An original Thinka practice paper modelled on the structure and difficulty of the May 2023 AP AP Statistics paper. Not affiliated with or reproduced from AP.
Section II Part A: Traditional Free-Response Questions
Answer Questions 1 through 5. Show all your work. Clearly indicate the methods you use, because you will be scored on the correctness of your methods as well as on the accuracy and completeness of your results and explanations.
5 Question · 20 marks
Question 1 · free-response
4 marks
An agricultural researcher is studying the post-harvest shelf life (in days) of a newly developed variety of organic strawberry under two different cold-storage methods: Method A (standard refrigeration) and Method B (modified-atmosphere cooling). Water loss and fungal growth are monitored until the fruit is no longer marketable.
(a) The researcher recorded the shelf life for a random sample of 25 containers of strawberries stored under Method A. The data are displayed in the stemplot below.
Based on the stemplot, describe the distribution of shelf life for strawberries stored under Method A.
(b) Summary statistics for the shelf life (in days) of a random sample of 25 containers of strawberries stored under Method B are given in the table below.
Determine whether there are any potential outliers in the distribution of shelf life for Method B according to the \(1.5 \times \text{IQR}\) rule. Show your work.
(c) A supermarket chain wants to choose the storage method that provides a longer shelf life for its strawberries. Which storage method, Method A or Method B, would you recommend to the supermarket chain? Using characteristics of the distributions of shelf life for both methods, justify your choice.
Show answer & marking schemeHide answer & marking scheme
Worked solution
(a) Description of the distribution of shelf life for Method A: - Shape: The distribution of shelf life is skewed to the right (unimodal with a tail extending toward higher values). - Center: The median shelf life is 27 days (the 13th observation in the ordered data of 25 values). - Spread: The shelf lives range from 11 days to 58 days, giving a range of \(58 - 11 = 47\) days. (Alternatively, \(Q_1 = 20\) days, \(Q_3 = 33.5\) days, giving an \(\text{IQR} = 13.5\) days). - Unusual features: There is a potential high outlier at 58 days, with a gap between 42 and 58 days. - Context: Units (days) and context (shelf life of strawberries under Method A) are included.
(b) Outlier calculation for Method B: - \(\text{IQR} = Q_3 - Q_1 = 42 - 31 = 11\) days. - Lower boundary (inner fence): \(Q_1 - 1.5(\text{IQR}) = 31 - 1.5(11) = 31 - 16.5 = 14.5\) days. - Upper boundary (inner fence): \(Q_3 + 1.5(\text{IQR}) = 42 + 1.5(11) = 42 + 16.5 = 58.5\) days. - Since the minimum value (18 days) is greater than 14.5 days and the maximum value (49 days) is less than 58.5 days, all observations lie within the interval \([14.5, 58.5]\). Therefore, there are no potential outliers in the distribution of shelf life for Method B.
(c) Recommendation and comparison: I recommend Method B. - Center: The median shelf life for Method B (38 days) is substantially greater than the median shelf life for Method A (27 days). Over 75% of strawberries in Method B have a shelf life of at least 31 days (since \(Q_1 = 31\)), whereas more than half of the strawberries in Method A have a shelf life of 27 days or fewer. - Spread: The spread of the two distributions is comparable, with Method B having an \(\text{IQR}\) of 11 days (range of 31 days) compared to Method A's \(\text{IQR}\) of 13.5 days (or range of 47 days). Because Method B typically yields a longer shelf life with higher average and median days before spoilage, it is the superior choice for the supermarket chain.
Marking scheme
Each part is scored as Essentially Correct (E), Partially Correct (P), or Incorrect (I).
Part (a): - Essentially Correct (E) if the response includes context (shelf life in days) AND correctly addresses all 4 features: 1. Shape: Skewed right 2. Center: Median is 27 days (or approximate mean around 28 days) 3. Spread: Range is 47 days (or values from 11 to 58 days) or IQR is approximately 13.5 days 4. Unusual features: Potential outlier at 58 days or gap from 42 to 58 days. - Partially Correct (P) if the response includes context and addresses 2 or 3 features, OR addresses all 4 features without context. - Incorrect (I) otherwise.
