Section 5.1: Probability and area

Section 5.1: Probability and area

Review Normal Distribution s z = x - m s Standard Normal Distribution s=1 m x m=0 z The area that falls in the interval under the nonstandard normal curve is the same as that under the standard normal curve within the corresponding z-boundaries. The area under the standard normal curve to the left of a z-score gives the probability that z is less than that z- score.

Example: Finding Probabilities for Normal Distributions A survey indicates that for each trip to the supermarket, a shopper spends an average of 45 minutes with a standard deviation of 12 minutes in the store. The length of time spent in the store is normally distributed and is represented by the variable x. A shopper enters the store. Find the probability that the shopper will be in the store more than 39 minutes.

Solution: Finding Probabilities for Normal Distributions Normal Distribution μ = 45 σ = 12 P(x > 39) x m 39 45 z = = = 05. s 12 Standard Normal Distribution μ = 0 σ = 1 P(z > 0.5) 39 45 x 0.50 0 z P(x > 39) = P(z > 0.5) =?

Example: Finding Probabilities for Normal Distributions If 200 shoppers enter the store, how many shoppers would you expect to be in the store more than 39 minutes? Solution: Recall P(x > 39) = 0.6915. 200(0.6915) =138.3 (or about 138) shoppers

Use technology to find Normal Probabilities TI-83/84 Plus: normalcdf(lower bound, upper bound, mean, standard deviation) If you want all the numbers greater than a certain value, your upper boundary will be positive infinity. Use a large positive number like 10000. If you want all the numbers less than a certain value, your lower boundary will be negative infinity. Use a large negative number like -10000. Do not ever use normalpdf! No need to computer z-scores using TI-83/84.

Example: Find Normal Probabilities IQ scores are normally distributed with a mean of 100 and a standard deviation of 15. Find the probability a randomly selected person has an IQ score between 100 and 120. Choose the closest answer. A. 0.9082 B. 0.0912 Use the table and the technology. Discuss with your classmates if you don t know how to get the answer. C. 0.6293 D. 0.4082

Finding values Given a Probability In Section 5.2, we were given a normally distributed random variable x and we were asked to find a probability. In this section, we will be given a probability and we will be asked to find the value of the random variable x. 5.2 x z probability 5.3

Example: Finding a z-score Given an Area Find the z-score that corresponds to a cumulative area of 0.3632. 0.3632 z 0 z

Solution: Finding a z-score Given an Area Locate 0.3632 in the body of the Standard Normal Table. The z-score is 0.35. The values at the beginning of the corresponding row and at the top of the column give the z-score.

Example: Finding a z-score Given an Area Find the z-score that has 10.75% of the distribution s area to its right. 1 0.1075 = 0.8925 0.1075 0 z z Because the area to the right is 0.1075, the cumulative area is 0.8925.

Solution: Finding a z-score Given an Area Locate 0.8925 in the body of the Standard Normal Table. The z-score is 1.24. The values at the beginning of the corresponding row and at the top of the column give the z-score.

Example: Finding a z-score Given an Area Find the z-score that has 96.16% of the distribution s area to its right. Find the z-score for which 95% of the distribution s area lies between z and z.

Review: Percentile P 83 : the 83 rd percentile, 83% of the data values below and 17% above P 21 : the 21 st percentile, 21% of the data values below and 79% above

Example: Finding a z-score Given a Percentile Find the z-score that corresponds to P 5. 0.05 z 0 The z-score that corresponds to P 5 is the same z- score that corresponds to an area of 0.05. z The areas closest to 0.05 in the table are 0.0495 (z = 1.65) and 0.0505 (z = 1.64). Because 0.05 is halfway between the two areas in the table, use the z-score that is halfway between 1.64 and 1.65. The z-score is -1.645.

Example: Finding a z-score Given a Percentile P 50 P 90

Transforming a z-score to an x-score To transform a standard z-score to a data value x in a given population, use the formula z = x μ σ x = μ + zσ

Example: Finding an x-value A veterinarian records the weights of cats treated at a clinic. The weights are normally distributed, with a mean of 9 pounds and a standard deviation of 2 pounds. Find the weights x corresponding to z-scores of 1.96, -0.44, and 0. z = 1.96: x = 9 + 1.96(2) = 12.92 pounds z = -0.44: x = 9 + (-0.44)(2) = 8.12 pounds z = 0: x = 9 + 0(2) = 9 pounds Notice 12.92 pounds is above the mean, 8.12 pounds is below the mean, and 9 pounds is equal to the mean.

Example: Finding a Specific Data Value Scores for the California Peace Officer Standards and Training test are normally distributed, with a mean of 50 and a standard deviation of 10. An agency will only hire applicants with scores in the top 10%. What is the lowest score you can earn and still be eligible to be hired by the agency? An exam score in the top 10% is any score above the 90 th percentile. Find the z-score that corresponds to a cumulative area of 0.9.

Solution: Finding a Specific Data Value From the Standard Normal Table, the area closest to 0.9 is 0.8997. So the z-score that corresponds to an area of 0.9 is z = 1.28. Using the equation x = μ + zσ x = 50 + 1.28(10) 62.8 The lowest score you can earn and still be eligible to be hired by the agency is about 63.

Example: Finding a Specific Data Value A researcher tests the braking distances of several cars. The braking distance from 60 miles per hour to a complete stop on dry pavement is measured in feet. The braking distances of a sample of cars are normally distributed, with a mean of 129 feet and a standard deviation of 5.18 feet. What is the longest braking distance one of these cars could have and still be in the bottom 1%?

Finding a Specific Data Value Using Graphing Calculator Inverse normal problem: 2 nd -> DISTR-> invnorm(area, μ, σ) If you use technology to find an x-value, it is not necessary first to find a z-score. For cumulative area of a specified z-score, μ = 0 and σ = 1. Practice with the previous example.

Example: Finding a Specific Data Value According to the United States Geological Survey, the mean magnitude of worldwide earthquakes in a recent year was about 3.98. The magnitude of worldwide earthquakes can be approximated by a normal distribution. Assume the standard deviation is 0.90. Between what two values does the middle 90% of the data lie?

Summary of Section 5.3 Calculate z from probability. Transform z to x.

Project 5 case study on P260 Each group has 4-5 people. Five groups in total. Work on all exercises in 30 minutes. One student will be assigned to present a solution of one question from each group. Each group submits one answer sheet.