L-DS-5 Introduction to Probability and Statistics

L-DS-5 · Position and Distribution Shape

Is 78 a good score?

A raw value cannot describe its own position.

What would you need to know before answering?

Anonymous check

Which claim is most defensible?

A student scores 78. Class North has mean 70 and standard deviation 4. Class South has mean 84 and standard deviation 3.

  1. In both classes, 78 has the same relative position.

  2. In both classes, 78 lies above its class mean.

  3. Relative position changes: above North, below South.

  4. Neither class can be compared from its mean and SD.

Opening reveal · same raw score

The comparison group flips the story.

The numerator changes sign because the reference mean changes.

Class North \(78-70=+8\)

The score is above this class mean.

Class South \(78-84=-6\)

The score is below this class mean.

Raw values become interpretable only after the comparison distribution is named.

Inquiry map

Five decisions turn a value into a claim.

Each concept answers a question left open by the one before it.

  1. C1 Percentile

    Where is a boundary or value in the ordered data?

  2. C2 Compute z

    How many standard deviations from the mean?

  3. C3 Interpret z

    Which direction, distance, and comparison are justified?

  4. C4 Read shape

    Which tail is long, and what evidence supports that label?

  5. C5 Check the Rule

    Does bell-shaped evidence justify 68–95–99.7?

Comparison group → position measure → signed distance → shape condition → bounded claim.

C1 · Percentiles, quartiles, and deciles

One ordered dataset supports four position questions.

Name the convention, select the boundary, then distinguish it from the actual rank count.

Percentile boundary \(L=\left\lceil\frac{k}{100}n\right\rceil\)

Nearest rank selects the ordered value at position L.

Percentile rank \(\%\text{ strictly below }x\)

Count the data; finite samples and ties matter.

Quartile checkpoints \(Q_1=P_{25},\;Q_2=P_{50},\;Q_3=P_{75}\)

Keep the L-DS-4 median-of-halves convention.

Decile checkpoints \(D_k=P_{10k},\quad k=1,\ldots,9\)

D1 through D9 mark ordered tenths.

A selected boundary does not guarantee exactly k% strictly below it.

C1 · Compute

Nearest rank selects an observed boundary.

Sorted scores: \(12,\,15,\,18,\,21,\,24,\,27,\,30,\,33\)

  1. 1 · Name the method
    Nearest rank

    A different software convention may interpolate.

  2. 2 · Locate
    \(L=\left\lceil\frac{k}{100}n\right\rceil\)

    Here, L = ⌈0.75(8)⌉ = 6.

  3. 3 · Select
    \(P_{75}=27\)

    Six of eight values are at or below 27.

  4. 4 · Qualify
    \(5/8=62.5\%\)

    Five of eight values are strictly below 27.

Nearest rank guarantees at least 75% at or below the selected value—not exactly 75% strictly below it.

Anonymous check

Which complete report is correct?

Sorted quiz scores: 42, 48, 55, 55, 55, 72, 78, 83, 89, 95. Use nearest rank to find P30.

  1. L = 3; P30 = 55; 20% are strictly below 55.

  2. L = 3; P30 = 55; 50% are strictly below 55.

  3. L = 5; P30 = 55; 20% are strictly below 55.

  4. L = 3; P30 = 48; 20% are strictly below 48.

C2 · Standardize

A z-score is a signed distance in standard-deviation units.

Choose notation from the data source, then divide a centred difference by a positive spread.

Population \(z=\frac{x-\mu}{\sigma}\)

Use population mean and SD together.

Sample \(z=\frac{x-\bar{x}}{s}\)

Use sample mean and SD together.

Direction \(\operatorname{sign}(z)\)

Positive is above; negative is below.

Boundary \(\sigma>0\;\text{or}\;s>0\)

Zero spread makes standardization undefined.

Subtract the centre first; the SD converts the result into a unitless relative distance.

C2 → C3 · Compute and interpret

The same 78 lands at opposite standardized positions.

Now complete the opening comparison in common units.

  1. 1 · North
    \(z=\frac{78-70}{4}=+2\)

    Two SDs above the North mean.

  2. 2 · South
    \(z=\frac{78-84}{3}=-2\)

    Two SDs below the South mean.

  3. 3 · Compare
    \(|+2|=|-2|\)

    Equal standardized distance; opposite direction.

The raw score stayed fixed. The comparison group changed both its direction and meaning.

C3 · Flex visual · FLEX

Read z on a signed ruler.

Zero anchors the mean; equal steps represent equal standard-deviation distances.

Signed z-score ruler from negative three to positive three Class South's score 78 is marked at z equals negative two, the mean is marked at zero, and Class North's score 78 is marked at z equals positive two. -3 -2 -1 0 +1 +2 +3 South · 78 2 SDs below mean North · 78 2 SDs above

Sign gives direction. Absolute value gives distance.