Part (b): - Essentially Correct (E) if the response correctly calculates the \(\text{IQR} = 11\), finds the correct lower fence (14.5) and upper fence (58.5), and correctly concludes there are no outliers based on the minimum (18) and maximum (49). - Partially Correct (P) if the fences are calculated correctly with minor arithmetic errors but the conclusion is consistent with the work, OR if the boundaries are calculated correctly but no explicit conclusion is drawn. - Incorrect (I) otherwise.
Part (c): - Essentially Correct (E) if the response chooses Method B AND provides a direct numerical comparison of the centers (e.g., median 38 vs 27) in context. - Partially Correct (P) if the response chooses Method B and compares centers without numerical support, OR compares other characteristics without adequately addressing center. - Incorrect (I) otherwise.
A horticulturist wants to investigate whether adding a biochar soil amendment to the potting mixture reduces moisture stress in young olive trees. The horticulturist has 40 potted young olive trees of the same age and variety placed in an outdoor nursery. Twenty of the trees will receive the standard potting mix with the biochar amendment added, and the other 20 trees will receive the standard potting mix without biochar.
After two months under controlled watering conditions, the horticulturist will record the moisture stress level of each tree using a leaf water potential meter, where lower readings indicate higher moisture stress.
(a) Based on the information provided, identify each of the following: - Experimental units - Treatments - Response variable
(b) Describe an appropriate method the horticulturist could use to randomly assign the 40 olive trees to the two potting mixtures so that each treatment group contains 20 trees.
(c) Suppose the horticulturist observes that trees on the south side of the outdoor nursery receive significantly more daily sun exposure than trees on the north side. (i) Explain why sun exposure could be a confounding variable in this study. (ii) Describe how the horticulturist could incorporate a randomized block design to control for the variation in sun exposure.
Show answer & marking schemeHide answer & marking scheme
Worked solution
(a) - Experimental units: The 40 potted young olive trees (or the individual potted olive trees). - Treatments: Standard potting mix with biochar amendment and standard potting mix without biochar (standard potting mix alone). - Response variable: Leaf water potential (or moisture stress level) of each tree measured after two months.
(b) An appropriate random assignment method using a random number generator: 1. Assign each of the 40 olive trees a unique integer label from 1 to 40. 2. Use a random number generator to produce random integers between 1 and 40 without replacement (ignoring any repeated numbers). 3. The trees corresponding to the first 20 unique integers generated will be assigned to receive the potting mix with the biochar amendment. 4. The remaining 20 trees will receive the potting mix without biochar.
(Alternative valid method: Write the numbers 1 to 40 on 40 identical slips of paper, mix them thoroughly in a hat, and draw 20 slips without replacement for the biochar treatment, assigning the remaining 20 to the control.)
(c) (i) Sun exposure is related to the location of the trees (south side vs. north side) and directly affects moisture stress, as trees exposed to more intense sunlight naturally lose more water through transpiration. If one treatment group happens to have more trees situated on the sunny south side, it would be impossible to determine whether a difference in moisture stress was caused by the biochar amendment or by the difference in sun exposure.
(ii) The horticulturist should create two blocks based on location: - Block 1: The 20 trees located on the south side (high sun exposure). - Block 2: The 20 trees located on the north side (lower sun exposure).
Within each block, randomly assign 10 trees to receive the potting mix with biochar and 10 trees to receive the potting mix without biochar using an appropriate random assignment procedure (e.g., flipping a coin or using a random number generator for each block independently).
Marking scheme
Each of parts (a), (b), and (c) is scored as Essentially Correct (E), Partially Correct (P), or Incorrect (I).
Part (a) is scored as follows: - Essentially Correct (E) if the response correctly identifies all three components: 1. Experimental units: 40 olive trees (or individual potted olive trees). 2. Treatments: Potting mix with biochar and potting mix without biochar. 3. Response variable: Moisture stress level / leaf water potential (after two months). - Partially Correct (P) if the response correctly identifies only two of the three components. - Incorrect (I) if the response correctly identifies at most one component.
