AP Statistics Chapter 2 Review 1. The SAT (Scholastic Aptitude Test) and ACT (American College Testing) mathematics scores vary normally with the SAT described by N(500,100) and ACT described by N(18,6) . What proportion of observations will have an SAT score a. higher than 675? .0401 b. less than 450? .3085 c. between 450 and 675? .6514 What proportion of observations will have an ACT score d. higher than 28? .0475 e. higher than 12? .8413 f. between 12 and 28? .7938 Charlie has an SAT math score of 740, while his friend James has an ACT score of 28. g. Who has made the better test score and why? ZCharlie = 2.4; ZJames = 1.67 ∴Charlie has the better score. Suppose a university chemistry program will only accept students who score in the top 10% on either ACT or SAT. What test scores are required for the h. ACT 26 i. SAT 629 (rounded up) Suppose a university chemistry program will only accept students who score in the top 25% on either ACT or SAT. What test scores are required for the j. ACT 23 (rounded up) k. SAT 567 Suppose a university chemistry program will exclude students who score in the bottom 40% on either ACT or SAT. What test scores are required to avoid exclusion based on the l. ACT 17 (rounded up) m. SAT 475 2. Physicians often measure fasting blood glucose when performing a routine health assessment to help rule out diabetes. If we assume that blood glucose after a 12 hour fast follows a normal distribution N(85,25) in units of mg glucose per dl of blood. What proportion of observations of a (12 hour) fasting adult will be a. more than 60? .8413 b. less than 110? .8413 c. between 60 and 110? .6827 d. greater than 140 (diabetes begins to be a concern)? .0139 We wish to find the blood glucose level of the e. highest 5% of the population. 126 f. lowest 5% of the population. 43.9 g. median of the population. 85 h. Q1 of the population. 68.1 3. Ecologists are interested in factors which may determine how far north or south a species may occur. As part of a study “Temperature and the Northern Distribution of Wintering Birds” from Ecology (1991):2274-2285 gave the following body masses in grams for 50 bird species found in a particular region. A shortened list is given: 7.7 10.1 21.6 8.6 12.0 9.4 11.5 9.0 8.2 20.2 26.1 6.2 19.1 21.0 28.1 21.6 16.6 10.6 48.5 11.4 a. Evaluate whether the given data originated from a normal distribution. The above data is entered into L1 and set up as shown. The Normal Probability Plot is given. This plot is nonlinear, suggesting that the data does not arise from a normal distribution. 4. In a certain normal distribution 50% of values are below 200. 20% of values are above 220. Find the normal distribution that fits these facts. invNorm(.8) = 0.8416 200 − µ 220 − µ invNorm(.5) = 0 = 0.8416 = σ σ ∴ µ = 0 0.8416σ = 220 − 200 σ ≈ 23.76 µ = 200, σ = 23.7636 5. In a certain normal distribution 10% of values are above 60. 10% of values are below 30. Find the normal distribution that fits these facts. 60 − µ 30 − µ invNorm(.9) = 1.28155 = invNorm(.1) = −1.28155 = σ σ Solving the system of 1.28155σ = 60 − µ −1.28155σ = 30 − µ equations: 1.28155σ = 60 − µ −1.28155σ = 30 − µ adding the two equations 0 = 90 − 2µ µ = 45 substituting into the first equation: 1.28155σ = 60 − 45 σ = 11.70 µ = 45, σ = 11.7045 6. The middle 50% of a normal distribution is from 130 to 145. Find the distribution. 130 − µ invNorm(.25) = −0.6744 = σ −0.6744σ = 130 − µ 145 − µ invNorm(.75) = 0.6744 = σ 0.6744σ = 145 − µ −0.6744σ = 130 − µ 0.6744σ = 145 − µ 0=275-2µ µ=137.5 −0.6744σ = 130 −137.5 σ ≈ 11.12 µ = 137.5, σ = 11.1195 .