Mr. Arbit's Classroom
Chapter 4 PracticeAP Statistics1. The graph below displays the scores of 32 students on a recent exam. The scores ranged from 64 to 95 points.(a) Describe the shape of this distribution.(b) In order to motivate her students, the instructor of the class wants to report that, overall, the class’s performance on the exam was high. Which summary statistic, the mean or the median, should the instructor use to report that overall exam performance was high? Explain.(c) The midrange is defined as: . (i) Compute this value using the data above. (ii) Is the midrange considered a measure of center or a measure of spread? Explain.(continued on back)2. Newborn boys at a certain hospital have a mean weight of 7.1 pounds and a standard deviation of 0.5 pounds. Newborn girls at this hospital have a mean weight of 6.8 pounds and a standard deviation of 0.3 pounds. Find the weight of a newborn girl with the same standardized score (z-score) as a newborn boy weighing 6.5 pounds.3. The department of agriculture at a university was interested in determining whether a preservative was effective in reducing discoloration in frozen strawberries. A sample of 50 ripe strawberries was prepared for freezing. Then the sample was randomly divided into two groups of 25 strawberries each. Each strawberry was placed into a small plastic bag.The 25 bags in the control group were sealed. The preservative was added to the 25 bags containing strawberries in the treatment group, and then those bags were sealed. All bags were stored at 0°C for a period of 6 months. At the end of this time, after the strawberries were thawed, a technician rated each strawberry’s discoloration from 1 to 10, with a low score indicating little discoloration.The dotplots below show the distributions of discoloration rating for the control and treatment groups.a) The standard deviation for the control group is 2.141. Explain what this value tells us about the variability in the control group.b) Based on the dotplots, comment on the effectiveness of the preservative in lowering the amount of discoloration in the strawberries. (No calculations are necessary.)“Chapter 4 Practice” Solutions1a) Solution: The distribution is skewed left (or negatively skewed)E: Correct shapeP: Say it’s skewed, but leave off the direction or put the wrong directionI: Any other shape1b) Solution: Since the distribution is skewed left, the mean will be pulled to the left – lower than the median. Therefore, the instructor should report the median.E: Choose the median and provide a rationale based on skewnessP: Choose the median but give a weak rationaleI: Choose the mean based on a flawed rationale OR Choose the median and give no rationale1c) Solution: Midrange = (64 + 95)/2 = 79.5. This is a measure of center, since it finds the halfway point between the two extreme ends of the data set.This solution has 3 steps: Correctly calculating it, calling it a measure of center, and providing a correct justification for that decision.E: All 3 steps are correctP: Two of the 3 steps are correctI: At most one of the 3 steps is correctNote: Acceptable justifications for the third step must somehow discuss the fact that the midrange is a “middle value,” or explain why it is not a measure of spread.Overall score on problem #1:4 = 3 E’s3 = 2 E’s and 1 P2 = 2 E’s and 1 I OR 1 E and 2 P’s1 = 1 E, 1 P, 1 I OR 1 E and 2 I’s OR 2 P’s and 1 I2) Solution: 6.44 pounds3a) Solution: 2.141 is the average distance between each strawberry’s discoloration and the mean discoloration of the control group.E: Correctly explain the meaning of standard deviation, in the context of the problemP: Correctly explain the meaning of standard deviation, without contextI: Incorrect explanation of standard deviation (simply copying the formula counts as an I as well)3b) Solution: The preservative appears to be effective in lowering the amount of discoloration. The center of the treatment group is lower than the center of the control group.E: Say that the preservative is effective, and link that decision to an explicit comparison of a measurement in both groups (like the center)P: Say that the preservative is effective, but do not explicitly link that decision to a comparison of a measurement in both groups (for example, “discoloration ratings are lower for the treatment group” does not discuss a specific measurement such as the center, which is important because not all treatment group strawberries have lower ratings than all control group strawberries)Note: Correctly comparing the two centers, but forgetting to conclude that the preservative is effective, also counts as a P.I: Say the preservative is effective, with incorrect or no justification OR Say the preservative is not effective because the centers are roughly the same(This problem had a part (c), but we haven’t covered it yet this year. Since we only did part of the problem, we can’t give it an overall score out of 4.) ................
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