Leigh currently teaches AP Statistics, Probability & Statistics and PreCalculus at Moravian Academy in Bethlehem, PA. Leigh has been an AP Statistics Reader since 2010 and she has written multiple choice and free response questions for the AP Statistics exam. Leigh has also served on the AP Statistics Instructional Design Team where she tagged items for use in AP Classroom. In addition to leading AP Statistics workshops, Leigh is a Desmos Certified Presenter and leads workshops on various Desmos calculator tools. Leigh can be found on Twitter/X at @mathteacher24.

Question:

A jewelry company uses a machine to apply a coating of gold on a certain style of necklace. The amount of gold applied to a necklace is approximately normally distributed. When the machine is working properly, the amount of gold applied to a necklace has a mean of 300 milligrams (mg) and standard deviation of 5 mg.

(a) A necklace is randomly selected from the necklaces produced by the machine. Assuming that the machine is working properly, calculate the probability that the amount of gold applied to the necklace is between 296 mg and 304 mg.

WOULD THIS GET CREDIT?

Response 1:

Response 2:

Response 3:

Response 4:

To get full credit in part (a), responses need to demonstrate knowledge of the distribution being used with its parameters, the event of interest and the probability. Each of the first three responses would earn full credit, because they demonstrate knowledge of the normal distribution being used with its parameters of m = 300 and s = 5. Plus, the correct event is shown by the boundaries of 296 and 304 with inequalities, shading on a diagram or a sentence describing the direction of the event. The correct probability is needed for full credit, but just the probability with unlabeled “calculator speak”, as shown in response 4, would receive no credit. Clear communication is a hallmark of a student response that earns full credit.

Teaching tips:

Have students draw a normal distribution and shade the region of interest. Often this gives them the sense of the reasonableness of their answer. For example, the bounds of 296 and 304 are within one standard deviation of the mean. By the empirical rule, a student could reason that the answer should be smaller than 0.68.

Labeled calculator syntax with the correct answer would have received full credit. However, responses that show more than just labeled calculator syntax, such as drawing a labeled diagram, using a correct probability statement and showing z-score calculations are viewed favorably if holistic scoring is required. (Note: Holistic scoring is when a student response earns a score that is not a whole number, like 2.5. Good statistical communication can help a response between two scores to go up, but poor or lacking statistical communication may lead to a response between two scores to go down.)

Encourage students to try part (a) of the investigative task. Typically, the first part or two of the investigative task will come directly from the curriculum, and the investigative portion of the question will begin on the second or third part. Part (a) of this question was a straightforward normal probability calculation.

The jewelry company wants to make sure the machine is working properly. Each day, Cleo, a statistician at the jewelry company, will take a random sample of the necklaces produced that day. Each selected necklace will be melted down and the amount of gold applied to that necklace will be determined. Because a necklace must be destroyed to determine the amount of gold that was applied, Cleo will use random samples of size n = 2 necklaces.

Cleo starts by considering the mean amount of gold being applied to the necklaces. After Cleo takes a random sample of n = 2 necklaces, she computes the sample mean amount of gold applied to the two necklaces.

(b) Suppose the machine is working properly with a population mean amount of gold being applied of 300 mg and a population standard deviation of 5 mg.

(i) Calculate the probability that the sample mean amount of gold applied to a random sample of n = 2 necklaces will be greater than 303 mg.

(ii) Suppose Cleo took a random sample of n = 2 necklaces that resulted in a sample mean amount of gold applied of 303 mg. Would that result indicate that the population mean amount of gold being applied by the machine is different from 300 mg? Justify your answer without performing an inference procedure.

WOULD THIS GET CREDIT?

Response 1:

Response 2:

Response 3:

Response 4:

To get full credit in part (b), the response must include: the use of a normal distribution with the correct parameter values (including work or a formula shown for the standard deviation), a response about whether a sample mean of 303 would provide convincing evidence, a justification for that response, and at least one of the following: correct event or correct probability. If a paper has all five components, this is viewed favorably for holistic scoring.

Response 1 is missing how the standard deviation of the sampling distribution is calculated. The response does not show how the value of 3.536 was obtained. This response would receive partial credit, P.

Response 2 has the correct mean and standard deviation for the sampling distribution. However, it does not directly answer the second question. It is true that the sample mean of 303 or more could be due to variability, but the response does not clearly state that the sample mean of 303 mg is not unusual. This response would receive partial credit, P.

Response 3 has the correct mean and standard deviation for the sum of the weights of two necklaces. This approach works and the response has the correct probability. The response also correctly states that the total weight (of 606 mg) is within one standard deviation of the mean of 600 mg. However, the conclusion of “yes” is not correct. This response would receive partial credit, P.

Response 4 does not have the correct notation of x-bar in the probability statement or in the labeling of the standard error. However, the probability statements show the correct z-score with the calculation of the standard deviation shown. The boundary value and direction are also shown. The response correctly states “no” and justifies the response with noting that the probability from part (b-i) is not unlikely. This response would receive full credit, E.

Teaching tips:

Be sure students know that the standard deviation of a sampling distribution is not the same as the standard deviation of the population. Many students used a standard deviation of 5 in part (b-i).