Part (b) is scored as follows: - Essentially Correct (E) if the response satisfies both of the following criteria: 1. Describes a valid random assignment process (e.g., uniquely labeling trees 1 to 40, using a random generator or slips in a hat without replacement) where all assignments are equally likely. 2. Ensures that exactly 20 trees are assigned to each treatment group. - Partially Correct (P) if the response describes a random process that satisfies only one of the two criteria (e.g., assigns treatments randomly by coin flip for each tree, which does not guarantee equal group sizes of 20, or omits 'without replacement' in sampling slips). - Incorrect (I) if the response does not describe a valid random assignment process.
Part (c) is scored as follows: - Essentially Correct (E) if the response satisfies both parts (i) and (ii): 1. Part (i): Explains how sun exposure is linked to both tree location/assignment and the response variable (moisture stress), making it impossible to isolate the effect of biochar. 2. Part (ii): Correctly forms two blocks based on sun exposure (south and north) and describes randomly assigning treatments within each block. - Partially Correct (P) if the response correctly satisfies only (i) or only (ii). - Incorrect (I) if the response fails to meet the criteria for P.
Composite Score: - 4 Points: All three parts scored E. - 3 Points: Two parts scored E and one part scored P. - 2 Points: Two parts scored E and no part P, OR one part scored E and two parts scored P, OR three parts scored P. - 1 Point: One part scored E and no parts P, OR no part scored E and two parts scored P. - 0 Points: Response does not meet criteria for 1 point.
Question 3 · Free-Response
4 marks
A specialty tea company produces holiday gift tins. Each tin contains a mystery discount voucher for a future order. Let the random variable, \(Y\), represent the dollar amount of the discount voucher found inside a randomly selected tin. The probability distribution of \(Y\) is displayed in the table below.
(a) Based on the probability distribution of \(Y\), answer the following. Show your work. (i) Calculate the proportion of gift tins that contain a $2 voucher. (ii) Calculate the proportion of gift tins that contain a voucher worth at least $10.
(b) Based on the probability distribution of \(Y\), calculate the probability that a randomly selected gift tin contains a $50 voucher, given that it contains a voucher worth at least $10. Show your work.
(c) Based on the probability distribution of \(Y\), calculate and interpret the expected value of the distribution of the voucher amount in the gift tins. Show your work.
(d) During a special promotional weekend, the tea company announces that each voucher value will be doubled, and an additional flat bonus of $1.50 will be credited. Let \(W = 2Y + 1.50\) represent the adjusted dollar value of a voucher. Using your expected value from part (c), calculate the expected value of \(W\). Show your work.
Show answer & marking schemeHide answer & marking scheme
Worked solution
(a)(i) The sum of probabilities for a valid discrete probability distribution must equal 1:
Each of the four parts (a, b, c, d) is scored as Essentially Correct (E), Partially Correct (P), or Incorrect (I).
Part (a) is scored as follows: - Essentially Correct (E) if the response correctly calculates the proportion in (a-i) with supporting work AND correctly calculates the proportion in (a-ii) with supporting work. - Partially Correct (P) if only one of the two sub-parts is correctly calculated with work shown. - Incorrect (I) if neither sub-part meets the criteria.
Part (b) is scored as follows: - Essentially Correct (E) if the response correctly identifies the conditional probability formula/ratio, substitutes the correct probabilities, and obtains the correct value of 0.16 (or a value consistent with (a-ii)). - Partially Correct (P) if the correct formula is stated but an arithmetic/substitution error is made, OR the numerator uses product of independent events $0.04 \times 0.25$. - Incorrect (I) otherwise.
Part (c) is scored as follows: - Essentially Correct (E) if the response correctly calculates the expected value ($6.95) with work shown AND provides a correct interpretation including (1) long-run/repeated trials and (2) mean/average in context. - **Partially Correct (P)** if the calculation is correct with work shown but the interpretation is incomplete/missing, OR if the calculation contains a minor arithmetic error but the interpretation is correct. - **Incorrect (I)** otherwise.
**Part (d)** is scored as follows: - **Essentially Correct (E)** if the response correctly calculates $15.40 (or consistent with part (c)) showing appropriate linear transformation work and includes units ($ or dollars). - Partially Correct (P) if the calculation is correct but work is missing or units are omitted. - Incorrect (I) otherwise.
