L-DS-5: Solutions — Position and Distribution Shape

A tag such as (C2 + C3) after a title names the Core Concepts parts the problem draws on, so you know where to look back.

Problem 1 — Relative Position and Z-Score Computation (C2 + C3)

This problem has two generated rounds: a guided calculation followed by a fresh solo calculation. The values change, but the work always follows the same sequence.

1. Compute the signed distance from the mean:

2. Divide that distance by the standard deviation:

3. Interpret both parts of the result:

   • The sign says whether the value is above or below the mean.
   • The magnitude says how many standard deviations separate the value from the mean.

For example, if , , and :

The value is standard deviations above the mean. The generated feedback attached to each round supplies the corresponding calculation for its values.

Problem 2 — Interpreting Percentile Rank and Score (C1)

Correct answer: Under the nearest-rank convention, 68 is the value, so at least 50% of observations are at or below 68.

The statement identifies 68 as the selected nearest-rank value at the 50th percentile. It does not determine the exact percentage strictly below 68 because finite sample size and ties can change that share.

The other statements confuse different quantities:

• The raw score 68 is not a percentile rank of 68.
• means the 75th-percentile score is 79; it does not mean a score of 75.
• identifies a percentile boundary; it does not say that exactly half of all test-takers scored 68.

Problem 3 — Application of the Empirical Rule (C5)

This problem also has a guided generated round followed by a fresh solo round. First identify how many standard deviations separate each boundary from the mean.

For a single, approximately symmetric bell-shaped distribution:

Standardized band	Approximate coverage

For example, suppose and .

The interval from 80 to 120 has boundaries:

The interval is therefore , so it contains approximately of the observations.

For a one-tail question below :

Approximately lies below . Use the feedback attached to each generated round for its numerical boundaries.

Problem 4 — Shape Classification and Measure Selection (C4)

Correct answer: Right-skewed; the median is a more appropriate measure of centre.

The mean salary of 112,000 dollars is substantially above the median salary of 84,000 dollars. Among the supplied choices, that pattern is consistent with a long upper tail or a few very high salaries pulling the mean upward. The median is resistant to those extreme values and is therefore the more defensible measure of a typical salary.

A mean–median comparison is evidence about shape, not a mathematical proof by itself. A histogram or the raw salaries would provide the direct shape evidence.

Problem 5 — Z-Score vs. Percentile Rank (C2 + C3 + C5)

The teacher computed the z-score correctly:

The percentile claim is not justified by the Empirical Rule. The rule’s describes the proportion within one standard deviation of the mean, from to . It does not say that a score at has percentile rank 68. Converting to an exact normal percentile requires a normal table or software.

A tag such as (C2 + C3) after a title names the Core Concepts parts the problem draws on, so you know where to look back.

Problem 1 — Z-Score Generator (C2 + C3)

The activity presents a generated measurement and offers a solo attempt before its step-by-step route. For the values in your version:

1. Compute .
2. Compute .
3. Round as requested.
4. Interpret the sign and the distance in standard deviations.

For example, if , , and :

The value is standard deviations above the mean. This standardized distance alone does not supply an exact probability or percentile.

Problem 2 — Percentile Generator (C1)

The dataset and target percentile change. The generator uses the nearest-rank convention:

For example, consider the sorted values

To find :

The eighth value is 46, so:

The rule targets and selects rank 8. In this 10-value dataset, exactly of the observations are at or below 46, while are strictly below it. The nearest-rank guarantee is that at least are at or below the selected value. Do not report the rank as though it were the data value.

Problem 3 — Empirical Rule (C3 + C4 + C5)

A. Central coverage

Within two standard deviations of the mean, the answer is approximately:

B. One-tail coverage

Within one standard deviation lies approximately , leaving:

Symmetry splits that outside proportion equally:

Therefore, approximately lies below .

C. Is the rule applicable?

Use the Empirical Rule only when the supplied evidence supports a single, approximately symmetric bell-shaped distribution.

• A bell-shaped, symmetric distribution with no prominent outliers supports the rule.
• A long tail, a strong skew, or prominent outliers does not.
• A large sample does not turn a skewed distribution of individual observations into a normal one.
• Replacing the mean with the median does not repair the shape condition.

