Contemporary Math Class Problems Chapter 16 1. 150 students in a math class take the final exam. The scores on the exam have an approximately normal distribution with mean μ = 65 and standard deviation σ = 10. The third quartile of the scores on the exam is approximately: Q3 = μ + (.675) σ Q3 = 65 + (.675)(10) Q3 = 65 + 6.75 Q3 = 71.75, which rounds to 72 2. For a population of 2000 students taking the SAT math exam, the scores on the exam have an approximately normal distribution with mean μ = 590 and standard deviation σ = 70. The third quartile of the scores on the exam is approximately: Q3 = μ + (.675) σ Q3 = 590 + (.675)(70) Q3 = 590 + 47.25 Q3 = 637.25, which rounds to 637 3. As part of a study on the metabolism of athletes, 400 college basketball players are randomly chosen and their weights taken. The distribution of the weights is approximately normal. The average weight is 215 pounds and the standard deviation is 15 pounds. A weight of 260 pounds corresponds to a standardized value of: μ =215 σ=15 x=260 Z=(x – μ)/ σ Z=(260 – 215)/15 Z=45/15 Z=3 A weight of 188 pounds corresponds to a standardized value of μ =215 σ=15 x=188 Z=(x – μ)/ σ Z=(188– 215)/15 Z=‐27/15 Z=‐1.8 Contemporary Math Class Problems Chapter 16 4. Refer to a normal distribution described by the following figure. The mean is μ and the standard deviation is σ. The mean μ =54
95% of the total area implies that the values shown represent 3σ
(84‐24)/2=30 24+30=54 or 84‐30=54 5. Refer to a normal distribution described by the following figure. The mean is μ and the standard deviation is σ. The mean μ =
16% of the total area implies that the values shown represent σ
(60.5 – 44.5)/2=8 44.5+8=52.5 or 60.5‐8=52.5 Contemporary Math Class Problems Chapter 16 6. 150 students in a math class take the final exam. The scores on the exam have an approximately normal distribution with mean μ = 65 and standard deviation σ = 10. The number of students scoring 65 points or more is approximately: Since 65 is the median and we are using a normal distribution, we can state that there are just as many students scoring above the median as there are below the median. Therefore, since 150 students took the exam, the only split within the data where there was an equal amount above and below the median is at 75. The average score on the exam was approximately: Taken from the context of the problem… μ = 65 Assuming there were no outliers, the lowest score on the exam was around: 99.7% of all data falls within 3 standard deviations above and below the mean. So 3 σ = 3*10=30 65‐30=35 35 was the lowest score. Approximately what percent of the students scored between 65 and 75 points? 68% of all values will fall between +‐ 1 standard deviation of the mean. Since σ = 10, one deviation in both directions will yield 55 to 75. Since this range is 68%, half of it in either direction is 34%, thus the range between 65 and 75. 7. Peterʹs score on the exam places him in the 20th percentile of the class. Peterʹs score on the exam is approximately: X= μ ‐ σ * z – use to unstandardize z and find the original value of x. μ=65 σ=10 X=65 – 0.84 * 10, thus 56 8. In 2007, a total of 1,494,531 college‐bound seniors took the SAT exam. The distribution of scores in the Critical Reading section of the SAT was approximately normal with a mean μ = 502 and a standard deviation σ=113. a. Estimate the 75th percentile score on the exam. (SAT scores are multiples of 10) 75th percentile has an approximate location of μ + .675σ So, 502 + (.675)(113) = 578.275 which rounds up to 580 b. Estimate the 70th percentile score on the exam 75th percentile has an approximate location of μ + .52σ So, 502 + (.52)(113) = 560 c. Estimate the percentile corresponding to a score of 530 502 + x(113) = 580…..x(113) = 28/113…..x=.247 or .25 which is 60th percentile. Contemporary Math Class Problems Chapter 16 9. 150 students in a math class take the final exam. The scores on the exam have an approximately normal distribution with mean μ = 65 and standard deviation σ = 10. a. A score of 85 corresponds to a standardized value of 2 From the mean of 65, we have a +‐ S.D. of 10. Thus 55 and 75 are both one S.D. from the median. If we go another 10 from 55 and 75, we get 45 and 85 respectively. That is then 2 S.D. from the median. B. A score of 50 corresponds to a standardized value of Z=(x –μ) / σ x=50 μ=65 σ=10 Z=(50‐65)/10 Z=(‐15)/10 Z= ‐ 1.5