Holistic Score Breakdown: - 4 points (Complete Response): 4 E's (or 3 E's and 1 P depending on overall quality) - 3 points (Substantial Response): 3 E's and 1 P, or 2 E's and 2 P's - 2 points (Developing Response): 2 E's, or 1 E and 2 P's, or 4 P's - 1 point (Minimal Response): 1 E, or 2 P's - 0 points: Response does not meet criteria for 1 point.
Question 4 · free_response
4 marks
An agricultural researcher wanted to investigate whether a newly developed organic soil amendment increases the yield of strawberry plants compared to the standard organic fertilizer. Forty strawberry plants of the same variety were randomly assigned to two treatment groups of 20 plants each. One group received the standard organic fertilizer, and the other group received the new organic soil amendment. All plants were grown under identical greenhouse conditions. At the end of the harvest season, the total yield of strawberries (in grams) was recorded for each plant.
The summary statistics for the strawberry yields in the two groups are given in the table.
Parallel boxplots of the yield data showed that both distributions are roughly symmetric with no apparent outliers.
Do the data provide convincing statistical evidence, at a significance level of \(\alpha = 0.05\), that strawberry plants grown with the new soil amendment have a greater mean yield than strawberry plants grown with the standard fertilizer? Complete the appropriate inference procedure to support your answer.
Show answer & marking schemeHide answer & marking scheme
Worked solution
Step 1: State the hypotheses and identify the inference procedure.
Let \(\mu_{\text{new}}\) represent the true mean yield (in grams) for all strawberry plants of this variety grown with the new soil amendment. Let \(\mu_{\text{standard}}\) represent the true mean yield (in grams) for all strawberry plants of this variety grown with the standard fertilizer.
The hypotheses to be tested are: $$H_0: \mu_{\text{new}} - \mu_{\text{standard}} = 0 \quad (\text{or } \mu_{\text{new}} = \mu_{\text{standard}})$$ $$H_a: \mu_{\text{new}} - \mu_{\text{standard}} > 0 \quad (\text{or } \mu_{\text{new}} > \mu_{\text{standard}})$$
The appropriate inference procedure is a two-sample \(t\)-test for the difference between two population means.
---
**Step 2: Check conditions and calculate the test statistic and \(p\)-value.**
Conditions: 1. Random assignment: The 40 strawberry plants were randomly assigned to the two treatment groups (20 plants in each group), which satisfies the condition for independent groups in an experiment. 2. Normality: Although both sample sizes (\(n_1 = 20, n_2 = 20\)) are less than 30, the problem states that parallel boxplots show both sample distributions are roughly symmetric with no apparent outliers. Therefore, it is reasonable to assume that the sampling distribution of \(\bar{x}_{\text{new}} - \bar{x}_{\text{standard}}\) is approximately normal.
*Degrees of Freedom and \(p\)-value:* Using the Satterthwaite approximation, \(\text{df} \approx 37.63\) (or conservative \(\text{df} = \min(19, 19) = 19\)): - With \(\text{df} \approx 37.63\), \(p\text{-value} = P(t > 2.889) \approx 0.0032\). - With conservative \(\text{df} = 19\), \(p\text{-value} = P(t > 2.889) \approx 0.0047\).
---
Step 3: State the conclusion in context.
Because the \(p\text{-value} \approx 0.0032\) is less than the significance level \(\alpha = 0.05\), we reject the null hypothesis \(H_0\).
There is convincing statistical evidence that the true mean yield of strawberry plants grown with the new soil amendment is greater than the true mean yield of strawberry plants grown with the standard fertilizer.
Marking scheme
Scored in three sections:
Section 1: Hypotheses and Procedure Essentially correct (E) if the response satisfies the following: 1. Identifies a two-sample \(t\)-test for a difference in means by name or formula. 2. States correct null and alternative hypotheses using parameters (\(\mu_1, \mu_2\) or \(\mu_d = \mu_1 - \mu_2\)). 3. States the correct direction (one-sided upper tail) for the alternative hypothesis. 4. Defines the parameters in context (mean yield of strawberry plants for each treatment). Partially correct (P) if 2 or 3 of these components are met. Incorrect (I) otherwise.
