L-INF-1: Solutions — Sampling Distributions and the CLT

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 — Sample-Mean Probability (C2 + C5)

This problem is generated, so its population, sample size, threshold, and tail change. The method does not:

1. Identify that the statistic is a sample mean.
2. Compute the standard error:

3. Standardize the observed sample mean:

4. Read the requested tail. The z-table gives a left-tail area, so a right-tail question requires a complement.

Problem 2 — Central Limit Theorem Conditions (C3)

• A random independent sample of from a moderately right-skewed population with no severe outliers supports a reasonable normal approximation for .
• If the population is normal, the sampling distribution of is normal for any sample size.
• A sample of from an extreme bimodal population is not enough information to trust the normal approximation. Use a larger sample or another justified method.
• Sixty consecutive wait times from one lunch rush are not independent: each wait depends on the queue before it. A larger does not repair dependence, so would understate how much the average really varies.

The number 30 is a practical introductory guideline, not a universal guarantee. The quality of the approximation also depends on population shape, independence, and the existence of a finite variance.

Problem 3 — Comparing Sample-Size Effects (C4 + C5)

This problem is generated. For the two sample sizes:

The larger sample has the smaller standard error. Use the standard error for the sample size named in the probability question:

Then take the requested left or right tail. Changing changes the spread of the sampling distribution, not its centre .

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 — Which Distribution Is It? (C1)

Ask what each statement is about:

• Population: every customer’s wait. It has mean and SD , it is right-skewed, and sampling again does not change it.
• One sample: the 25 recorded waits. Its shape copies the population’s skew with random bumps, and its mean changes from sample to sample.
• Sampling distribution: the distribution of over all possible samples of 25. Each value in it is an average, its SD is , it is roughly normal, and no one ever observes it directly.

Problem 2 — When Does the CLT Apply? (C3)

Check the conditions in this order:

1. Independence. Repeated readings on one person or one phone, consecutive customers in a queue, and whole classrooms are linked observations. No sample size repairs that: these cards belong in “more data will not fix it”.
2. A normal population gives exactly normal sample means at any .
3. Shape against . Moderate skew is fine around ; strong skew needs more (about 100 is enough without extreme outliers); extreme outliers or income-like skew need far more than 30.

Problem 3 — Sampling Distribution Probability (C2 + C5)

The generated solution should show these three computations vertically:

For an upper-tail question, use . From this problem on, can be any two-decimal value; read it from the course z-table.

Problem 4 — Three Inspection Plans (C2 + C4 + C5)

For each plan compute , then where is the distance from the target to the cutoff, then the requested tail. The cutoff stays the same, but each larger sample has a smaller SE, so the same shortfall is more standard errors away and its probability shrinks. Quadrupling halves the SE.

Problem 5 — Probability for a Non-Normal Population (C5)

The stem states that the normal approximation is adequate, so this problem practises the calculation. Compute and standardize every boundary. For an interval,

Problem 6 — One Box or the Average of 36? (C1 + C2)

The only difference between the cases is the random object. One box varies with :

An average of 36 boxes varies with :

Problem 7 — Where Do Most Sample Means Land? (C2 + C5)

The middle 95% of sample means lies within standard errors of :

Common slips: using (that describes single items), using (only about 68%), or (the middle 90%).

Problem 8 — Diagnose a Sampling-Distribution Error (C2 + C3 + C5)

The generator varies the scenario and rotates among four named error types: uses the population SD instead of the standard error (Step 2), divides by instead of (Step 2), drops the sign of (Step 3), and uses the opposite tail (Step 4). Check the analysis in this order:

1. Model and conditions: Does the situation justify a normal sampling model for ?
2. Standard error: Was used rather than or ?
3. Standardization: Does have the correct sign?
4. Tail: Does the probability match the inequality in the question?

Correct the first flawed step, then carry that correction through the remaining work. The exact corrected number depends on the generated case.