D. Three-standard-deviation bounds

For the generated and :

An approximately normal distribution has about of its observations between those bounds.

E. An unusual value

The generated value is four standard deviations above the mean:

Because the distribution is stated to be approximately normal, a value more than three standard deviations from the mean is statistically unusual. “Unusual” is a flag for investigation, not proof that a measurement or record is wrong.

Problem 4 — Find the Error (C2 + C3)

The generated error-analysis item draws from several z-score misconceptions. Identify the first incorrect step, then repair it.

Possible repairs include:

• Reversed subtraction: replace with .
• Variance used as the denominator: divide by , not .
• Sign misinterpreted: a negative z-score is below the mean and a positive z-score is above it.

For example, if , , and :

The score is standard deviations above the mean.

Problem 5 — Multi-Step Synthesis (C2 + C3 + C5)

This generator places specification limits symmetrically around the original mean. Complete the work in this order:

1. Standardize the lower specification limit.
2. Standardize the upper specification limit.
3. Match the resulting band to the Empirical Rule.
4. Standardize the measured item and compare it with the specification z-scores.
5. Recalculate both specification z-scores after the process mean shifts.

If the original limits are and :

The corresponding approximate central coverages are for , for , and for .

After a mean shift to , the physical limits stay fixed but their z-scores change:

They are no longer a symmetric integer-standard-deviation band, so the clean Empirical Rule percentage no longer applies.

Mixed Review — Retrieval from Earlier Lessons

Review Problem 1 — Mean, Median, and Outlier Sensitivity

For the generated ten-value dataset:

1. Add all ten values and divide by 10 to find the mean.
2. Sort the values and average positions 5 and 6 to find the median.
3. Prefer the median when one unusually high value pulls the mean upward.
4. If only the largest value becomes more extreme, the mean increases while the median stays fixed.

Review Problem 2 — IQR, Outlier Detection, and Spread Comparison

Using the stated median-of-halves convention:

1. Find from the lower half.

2. Find from the upper half.

3. Compute:

4. Compute the upper fence:

5. A value above that fence is a potential high outlier. A larger IQR means the middle of the observations are more spread out.

A tag such as (C2 + C3) after a title names the Core Concepts parts the problem draws on, so you know where to look back.

Question 1 — Feynman Test

“Below average” gives only direction. A z-score of gives direction and standardized distance: the value is standard deviations below the mean. Because z-scores are unitless, that distance can also be compared with standardized positions from other datasets.

The z-score alone does not give an exact percentile or probability. Those conclusions require information about the distribution’s shape and, for an exact normal percentile, a normal table or software.

Question 2 — Novel Application

Amira earned 90 on the North assessment, whose scores are approximately normal with mean 76 and standard deviation 8. She also recorded 60 service hours in the Arts comparison group, whose hours are strongly right-skewed with mean 42, median 30, and standard deviation 12.

The fair plan is to compare standardized distances for both results, while reserving the Empirical Rule for an approximately normal distribution.

North:

Arts:

Amira stands farther above the North mean than the Arts mean because . The raw values 90 and 60 were not directly comparable because the programs use different scales.

No exact tail probability follows from the supplied summaries. North is approximately normal, but is not one of the Empirical Rule’s , , or standard-deviation boundaries. Arts is right-skewed, so its z-score compares standardized distance but does not justify an Empirical Rule probability. Percentile ranks, raw counts, a normal table, or software would be needed for a more specific claim, depending on the distribution and question.

Question 3 — Error Analysis

The error-analysis generator can present different incorrect z-score methods. In every version:

1. Locate the first incorrect step.

2. State the misconception precisely.

3. Recompute with:

4. Interpret the repaired sign and magnitude in context.

The three possible errors are:

• Dividing the raw score by the mean instead of centring and scaling.
• Reversing while substituting.
• Computing the z-score correctly but interpreting its sign backward.

The generated feedback identifies the exact repair for your version.

A tag such as (C2 + C3) after a title names the Core Concepts parts the problem draws on, so you know where to look back.

Case A — Exam Review

The exam scores are approximately normal with and . The three flagged scores are 89, 45, and 71.

Task 1 — Select the evidence

Use z-scores to place each record on the same standardized exam scale. A z-score measures relative position; it does not establish that a grade is erroneous.