Section 2: Conditions and Calculations Essentially correct (E) if the response satisfies the following: 1. Mentions random assignment of treatments to satisfy independence/randomness. 2. Checks normality using the symmetry/lack of outliers in the boxplots of both samples. 3. Calculates the correct \(t\)-statistic (\(t \approx 2.89\)). 4. Reports the correct \(p\)-value consistent with the alternative hypothesis (\(p \approx 0.003\) to \(0.005\)). Partially correct (P) if 2 or 3 of these components are met. Incorrect (I) otherwise.
Section 3: Conclusion Essentially correct (E) if the response satisfies the following: 1. Correctly compares the \(p\)-value to \(\alpha = 0.05\) and makes a correct decision regarding \(H_0\) (reject \(H_0\)). 2. States a non-deterministic conclusion in context consistent with the alternative hypothesis. Partially correct (P) if only 1 of the 2 components is met. Incorrect (I) otherwise.
Overall Question Score: - 4 points (Complete): EEE - 3 points (Substantial): EEP - 2 points (Developing): EEI, EPP, or PPP - 1 point (Minimal): EPI, PPI, or EII - 0 points: PII or III
Question 5 · free-response
4 marks
Marine biologists are studying the growth patterns of adult Atlantic blue crabs in a coastal estuary. Measuring the total body mass of a crab in the field can be difficult and inaccurate due to wind and vessel movement, whereas carapace width across the shell can be measured quickly and precisely using digital calipers. A study was conducted to investigate whether carapace width, measured in centimeters (cm), could be used to accurately estimate the body mass, measured in grams (g), of adult blue crabs.
The biologists captured a random sample of 22 adult blue crabs, recorded their carapace width and body mass, and returned them to the estuary. A scatterplot of body mass versus carapace width reveals a strong, positive, linear relationship with no apparent outliers.
Following is the equation of the least-squares regression line relating carapace width and body mass for the sample of blue crabs: \[\widehat{\text{mass}} = -142.5 + 21.80(\text{carapace width})\]
(a) Describe the relationship between carapace width and body mass of adult Atlantic blue crabs in context.
(b) One blue crab in the sample had a carapace width of \(14.0\text{ cm}\) and an actual body mass of \(175.0\text{ g}\). (i) Using the equation of the least-squares regression line, calculate the predicted body mass for this blue crab. Show your work. (ii) Calculate the residual for this blue crab. Show your work.
(c) Interpret the slope of the least-squares regression line in context.
(d) A previous baseline study suggested that for a related sub-species, the population slope relating carapace width and body mass is \(18.0\text{ grams per centimeter}\). A biologist wants to test whether the population slope for this estuary's adult blue crabs differs from \(18.0\text{ g/cm}\). Let \(\beta\) represent the slope of the true population regression line.
The biologist conducted a hypothesis test of: \[H_0: \beta = 18.0\] \[H_a: \beta \neq 18.0\]
The test statistic was calculated to be \(t = 2.845\). Assume all conditions for inference are met. (i) Determine the \(p\)-value of the test. (ii) At a significance level of \(\alpha = 0.05\), what conclusion should the biologist reach regarding the slope of the population regression line? Justify your answer.
Show answer & marking schemeHide answer & marking scheme
Worked solution
(a) The relationship between carapace width and body mass for adult Atlantic blue crabs is strong, positive, and linear (or approximately linear), with no distinct outliers or influential points.
(b)(i) Using the regression model: \[\widehat{\text{mass}} = -142.5 + 21.80(14.0) = -142.5 + 305.2 = 162.7\text{ g}\]
(b)(ii) The residual is calculated as: \[\text{Residual} = y - \hat{y} = 175.0 - 162.7 = 12.3\text{ g}\]
(c) The slope is \(21.80\). This means that for each additional 1-centimeter increase in the carapace width of an adult Atlantic blue crab, the predicted (or estimated/expected) body mass increases by \(21.80\text{ grams}\).
(d)(i) The degrees of freedom for the test of slope are: \[df = n - 2 = 22 - 2 = 20\] From the \(t\)-distribution table with \(df = 20\), the one-sided tail probability for \(t = 2.845\) is \(0.005\). Because the alternative hypothesis is two-sided (\(H_a: \beta \neq 18.0\)): \[p\text{-value} = 2 \times P(t \ge 2.845) = 2(0.005) = 0.010\]
(d)(ii) Because the \(p\)-value (\(0.010\)) is less than the significance level \(\alpha = 0.05\), we reject the null hypothesis \(H_0\). There is convincing statistical evidence that the population slope of the linear regression line relating body mass to carapace width for adult Atlantic blue crabs is significantly different from \(18.0\text{ g/cm}\).
