Section 12.7 Applications of the Normal Distribution Math in Our World

Learning Objectives Use the normal distribution to find percentages. Use the normal distribution to find probabilities. Use the normal distribution to find percentile ranks.

EXAMPLE 1 Solving a Problem Using the Normal Distribution If the weights of Oreos in a package are normally distributed with mean 518 grams and standard deviation 4 grams, find the percentage of packages that will weigh less than 510 grams.

EXAMPLE 1 Solving a Problem Using the Normal Distribution Step 1 Draw the picture and represent the area. SOLUTION Step 2 Find the z score for data value 510. This tells us that 510 grams is 2 standard deviations below the mean.

EXAMPLE 1 Solving a Problem Using the Normal Distribution Step 3 Find the shaded area using Table 12-3. SOLUTION The area between z = 0 and z = – 2 is 0.477 (using z = + 2 from the table). Area = 0.500 – 0.477 = 0.023. The shaded area is the area of the left half (0.5) minus the area between z = 0 and z = – 2. Step 4 Interpret the area. In this case, an area of 0.023 tells us that only 2.3% of packages will have weights less than 510 grams.

Probability vs. Area The area under a normal distribution between two data values is the probability that a randomly selected data value is between those two values.

EXAMPLE 2 Using Area under a Normal Distribution to Find Probabilities Based on data in the 2009 Statistical Abstract of the United States, the average American generates 1,679 pounds of garbage per year. Let’s assume that the number of pounds generated per person is approximately normally distributed with standard deviation 200 pounds. Find the probability that a randomly selected person generates (a) Between 1,300 and 2,000 pounds of garbage per year. (b) More than 2,000 pounds of garbage per year.

EXAMPLE 2 Using Area under a Normal Distribution to Find Probabilities (a) Step 1 Draw the picture and represent the area. SOLUTION Step 2 Find the z score for the two given weights.

EXAMPLE 2 Using Area under a Normal Distribution to Find Probabilities SOLUTION Step 3 Find the shaded area. Using Table 12-3 the area between z = 0 and z = 1.6 is 0.445 and the area between z = 0 and z = – 1.9 is 0.471. Shaded Area = 0.445 + 0.471 = 0.916. Step 4 Interpret the area. This tells us that if a person is selected at random, the probability that he or she generates between 1,300 and 2,000 pounds of garbage per year is 0.916, or 91.6%.

EXAMPLE 2 Using Area under a Normal Distribution to Find Probabilities SOLUTION (b) Most of the work was already done in part a. We can adapt the graph a bit. The shaded area is the area between z = 0 and z = 1.6 subtracted from the area of the entire right half (0.5). 0.500 – 0.445 = 0.055. The probability that a randomly selected person generates more than 2,000 pounds of garbage per year is 0.055, or 5.5%.

EXAMPLE 3 Finding Number in a Sample and Percentile Rank The American Automobile Association reports that the average time it takes to respond to an emergency call is 25 minutes. Assume the response time is approximately normally distributed and the standard deviation is 4.5 minutes. (a) If 80 calls are randomly selected, approximately how many will have response times less than 15 minutes? (b) In what percentile is a response time of 30 minutes?

EXAMPLE 3 Finding Number in a Sample and Percentile Rank (a) Step 1 Draw the picture and represent the area. SOLUTION Step 2 Find the z for 15. Step 3 Find the shaded area. Using Table 12-3 the area between z = 0 and z = – 2.2 is 0.486. Area = 0.500 – 0.486 = 0.014.

EXAMPLE 3 Finding Number in a Sample and Percentile Rank SOLUTION 1.4% of 80 calls = (0.014)(80) = 1.12 Approximately one call in 80 will have response time less than 15 minutes. Step 4 Interpret the area. An area of 0.014 tells us that 1.4% of calls will have a response time of less than 15 minutes. Multiply the percentage by the number of calls in our sample:

EXAMPLE 3 Finding Number in a Sample and Percentile Rank (b) Step 1 Draw the picture and represent the area. We are interested in the percentage of times that are less than 30 minutes. SOLUTION Step 2 Find the z value for 30 minute response time. Step 3 Find the shaded area. Using Table 12-3 the area between z = 0 and z = 1.1 is 0.364. Area = 0.500 + 0.364 = 0.864.

EXAMPLE 3 Finding Number in a Sample and Percentile Rank SOLUTION Step 4 Interpret the area. An area of 0.864 means that 86.4% of calls will get a response time of less than 30 minutes, so the time of 30 minutes is in the 86th percentile.