Calculating values from probabilitiesAs we saw in the previous lesson, to find the probability of something that follows a normal distribution, we simply transform the data into the standard normal distribution, whose probabilities are in the table below (the body of the table gives the area between 0 and z; it uses the Brazilian decimal comma, so 1,2 means 1.2):Normal distributionThe transformation is given by the formula:\[ z = \frac{x - \mu}{\sigma} \]Often, however, the problem is the reverse: we know the probability and want the value of X that produces it (for example, the time that marks off the slowest 10%). To calculate values from probabilities, we follow these steps:Remember that percentages and probabilities are areas under the curve, not z values.Read the table backwards: look for the area in the body of the table and read the corresponding z from the row and column.Choose the correct side of the graph (in a time problem, for example, the longer times, of the slowest, are on the right, with positive z).Apply the standardization formula solved for x.Solving the standardization formula for x is simple:\[ z = \frac{x - \mu}{\sigma} \]\[ z \cdot \sigma = x - \mu \]\[ x = \mu + z \cdot \sigma \]In other words, knowing the mean and the standard deviation, we can calculate the value of x corresponding to any probability, because the probability gives us z.Since the table gives the area between 0 and z, for a 10% tail we look for the area 0.5 − 0.10 = 0.40, which corresponds to z ≈ 1.28; for a 5% tail we look for 0.45, which lies between 1.64 and 1.65, that is, z ≈ 1.645.ExamplesIn a factory, workers take an average of 75 minutes to complete a task, with a standard deviation of 6 minutes. How much time separates the fastest 10% from the slowest 10%?The fastest 10% (shortest times) are on the left side of the graph: the z that leaves 10% of the area to its left is −1.28.The slowest 10% (longest times) are on the right side of the graph: the z that leaves 10% of the area to its right is 1.28.So the time that marks off the fastest 10% is:\[ x = \mu + z \cdot \sigma \]\[ x = 75 - 1.28 \cdot 6 = 67.32 \text{ minutes} \]And the time that marks off the slowest 10%:\[ x = \mu + z \cdot \sigma \]\[ x = 75 + 1.28 \cdot 6 = 82.68 \text{ minutes} \]The difference between the two times is:\[ \text{difference} = 82.68 - 67.32 = 15.36 \text{ minutes} \]In the same factory (mean of 75 minutes and standard deviation of 6 minutes), what is the minimum time taken by the slowest 5%?The slowest 5% are on the right side of the graph: the z that leaves 5% of the area to its right is 1.645.So the minimum time of the slowest 5% is:\[ x = \mu + z \cdot \sigma \]\[ x = 75 + 1.645 \cdot 6 = 84.87 \approx 84.9 \text{ minutes} \]In a factory, a task takes 75 minutes on average, with a standard deviation of 6 minutes. What is the maximum time taken by the fastest 5%?Answer: 65.1 minutes — the fastest have the shortest times, on the left, with z = −1.645: x = 75 − 1.645 × 6 ≈ 65.1. The 84.9 is the trap of using the right side, which belongs to the slowest; 67.3 uses the z for 10%.Service time follows a normal distribution with mean 8 minutes and standard deviation 2 minutes. Which time is exceeded by only 2% of calls? (Use z ≈ 2.05.)Answer: 12.1 minutes — the 2% tail is on the right, so x = 8 + 2.05 × 2 = 12.1. The 3.9 uses the wrong side; 10.05 adds z to the mean without multiplying by the standard deviation.Symmetry in the distributionWe have already seen that the normal distribution is symmetric around the mean. Many exercises exploit this feature by splitting the population into two halves:richer or poorer;faster or slower;taller or shorter.A word of warning: when the exercise splits the population into two halves, each half represents 50% of the distribution. So "10% of the fastest" means 10% of the faster half, that is, 5% of the whole distribution.ExampleIn a factory, workers take an average of 75 minutes to complete a task, with a standard deviation of 6 minutes. What time separates the fastest 10% among the fast from the slowest 10% among the slow?In this case, 10% among the fastest (who make up 50% of the distribution) corresponds to 5% of the distribution, on the left side. Likewise, 10% among the slowest corresponds to 5% of the distribution, on the right side.The z that leaves 5% to the left, for the fastest, is −1.645:\[ x = \mu + z \cdot \sigma \]\[ x = 75 - 1.645 \cdot 6 = 65.13 \text{ minutes} \]The z that leaves 5% to the right, for the slowest, is 1.645:\[ x = \mu + z \cdot \sigma \]\[ x = 75 + 1.645 \cdot 6 = 84.87 \text{ minutes} \]The difference:\[ \text{difference} = 84.87 - 65.13 = 19.74 \text{ minutes} \]The heights of a class follow a normal distribution with mean 170 cm and standard deviation 10 cm. From what height on are the tallest 20% among those above the mean?Answer: 182.8 cm — 20% of the upper half equals 10% of the whole distribution, right tail, with z ≈ 1.28: x = 170 + 1.28 × 10 = 182.8. The 178.4 is the trap of using 20% of the whole distribution (z ≈ 0.84).Exercises1) Suppose that the time needed to serve customers at a call center follows a normal distribution with a mean of 8 minutes and a standard deviation of 2 minutes. 