Marking scheme
Each part (a, b, c, d) is scored as Essentially Correct (E), Partially Correct (P), or Incorrect (I).
- Part (a): - Satisfies at least 3 of 4 characteristics (Direction: positive; Strength: strong; Form: linear; Unusual features: no outliers) AND includes Context (carapace width and body mass of blue crabs) -> E - Satisfies only 1-2 characteristics with context OR satisfies characteristics without context -> P - Otherwise -> I
- Part (b): - Component 1: Correct predicted value (162.7 g) with work shown. - Component 2: Correct residual (12.3 g) with work shown. - Both components correct -> E; Exactly one component correct -> P; Neither -> I
- Part (c): - Component 1: Identifies slope value as 21.80. - Component 2: States an increase of 21.80 g in body mass per 1 cm increase in carapace width (correct units for both). - Component 3: Uses non-deterministic language (e.g., 'predicted', 'estimated', 'expected', 'on average'). - All 3 components satisfied -> E; 2 components satisfied -> P; 0 or 1 component satisfied -> I
- Part (d): - Component 1: Correctly calculates \(p\text{-value} = 0.010\) using \(df = 20\) and doubling the tail area. - Component 2: Compares \(p\text{-value}\) to \(\alpha = 0.05\) and states a correct decision regarding \(H_0\). - Component 3: States a conclusion in context in terms of the alternative hypothesis using non-deterministic language. - All 3 components satisfied -> E; 2 components satisfied -> P; 0 or 1 component satisfied -> I
Overall Question Score: - 4 (Complete Response): 4 E's (or 1 point per E, 0.5 per P totaling 3.5-4 points) - 3 (Substantial Response): 3 E's or total 2.5-3 points - 2 (Developing Response): 2 E's or total 1.5-2 points - 1 (Minimal Response): 1 E or total 0.5-1 point
Ready to test yourself?
Turn these notes into exam-style practice. Get unlimited AI questions on this topic with instant marking and explanations.
Answer Question 6. Show all your work. This question requires you to integrate skills and content from multiple units and apply them in an unfamiliar, investigative context.
1 Question · 4 marks
Question 1 · Investigative Task
4 marks
A tech hardware company manufactures high-performance microchips for electric vehicle batteries. An important measure of quality is the operating temperature, in degrees Celsius (\(^\circ\text{C}\)), of a microchip under peak operational load. When the manufacturing process is operating under control, the peak operating temperature of an individual microchip is approximately normally distributed with mean \(\mu = 65^\circ\text{C}\) and standard deviation \(\sigma = 4^\circ\text{C}\).
(a) A microchip is randomly selected from a production run operating under control. Calculate the probability that the peak operating temperature of this microchip is greater than \(71.0^\circ\text{C}\). Show your work.
(b) To monitor daily production quality, a quality control technician takes a random sample of \(n = 4\) independent microchips produced that day and calculates the sample mean operating temperature, \(\bar{X}\).
(i) Assuming the process is operating under control, calculate the probability that the sample mean operating temperature \(\bar{X}\) of the \(4\) microchips is greater than \(68.0^\circ\text{C}\). Show your work.
(ii) A random sample of \(n = 4\) microchips yielded a sample mean of \(\bar{X} = 68.0^\circ\text{C}\). Does this sample result provide convincing statistical evidence that the true mean operating temperature of the production run has shifted above \(65.0^\circ\text{C}\)? Justify your answer without performing a formal hypothesis test.
(c) The company's automated safety monitor immediately flags a batch if the maximum peak operating temperature among the \(4\) sampled microchips, denoted by \(M = \max(X_1, X_2, X_3, X_4)\), exceeds a threshold of \(71.0^\circ\text{C}\). Assuming the process is operating under control:
(i) Explain why the probability that all \(4\) sampled microchips have an operating temperature less than or equal to \(71.0^\circ\text{C}\) is equal to \([P(X \le 71.0)]^4\), and calculate this probability.
