norm_areas_key

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PHW142 Practice Problems Normal Curve Areas KEY Baldi & Moore (B&M) 4th edition Chapter references: Chapter 11 Tools: R pnorm function, OOMPH Stat R pnorm function syntax: pnorm uses the default values mean = 0, sd = 1 but users can specify other values. The default for areas is lower.tail = TRUE This assumes that the area the user is asking about is the area below (or less than) the specified X or z value. For “upper tail” areas, use lower.tail = FALSE . pnorm either takes a z value as its first argument and defaults to a mean = 0 and sd = 1 or an X value as its first argument, with the mean and sd arguments specified. 1 Baldi and Moore exercise 11.25 (3rd edition), exercise 11.27 (4th edition) Use OOMPH Stat or R pnorm function to find the proportion of observations from a standard Normal distribution that fall in each of the following regions. 1a z -2.25 1a answer pnorm ( - 2.25 , lower.tail = TRUE ) ## [1] 0.01222447 1
1b z -2.25 1b answer pnorm ( - 2.25 , lower.tail = FALSE ) ## [1] 0.9877755 1 - 0.0122 = 0.9878 1c z > 1.77 2
1c answer pnorm ( 1.77 , lower.tail = TRUE ) ## [1] 0.9616364 pnorm ( 1.77 , lower.tail = FALSE ) ## [1] 0.03836357 1 – 0.9616 = 0.0384 1d -2.25 < z < 1.77 1d answer This is P( z < 1.77 ) - P( z < -2.25 ) We have these values from part (c) and part (a). 0.9616 – 0.0122 = 0.9494 pnorm ( 1.77 , mean = 0 , sd = 1 , lower.tail = TRUE ) - pnorm ( - 2.25 , mean = 0 , sd = 1 , lower.tail = TRUE ) ## [1] 0.949412 3
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2: newborn respirations Baldi and Moore exercises 11.16 - 11.17 (3rd edition), exercises 11.18 - 11.19 (4th edition) The respiratory rate per minute in newborns varies according to a distribution that is approximately normal with mean of 50 and standard deviation of 5. 2a Find the probability that a randomly chosen newborn has a respiratory rate of 55 per minute or more. 2a answer x 55 corresponds to z 55 50 5 which is z 1 pnorm ( 55 , mean = 50 , sd = 5 , lower.tail= FALSE ) ## [1] 0.1586553 2b Find the probability that a randomly chosen newborn has a respiratory rate per minute between 40 and 55. 2b answer pnorm ( 55 , mean = 50 , sd = 5 , lower.tail= TRUE ) - pnorm ( 40 , mean = 50 , sd = 5 , lower.tail= TRUE ) ## [1] 0.8185946 5
Step 1: Calculate the z scores for 40 and 55: x = 40 , z = 40 50 4 = 2 . 0 x = 55 , z = 55 50 4 = 1 . 0 Step 2: For P(z < 1), choose Left-Tailed, input z-score of 1. OOMPH Stat returns auc: 0.8413 Step 3: For P(z < -2.00), choose Left-Tailed, input z-score of -2.00, OOMPH Stat returns auc: 0.0227 Step 4: The area we need is P( z < 1 ) - ( z < -2 ) Subtract 0.0227 from 0.8413 = 0.8186 6
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3: Drosophilia Baldi and Moore exercise 11.20 (3rd edition), exercise 11.22 (4th edition) The common fruit fly, Drosophila melanogaster, is the most studied organism in genetic research because it is small, easy to grow, and reproduces rapidly. The length of the thorax (where the wings and legs attach) in a population of male fruit flies is approximately normal with mean 0.800 millimeters (mm) and standard deviation 0.078 mm. Choose a male fruit fly at random. What is the probability that the fly you choose has a thorax longer than 1 mm? 3 answer pnorm ( 1 , mean = 0.8 , sd = 0.078 , lower.tail= FALSE ) ## [1] 0.005172149 OOMPH Stat Step 1: Calculate the Z-score: (1-0.800) / 0.078 = 2.564 Step 2: Choose Right-Tailed for the Type, Curve Attribute is z-score of 2.564 Step 3: OOMPH Stat returns the Area Under the Curve of 0.0517 7
