tableWe look up row 1,2 and column 5 in the table (the table uses the Brazilian decimal comma, so 1,2 means 1.2; the column is the second decimal place of z) and find 39435, that is, 0.39435. Therefore, the probability that a screw has a diameter between 2 and 2.05 cm is approximately 39.44%.\[ P(2 < X < 2.05) = P(0 < Z < 1.25) = 0.39435 \approx 39.44\% \]Using the symmetry of the curveThe table only gives areas between 0 and a positive z. All the others follow from two facts: the curve is symmetric around 0, and each half has an area of 0.5, that is, P(Z > 0) = 0.5 = P(Z < 0). Let's calculate a few more probabilities:a) P(−1.73 < Z < 0) = P(0 < Z < 1.73) = 0.4582, by the symmetry of the curve.b) P(Z > 1.73) = 0.5 − P(0 < Z < 1.73) = 0.5 − 0.4582 = 0.0418.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 random variable N(μ, σ²), with μ = 3 and σ² = 16 (so σ = 4), and that we want to calculate P(2 < X < 5). Standardizing both limits:\[ P(2 < X < 5) = P\left(\frac{2 - \mu}{\sigma} < \frac{X - \mu}{\sigma} < \frac{5 - \mu}{\sigma}\right) \]\[ = P\left(\frac{2 - 3}{4} < Z < \frac{5 - 3}{4}\right) = P\left(-\frac{1}{4} < Z < \frac{2}{4}\right) \]\[ P(-0.25 < Z < 0.5) = P(-0.25 < Z < 0) + P(0 < Z < 0.5) = 0.0987 + 0.1915 = 0.2902 \]If Z follows the standard normal and the table gives P(0 < Z < 1.25) = 0.3944, what is P(Z < −1.25)?Answer: 0.1056 — by symmetry, P(Z < −1.25) = P(Z > 1.25) = 0.5 − 0.3944 = 0.1056. The 0.3944 is the area between −1.25 and 0, not the tail; 0.8944 is P(Z < 1.25).A variable X is normal with mean 50 and standard deviation 8. Which value of z corresponds to x = 38?Answer: −1.5 — z = (38 − 50)/8 = −12/8 = −1.5. The negative sign means that 38 is below the mean; −12 is the difference without dividing by the standard deviation.ExampleDeposits made at Banco da Ribeira in January are normally distributed, with a mean of R\$ 10,000.00 (Brazilian reais) and a standard deviation of R\$ 1,500.00. A deposit is selected at random from all of that month's deposits. Find the probability that the deposit is:a) R\$ 10,000.00 or less;b) at least R\$ 10,000.00;c) an amount between R\$ 12,000.00 and R\$ 15,000.00;d) greater than R\$ 20,000.00.Solution:a)\[ P(X \leq 10{,}000) = P\left(Z \leq \frac{10{,}000 - 10{,}000}{1{,}500}\right) = P(Z \leq 0) = 0.5 = 50\% \]b)\[ P(X \geq 10{,}000) = P\left(Z \geq \frac{10{,}000 - 10{,}000}{1{,}500}\right) = P(Z \geq 0) = 0.5 = 50\% \]c)\[ P(12{,}000 < X < 15{,}000) = P\left(\frac{12{,}000 - 10{,}000}{1{,}500} < Z < \frac{15{,}000 - 10{,}000}{1{,}500}\right) = P(1.33 < Z < 3.33) \]\[ = 0.49957 - 0.40824 = 0.09133 \approx 9.1\% \](With unrounded z, 1.333 and 3.333, the exact value is 9.08%; the small difference comes from rounding z for the table.)d)\[ P(X > 20{,}000) = P\left(Z > \frac{20{,}000 - 10{,}000}{1{,}500}\right) = P(Z > 6.67) \approx 0.000000000013 \approx 0\% \]The table ends at z = 4.5, where the area between 0 and z is already 0.5000: a deposit above R\$ 20,000.00 is practically impossible under this model.Test scores follow a normal distribution with mean 100 and standard deviation 10. Given that P(0 < Z < 1.5) = 0.4332, what is the probability that a student scores more than 115?Answer: 6.68% — z = (115 − 100)/10 = 1.5 and P(Z > 1.5) = 0.5 − 0.4332 = 0.0668. The 43.32% is the area between the mean and 115, not above 115; 93.32% is P(Z < 1.5).Exercises1) Given that Z is a variable with a standard normal distribution, calculate:a) P(0 < Z < 1.44)b) P(−0.85 < Z < 0)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 school test has a normal distribution with mean 100 and standard deviation 10. Find the probability that a person taking the test scores:a) more than 120;b) more than 80;c) between 85 and 115;d) more than 100.6) The weights of 600 students are normally distributed with mean 65.3 kg and standard deviation 5.5 kg. Find 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 lifetime of a certain electronic component has a mean of 850 days and a standard deviation of 40 days. Given that the lifetime 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 a 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 that Z is a variable with a standard normal distribution, calculate:a) P(−1.25 < Z < 0)b) P(−0.5 < Z < 1.48)10) Given that Z is a variable with a standard normal distribution, calculate:a) P(0.8 < Z < 1.23)b) P(Z > 0.6)Answer key1) 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%, about 380 students; b. 64.80%, about 389 students; c. 68.79%, about 413 students;7) a. 99.98%; b. 89.44%; c. 0.62%;8) 29.02%9) a. 39.44%; b. 62.20%10) a. 10.25%; b. 27.43%
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