25% of phone calls require at least how much service time?2) Suppose that the time needed to serve customers at a call center follows a normal distribution with a mean of 8 minutes and a standard deviation of 2 minutes. 75% of phone calls require at least how much service time?3) Suppose that the time needed to serve customers at a call center follows a normal distribution with a mean of 8 minutes and a standard deviation of 2 minutes. What is the probability that a call lasts less than 5 minutes? More than 9.5 minutes? Between 7 and 10 minutes?4) The weights of rabbits raised on a farm can be well represented by a normal distribution with mean 5 kg and standard deviation 0.9 kg. A slaughterhouse will buy 5,000 rabbits and intends to classify them by weight as follows: the lightest 15% as small, the next 50% as medium, the next 20% as large and the heaviest 15% as extra. What are the weight limits for each class?5) An automatic soft-drink filling machine is set so that the mean volume of liquid in each bottle is 1000 cm³, with a standard deviation of 10 cm³. Assume the volume follows a normal distribution. What percentage of bottles contain less than 990 cm³ of liquid? What percentage of bottles have a volume that does not deviate from the mean by more than two standard deviations?6) A company produces 2 types of television sets, type A (standard) and type B (luxury), and guarantees a refund of the price paid if any set has a serious defect within six months. The time until a serious defect occurs is normally distributed: for type A, with a mean of 10 months and a standard deviation of 2 months; for type B, with a mean of 11 months and a standard deviation of 3 months. Type A and B sets are produced with a profit of R\$ 1200 and R\$ 2100 (Brazilian reais), respectively, and, if there is a refund, with a loss of R\$ 2500 and R\$ 7000, respectively. Calculate the probabilities of a refund for type A and type B sets and their average profits.7) A car manufacturer knows that the engines it makes have a normally distributed lifetime with mean 150,000 km and standard deviation 5,000 km. What is the probability that a car chosen at random has an engine that lasts less than 170,000 km? Between 140,000 km and 165,000 km? If the manufacturer replaces any engine whose lifetime is shorter than the warranty, what should the warranty be so that the percentage of replaced engines is below 0.2%?8) The mean inner diameter of a sample of 200 washers produced by a certain machine is 0.502 cm and the standard deviation is 0.005 cm. The purpose for which these washers are made allows a diameter tolerance of 0.496 to 0.508 cm; outside this range, the washers are considered defective. Find the percentage of defective washers produced by the machine, assuming the diameters are normally distributed.9) Suppose that measurements of the electric current in a piece of wire follow a normal distribution, with a mean of 10 milliamperes and a variance of 4 milliamperes² (standard deviation of 2 milliamperes). What is the probability that a measurement exceeds 13 milliamperes? What is the probability that the current measurement is between 9 and 11 milliamperes?10) With the same current measurements as in the previous exercise (mean of 10 milliamperes and variance of 4 milliamperes²), find the value below which the current measurement lies with probability 98%.11) The diameter of the main shaft of a hard disk follows a normal distribution with mean 25.08 cm and standard deviation 0.05 cm. If the specifications for this shaft are 25.00 ± 0.15 cm, find the percentage of units produced that meet the specifications.12) The concentration of a pollutant in water released by a factory has distribution N(8; 1.5), that is, mean 8 ppm and variance 1.5. What is the chance that, on a given day, the pollutant concentration exceeds the regulatory limit of 10 ppm?Answer key1) 9.35 minutes2) 6.65 minutes3) 6.68%; 22.66%; 53.28%;4) small < 4.07 kg < medium < 5.35 kg < large < 5.93 kg < extra5) 15.87%; 95.44%;6) Type A: 2.28% and average profit of R\$ 1115.64; Type B: 4.75% and average profit of R\$ 1667.757) 99.997%; 97.59%; a warranty of at most about 135,600 km.8) 23%;9) 6.68%; 38.29%;10) 14.1 milliamperes;11) 91.92%12) 5.2%
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