4: Boys’ Growth Baldi and Moore exercise 11.5 (both editions) Boys’ growth charts indicate that the heights of five-year-old boys are approximately normal with mean 43 inches and standard deviation 1.8 inches, whereas those of seven-year-old boys are approximately normal with mean 48 inches and standard deviation 2.1 inches. John measured 44 inches on his fifth birthday, and now, on his seventh birthday, he measures 49 inches. Find John’s standardized scores for both ages and compare them. Has he been growing faster or more slowly than the general population of boys his age? 4 answer z score when John is 5 years old: 44 43 1 . 8 = 0 . 556 z score when John is 7 years old: 49 48 2 . 18 = 0 . 476 John has been growing slightly more slowly than average because his z-score has gone down a little bit. However, children tend to grow by bursts, and the decreased z-score may simply reflect the natural variations in growth patterns. A third measurement one or two years later would help our conclusion. 8
5: Cholesterol in Middle-Aged Men exercise 11.36 in both 3rd and 4th editions The blood cholesterol levels of men aged 55 to 64 are approximately normal with mean 222 milligrams per deciliter (mg/dL) and standard deviation 37 mg/dL. 5a What percent of these men have high cholesterol (levels above 240 mg/dL)? 5a answer pnorm ( 240 , mean = 222 , sd = 37 , lower.tail = FALSE ) ## [1] 0.3133111 OOMPH Stat Step 1: Calculate the Z-score: (240 - 222) / 37 = 0.4865 Step 2: Choose Right-Tailed for the Type, Curve Attribute is z-score of .4865 Step 3: OOMPH Stat returns the Area Under the Curve of 0.3133 5b What percent of these men have elevated cholesterol (between 200 and 240 mg/dL)? 5b answer 9
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pnorm ( 240 , mean = 222 , sd = 37 , lower.tail = TRUE ) - pnorm ( 200 , mean = 222 , sd = 37 , lower.tail = TRUE ) ## [1] 0.4106316 = 41.06% OOMPH Stat: Z-score for 200 = (200-222)/37 = 0.5946 Z-score for 240 = (240-222)/37 = 0.4865 P(z < 0.4865) = 0.6867 P(z < -0.5946 = 0.2761 0.6867 – 0.2761 = 0.4106 = 41.06% 10
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6: Body Mass Index (BMI) in American Young Women adapted from Baldi and Moore exercise 11.39 in the 4th edition Your body mass index (BMI) is our weight in kilograms divided by the square of your height in meters. High BMI is a common but controversial indicator of overweight or obesity. A study by the National Center for Health Statistics found that the BMI values of American young women aged 20 to 29 follow a normal curve, approximately, with mean 26.8 kg m 2 and standard deviation 6.4 kg m 2 . 6a People with BMI less than 18.5 are considered underweight. According to this criterion, what percent of this population is underweight? 6a answer z score for 18.5 = 18 . 5 26 . 8 6 . 4 = 1 . 297 pnorm ( - 1.297 , mean = 0 , sd = 1 , lower.tail = TRUE ) ## [1] 0.09731559 # or more directly: pnorm ( 18.5 , mean = 26.8 , sd = 6.4 , lower.tail = TRUE ) ## [1] 0.0973371 6b People with BMI 25 and less than 30 are considered overweight. According to this criterion, what percent of this population is overweight? 6b answer # P ( BMI < 30 ) - P( BMI < 25 ) pnorm ( 30 , mean = 26.8 , sd = 6.4 , lower.tail = TRUE ) - pnorm ( 25 , mean = 26.8 , sd = 6.4 , lower.tail = TRUE ) ## [1] 0.3022031 6c What percent of this population is obese, with a BMI 30? 6c answer 12
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pnorm ( 30 , mean = 26.8 , sd = 6.4 , lower.tail = FALSE ) ## [1] 0.3085375 6d What BMI value the 95th percentile of this distribution? 6d answer z score for the 95th percentile is 1.645 26 . 8 + 1 . 645 6 . 4 = 26 . 8 + 10 . 5 = 37 . 3 kg m 2 13