When a question is a “yes” or “no” question, students must clearly state “yes” or “no”.

Often justifications to “yes” or “no” questions come from previously completed parts of the problem. When a justification involves a probability, students should comment on the likelihood of that probability as part of their justification.

Now Cleo will consider the variation in the amount of gold the machine applies to the necklaces. Because of the small sample size, n = 2, Cleo will use the sample range of the data for the two randomly selected necklaces, rather than the sample standard deviation.

Cleo will investigate the behavior of the range for samples of size n = 2. She will simulate the sampling distribution of the range of the amount of gold applied to two randomly sampled necklaces. Cleo generates 100,000 random samples of size n = 2 independent values from a normal distribution with mean μ = 300 and standard deviation σ = 5. The range is calculated for two observations in each sample. The simulated sampling distribution of the range is shown in Graph I. This process is repeated using σ = 8, as shown in Graph II, and again using σ = 12, as shown in Graph III.

(c) Use the information in the graphs to complete the following.

(i) Describe the sampling distribution of the sample range for random samples of size n = 2 from a normal distribution with standard deviation s = 5, as shown in Graph I.

(ii) Describe how the sampling distribution of the sample range for samples of size n = 2 changes as the value of the population standard deviation σ increases.

WOULD THIS GET CREDIT?

To get full credit in part (c), the response has to include 4 or 5 of the following five components: (1) a discussion of shape, (2) reasonable values for center and (3) spread, and an indication that the (4) mean (or center) increases and the (5) spread increases as the standard deviation increases.

Response 1 only has one of the five components, a discussion of shape. Although the mode is at 0 and the shape becomes less skewed, these statements do not satisfy any of the other components.

Response 2 has a reasonable value for center and the “spread of the sample ranges increases” is also correct. However, the description of the distribution shape as both normal and skewed to the right is considered parallel solutions. These descriptions are not possible at the same time within a single distribution and the weaker answer (normal) is the one that is scored. This response only has two of the five components.

Responses 3 and 4 are both given full credit. Response 3 has correct statements about the shape, center and spread in part (i) and correctly describes the increase of the mean in part (ii). Response 4 has correct statements for all five components.

Teaching tips:

Remind students to discuss shape, center and spread when asked to describe any distribution of a quantitative variable.

To describe the center, students should reference the mean or median. Mode is not considered a measure of center.

Investigating the sampling distribution of a sample range requires students to apply skills from Skill Category 2 (Data Analysis) and Skill Category 3 (Using Probability and Simulation) from the Course and Exam Description. Have students practice these skills on distributions beyond the types of sampling distributions that are part of the AP Statistics course content.

Recall that Cleo needs to consider both the mean and standard deviation of the amount of gold applied to necklaces to determine whether the machine is working properly. Suppose that one month later, Cleo is again checking the machine to make sure it is working properly. Cleo takes a random sample of 2 necklaces and calculated the sample mean amount of gold applied is 303 mg and the sample range is 10 mg.

(d) Recall that the machine is working properly if the amount of gold applied to the necklaces has a mean of 300 mg and standard deviation of 5 mg.

(i) Consider Cleo’s range of 10 mg from the sample of size n = 2. If the machine is working properly with a standard deviation of 5 mg, is a sample range of 10 mg unusual? Justify your answer.

(ii) Do Cleo’s sample mean of 303 mg and range of 10 mg indicate that the machine is not working properly? Explain your answer.

WOULD THIS GET CREDIT?

In order to get full credit in part (d), the response has to include 4 or 5 of the following five components: (1) the sample range of 10 mg is not unusual, (2) justification based on the graph of the sampling distribution, (3) there is not convincing evidence the machine is not working properly, (4) justification for the sample mean 303 mg not being unusual and (5) justification that the sample range of 10 mg is not unusual.

Response 1 has components 1 and 3, but the justifications are not correct. The justification for the range is based on an argument of normality, but the distribution of the sample ranges is not approximately normal. Although n = 2 is a small sample size, a valid conclusion can be drawn for the sample mean, because the original population is normally distributed.

Response 2 has an incorrect conclusion for component 1 and bases the argument on normality. It also has an incorrect conclusion for part (ii). Because the response does not explicitly reference either the sample mean or sample range, what is meant by “this” in the response is unclear.

Responses 3 and 4 are both given full credit. Response 3 has all 5 components. The response correctly uses the histogram to estimate the probability that the range is 10 mg or more and concludes this is not unusual. The response also has components 3, 4 and 5. Response 4 has components 1, 2, 3 and 4. The argument that 303 is less than one standard deviation from 300 is an argument based on probabilities of normal distributions.

Teaching tips:

The investigative task often tells a story with sequentially connected parts. For example, in this investigative task, part (d) could not be answered without referencing earlier parts of the question.

Work with problems where the distribution is not approximately normal. For example, if the wait times at a grocery store checkout counter have a mean of 7 minutes and a standard deviation of 5 minutes, would it be possible for the wait times to be distributed normally? Explain.

The investigative task involves extended reasoning and assessing Skill Category 4: Statistical Argumentation. Have students practice statistical argumentation early in the course to perfect their reasoning skills.

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