Mixed Review — Retrieval from Earlier Lessons

Review Problem 1 — Variability and Spread (L-DS-4)

For a sample, subtract the sample mean, square and sum the deviations, divide by , and take the square root:

Review Problem 2 — Normal Distribution Probability (L-PR-6)

Standardize both endpoints with . For a central interval, subtract the two left-tail areas:

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

A strong explanation includes these ideas:

1. A sampling distribution describes the values of across all possible random samples of the same size.
2. Its centre is the population mean, and its variance and standard error are

when the normal model is justified. 3. If the population is normal, the sampling distribution is normal for any . For a non-normal population with finite variance, the CLT gives an increasingly good normal approximation as grows; the required size depends on population shape. 4. Random independent sampling matters. When sampling without replacement, checking that the sample is no more than about 10% of the population is a common way to support approximate independence. 5. The result matters because it lets us calculate how unusual a sample mean is and supports later inference.

Question 2 — Which Distribution?

The situation is generated. First decide what is random: one observation or the average of a random sample. One observation has centre and spread . A sample mean has the same centre but spread . Neither centre is ever divided by , and the centre and standard error hold whatever the population’s shape.

Question 3 — Is a Normal Model Justified?

The study is generated and falls into one of four cases:

1. moderate skew, no severe outliers, random independent sample, large : a normal approximation is reasonable (not exact);
2. a normal population: is exactly normal for any ;
3. an extreme shape with a small : proceed with caution, because the approximation may be poor;
4. values that depend on one another (a sequence of related readings): the observations are not independent, and no sample size fixes that.

Question 4 — Scaling the Sample

The numbers are generated. The standard error follows , not : multiplying the sample size by divides the SE by , and dividing the SE by requires times as many observations. Solve for the requested value, or scale the current by .

Question 5 — A Quality-Control Decision

The numbers and direction are generated. Your first attempt asks for the one-sided probability; the diagnostic steps then lead to the recommendation. A complete solution should:

1. state the sampling model and check the supplied conditions;
2. compute ;
3. compute ;
4. use the tail matching the observed direction;
5. apply the displayed 5% flagging rule and explain the decision in the scenario’s units.

If , the displayed policy flags the line when the lower-tail probability is below 5%. If , apply the same 5% rule to the upper-tail probability. This is a task-specific action policy, not a universal statistical convention. A generated answer must never describe an above-target result as “unusually low,” or vice versa.

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.

Path A — 🔬 The Quality Analyst

The numbers are generated; the structure is always the same.

1. The random quantity is for samples of boxes, centred at the target with .
2. Standardize: , which is negative because the mean is below target.
3. The lower-tail probability is above 5% for the first sample, so under the plant’s rule the team does not investigate yet. Being below target is not by itself evidence of a problem.
4. With boxes, halves and doubles. The same average now has a lower-tail probability well below 5%, so the team investigates.

Reflection: the yardstick changed, not the average. A small probability makes the on-target model hard to believe; it does not prove underfilling. Check the scale, the sampling, and whether the shortfall matters.

Path B — 🏫 The School Board Analyst

Scores have and . Each school’s average is a sample mean.

1. Small school: (20 or 25). Large school: (4 or 5).
2. The small school’s higher average is only about 1.1 to 1.7 standard errors above 500, so a typical school of that size reaches it fairly often (a probability of roughly 0.05 to 0.14).
3. The large school’s lower average is about 2.3 to 2.9 standard errors above 500, which a typical large school reaches rarely (below 0.011).
4. Conclusion: the large school’s result is the harder one to explain by chance. The small school’s high average is weak evidence, and small schools also fill the bottom of rankings, so the ranking does not support splitting large schools.

Reflection: compare schools on the same yardstick (standard errors), follow the same schools over several years, or compare small and large schools serving similar students.

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 — Precision by Design

This challenge is generated. If the target is , then

Round the result up to the next whole number:

This calculation guarantees the requested standard-error bound. It does not, by itself, establish the conditions for a later normal-probability calculation.

Challenge 2 — The Finite Population Correction

For sampling without replacement from a finite population,

With , , and :

The correction reduces the standard error because sampling 25% of a finite population leaves less uncertainty than sampling the same number from an effectively infinite population.

The generated problem uses a different population each time. Compute , multiply by (the square root, not the fraction itself), and round to three decimals at the end.

Challenge 3 — Why Averaging Reduces Variability

Let be independent observations with .

Independence lets the variances add:

Therefore,

Taking the square root gives