(ii) Use your result from part (c)(i) to determine the probability that the automated monitor flags the batch (i.e., \(P(M > 71.0)\)).
(d) The lead engineer wants to evaluate whether the sample maximum \(M\) could also be used to detect changes in the process variability \(\sigma\). The engineer simulates the sampling distribution of \(M\) for samples of size \(n = 4\) drawn from normal populations with mean \(\mu = 65^\circ\text{C}\) across three different values of \(\sigma\). The table below summarizes the simulated sampling distributions:
(i) Based on the simulation results, describe how the sampling distribution of \(M\) changes in both center and variability as the population standard deviation \(\sigma\) increases.
(ii) In a subsequent quality audit of \(4\) microchips, the observed sample maximum was \(M = 78.5^\circ\text{C}\). Is a sample maximum of \(78.5^\circ\text{C}\) unusual if the process is operating under control with \(\mu = 65^\circ\text{C}\) and \(\sigma = 4^\circ\text{C}\)? Justify your answer using the simulation results.
Show answer & marking schemeHide answer & marking scheme
Worked solution
Part (a): Let \(X\) denote the peak operating temperature of a randomly selected microchip. \(X\) follows a normal distribution with \(\mu = 65^\circ\text{C}\) and \(\sigma = 4^\circ\text{C}\). \[ z = \frac{71.0 - 65.0}{4.0} = 1.50 \] \[ P(X > 71.0) = P(Z > 1.50) = 1 - P(Z \le 1.50) = 1 - 0.9332 = 0.0668 \] (Using a calculator: \(\text{normalcdf}(71, \infty, 65, 4) \approx 0.0668\))
---
Part (b): (i) Because the population is approximately normally distributed, the sampling distribution of \(\bar{X}\) for a sample of size \(n = 4\) is normally distributed with: \[ \mu_{\bar{X}} = \mu = 65.0^\circ\text{C} \quad \text{and} \quad \sigma_{\bar{X}} = \frac{\sigma}{\sqrt{n}} = \frac{4.0}{\sqrt{4}} = 2.0^\circ\text{C} \] \[ z = \frac{68.0 - 65.0}{2.0} = 1.50 \] \[ P(\bar{X} > 68.0) = P(Z > 1.50) = 1 - 0.9332 = 0.0668 \]
(ii) No. Under the assumption that the process is operating under control (\(\mu = 65.0^\circ\text{C}\)), the probability of obtaining a sample mean of \(68.0^\circ\text{C}\) or higher is approximately \(0.0668\). Because this probability is relatively large (greater than standard significance thresholds like \(0.05\) or \(0.01\)), observing a sample mean of \(68.0^\circ\text{C}\) could reasonably occur due to chance variation alone and does not provide convincing statistical evidence that the true mean has shifted.
---
Part (c): (i) Because the \(4\) microchips are independently sampled, the probability that all \(4\) microchips simultaneously have an operating temperature less than or equal to \(71.0^\circ\text{C}\) is the product of their individual probabilities: \[ P(X_1 \le 71.0 \cap X_2 \le 71.0 \cap X_3 \le 71.0 \cap X_4 \le 71.0) = [P(X \le 71.0)]^4 \] From part (a), \(P(X \le 71.0) = 1 - 0.0668 = 0.9332\). \[ [P(X \le 71.0)]^4 = (0.9332)^4 \approx 0.7601 \]
(ii) The maximum \(M = \max(X_1, X_2, X_3, X_4)\) exceeds \(71.0^\circ\text{C}\) if and only if not all microchips are \(\le 71.0^\circ\text{C}\) (i.e., at least one microchip exceeds \(71.0^\circ\text{C}\)): \[ P(M > 71.0) = 1 - P(M \le 71.0) = 1 - [P(X \le 71.0)]^4 = 1 - 0.7601 = 0.2399 \]
---
Part (d): (i) As the population standard deviation \(\sigma\) increases from \(4^\circ\text{C}\) to \(8^\circ\text{C}\): - Center: The mean of the sampling distribution of \(M\) increases (from \(69.1^\circ\text{C}\) to \(73.2^\circ\text{C}\)). - Variability: The spread/variability of the sampling distribution of \(M\), as measured by the standard deviation of \(M\), increases (from \(2.4^\circ\text{C}\) to \(4.8^\circ\text{C}\)).