Task 2 — Compute Student 1’s standardized position

Student 1:

Task 3 — Compute Student 2’s standardized position

Student 2:

Task 4 — Compare all three records and communicate a responsible conclusion

Student 3:

Compare absolute z-scores:

Student 2 is slightly farther from the mean than Student 1. Student 3 is close to the mean. The signs show that Students 1 and 2 lie on opposite sides of the mean.

The Empirical Rule says approximately of scores lie within two standard deviations of the mean. Student 2 lies just outside that band; Student 1 lies just inside it. The rule does not provide exact percentile ranks for either score.

A defensible memo would say:

Student 2’s score is the farthest from the exam mean in standardized terms, Student 1’s score is nearly as far in the opposite direction, and Student 3’s score is close to the mean. These positions justify a routine review of scoring, data entry, or documented circumstances under the same policy used for every student. They do not justify changing anyone’s grade automatically.

The institutional safeguard matters: a statistical flag must not become a discretionary grade adjustment or unequal treatment. Apply a documented, consistent review process and retain a human check of the underlying record.

Case B — Production Review

Line A has mm, mm, and an approximately bell-shaped symmetric histogram. Its specification limits are 7.1 mm and 8.9 mm.

Line B has mm, mm, median 10.5 mm, and a long right tail. Its specification limits are 10.0 mm and 14.0 mm.

Task 1 — Select the method for each line

The Empirical Rule is justified as an approximation for Line A because its distribution is described as approximately bell-shaped and symmetric.

The Empirical Rule is not justified for Line B because its distribution is strongly right-skewed. The mean and standard deviation can still be computed, but the –– coverage pattern is not supported.

Task 2 — Standardize Line A’s limits

Lower limit:

Upper limit:

Task 3 — Estimate Line A’s nonconforming proportion

The specifications are exactly . Under the approximately normal model:

The approximately is the total across both tails, or approximately per tail by symmetry.

Task 4 — State what can be concluded for Line B

The supplied summaries do not determine Line B’s proportion outside 10.0–14.0 mm. Its right-skewed shape rules out an Empirical Rule estimate, and the mean, median, and standard deviation do not reveal the required counts.

A defensible recommendation is to inspect the raw measurements or a well-labelled frequency distribution, count observations below 10.0 mm and above 14.0 mm directly, report those observed proportions, and investigate the process conditions producing the long upper tail.

A tag such as (C2 + C3) after a title names the Core Concepts parts the problem draws on, so you know where to look back.

Challenge 1 — Why Can Mean Exceed Median?

A. Explain an unfamiliar dataset

For the values :

The median is the middle value:

The value 93 is far from the other four values. Increasing it changes the sum and therefore increases the mean, while the middle rank remains 5 and the median stays fixed.

B. Make the general argument

Let . Hold the sorted values fixed, and vary an already-largest value while preserving . Let:

Define the median by parity:

Because , every order statistic defining lies among . Those values are fixed, so remains fixed while varies.

The mean is:

Now solve the condition :

Therefore, among values of that preserve the sorted order, the mean is greater than the median exactly when . This proves the claim for this constructed family; it does not make mean greater than median a definition of every right-skewed distribution.

C. Explain the location of the centre

The mean acts like the balance point of equal weights placed along a number line, so the distance of every value matters. A few values far to the right exert a strong pull on that balance point. The median depends on rank rather than distance, so moving an already-largest value farther right need not move the median at all.

D. Generalize with a new dataset

The generated values change, but the method does not. Let the first four sorted values have sum , and let their third value be the median after a larger fifth value is appended.

For five observations:

The condition becomes:

Because the generated activity asks for an integer, the smallest valid value is the first integer strictly greater than . This establishes the threshold for that generated family. It does not make a definition of right skew, because distribution shape is determined from the full pattern and especially the tail.

Challenge 2 — Z-Scores Preserve Relative Order

Suppose and .

Part (a)

Because :

Because :

Therefore:

Part (b)

If standardizing could reverse the order of two values, z-scores could not reliably compare relative position within or across datasets.

Part (c)

If , all observations are identical to the mean. The expression:

would require division by zero, so the z-score is undefined. There is no spread against which to standardize relative position.