(ii) Yes, a sample maximum of \(M = 78.5^\circ\text{C}\) is unusual if the process is under control with \(\sigma = 4^\circ\text{C}\). Under the controlled condition (\(\sigma = 4\)), the sampling distribution of \(M\) has a mean of \(69.1^\circ\text{C}\) and a standard deviation of \(2.4^\circ\text{C}\). The standardized value of \(M = 78.5^\circ\text{C}\) is: \[ z = \frac{78.5 - 69.1}{2.4} = \frac{9.4}{2.4} \approx 3.92 \] Because \(78.5^\circ\text{C}\) is nearly \(4\) standard deviations above the mean (well beyond \(2\) or \(3\) standard deviations), observing such a large maximum is extremely rare under proper calibration.
Marking scheme
Scoring Guidelines for Question 6 (Investigative Task)
This question is scored in four sections: Part (a), Part (b), Part (c), and Part (d). Each section is scored as Essentially Correct (E), Partially Correct (P), or Incorrect (I).
---
### Part (a) Essentially Correct (E) if the response satisfies the following two components: 1. Identifies the normal distribution with mean \(65\) and standard deviation \(4\), with correct boundary \(71.0\) and direction (\(X > 71.0\) or \(Z > 1.50\)). 2. Calculates the correct probability of approximately \(0.0668\).
Partially Correct (P) if the response satisfies only one of the two components.
Incorrect (I) if the response does not meet the criteria for E or P.
---
### Part (b) Essentially Correct (E) if the response satisfies all three components: 1. In (b)(i), correctly calculates \(\sigma_{\bar{X}} = \frac{4}{\sqrt{4}} = 2.0\) and finds the probability \(P(\bar{X} > 68.0) \approx 0.0668\). 2. In (b)(ii), correctly states that observing \(\bar{X} = 68.0\) does not provide convincing evidence that the true mean has increased. 3. In (b)(ii), provides a valid justification based on the probability from (b)(i) not being unusually small (e.g., \(0.0668 > 0.05\)) or noting \(68.0\) is within \(1.5\) standard errors of \(65.0\).
Partially Correct (P) if the response satisfies two of the three components.
Incorrect (I) if the response satisfies at most one component.
---
### Part (c) Essentially Correct (E) if the response satisfies all three components: 1. In (c)(i), justifies the product rule by appealing to the independence of the \(4\) microchips. 2. In (c)(i), computes \([P(X \le 71.0)]^4 = (0.9332)^4 \approx 0.7601\) (or consistent with part (a)). 3. In (c)(ii), computes \(P(M > 71.0) = 1 - 0.7601 = 0.2399\) (or consistent with (c)(i)).
Partially Correct (P) if the response satisfies two of the three components.
Incorrect (I) if the response satisfies at most one component.
---
### Part (d) Essentially Correct (E) if the response satisfies all three components: 1. In (d)(i), states that both the center (mean) and variability (standard deviation/spread) of the sampling distribution of \(M\) increase as \(\sigma\) increases. 2. In (d)(ii), concludes that \(M = 78.5^\circ\text{C}\) is unusual. 3. In (d)(ii), justifies this conclusion by referencing the mean (\(69.1\)) and standard deviation (\(2.4\)) when \(\sigma = 4\) (e.g., showing \(z \approx 3.92\) or noting \(78.5 > 69.1 + 3(2.4) = 76.3\)).
Partially Correct (P) if the response satisfies two of the three components.
Incorrect (I) if the response satisfies at most one component.
---
### Composite Score Conversion: - 4 Points (Complete Response): 4 E's (or 3 E's and 1 P with holistic strength) - 3 Points (Substantial Response): 3 E's and 1 I, OR 2 E's and 2 P's - 2 Points (Developing Response): 2 E's and 0–1 P, OR 1 E and 2–3 P's, OR 4 P's - 1 Point (Minimal Response): 1 E and 0–1 P, OR 2 P's and 2 I's - 0 Points: Does not meet criteria for 1 point.
Wondering how well you actually know this?
thinka is an AI practice app for IGCSE & IB students: unlimited questions, instant auto-marking, and detailed step-by-step solutions. 100,000+ students use it to confirm they actually know it, not just think they do.