C3 · Compare across scales

Unitless does not mean contextless.

Two unlike measurements can share a standardized position.

Discuss: What is legitimately equal in these two cases, and what would be an overclaim?

Temperature \(x=32^\circ\text{C},\;\mu=26,\;\sigma=4\)

\(z=+1.5\)

Reaction time \(x=410\text{ ms},\;\mu=350,\;\sigma=40\)

\(z=+1.5\)

Both values are 1.5 SDs above their means; “better,” percentile, and probability still require context and shape.

Anonymous check

Which conclusion is defensible?

Applicant A: x = 82, mean = 74, SD = 4. Applicant B: x = 91, mean = 85, SD = 2. Both SDs are positive; distribution shapes are not supplied.

  1. Applicant A: z = +2 is farther above its mean.

  2. They are tied: both raw scores exceed their means.

  3. Applicant B: z = +3, so its exact percentile is 99.7.

  4. Applicant B: z = +3 exceeds +2; percentiles remain unknown.

C4 · Distribution shape

Skew is named by the long tail.

Read the picture first; use centre order only as supporting evidence.

Left-skewed, symmetric, and right-skewed distribution shapes Three labelled curves emphasize that skew direction is named by the direction of the long tail. Left-skewed long tail ← Approximately symmetric no dominant tail Right-skewed long tail →

Tail right → right-skewed. Tail left → left-skewed. No dominant tail → approximately symmetric.

C4 · Supporting evidence

The mean is pulled toward the tail.

Extreme distances affect the arithmetic balance point more than the middle rank.

  1. Right tail
    Commonly mean > median

    Large values pull the mean right.

  2. Left tail
    Commonly mean < median

    Small values pull the mean left.

  3. Symmetric
    Mean ≈ median

    No dominant tail pulls one side.

Use the graph to establish skew; use mean–median order to check whether the story is coherent.

C4 · Discuss, then reveal

Which evidence establishes the shape?

A histogram has a long left tail. Its mean is 63 and its median is 71.

Discuss: Classify the shape, name the primary evidence, and choose the more representative centre.

  1. Primary evidence
    The long left tail

    This establishes left skew.

  2. Supporting clue
    Mean 63 < median 71

    Low values pull the mean toward the tail.

  3. Reporting choice
    Median for a typical value

    The median resists the low tail.

Left-skewed is a statement about the tail—not the peak and not centre order alone.

C3 + C4 · Flex challenge · FLEX

The same z-score can live in different shapes.

Both marked values have z = +1, but the two distributions are not shaped alike.

The same z-score shown in two differently shaped distributions Both marked values are one standard deviation above their means, but one distribution is bell-shaped and the other is right-skewed, so their percentile ranks need not match. Bell-shaped mean z = +1 Right-skewed mean z = +1

Standardized distance matches; exact percentile rank need not.

C5 · Applicability gate

Check the shape before using the Rule.

A raw value, centre, and positive SD support z-score computation; they do not by themselves justify 68–95–99.7.

Use approximately Symmetric + bell-shaped

A single mound with no strong skew or influential outliers.

Do not use Skewed, bimodal, or outlier-dominated

Request direct counts, the raw data, or a more suitable model.

Normal-shape evidence is a condition, not decoration.

C5 · Empirical Rule

Build 68–95–99.7 from the centre outward.

For an approximately normal distribution, the bands are symmetric around the mean.

Empirical Rule central standard-deviation bands A symmetric bell curve shows approximately 68 percent within one standard deviation, 95 percent within two, and 99.7 percent within three standard deviations of the mean. within 3 SDs · ≈99.7% within 2 SDs · ≈95% within 1 SD · ≈68% −3σ −2σ −1σ μ +1σ +2σ +3σ

Within 1 SD: ≈68%. Within 2 SDs: ≈95%. Within 3 SDs: ≈99.7%.

C5 · Worked example

Translate standard-deviation bands back to scores.

Exam scores are approximately normal with \(\mu=72\) and \(\sigma=8\).

  1. Within 1 SD
    \(72\pm8=[64,80]\)

    Approximately 68% of scores.

  2. Within 2 SDs
    \(72\pm16=[56,88]\)

    Approximately 95% of scores.

  3. Within 3 SDs
    \(72\pm24=[48,96]\)

    Approximately 99.7% of scores.

State the interval, percentage, approximation, and shape condition together.

Anonymous check

Which conclusion is defensible?

Line A is symmetric and bell-shaped with no influential outliers. Line B has a sharp peak and a long right tail. Both lines report a mean and standard deviation.

  1. Use the Rule for both; each line reports a mean and SD.

  2. Use the Rule for A; use direct counts or raw data for B.

  3. Use the Rule for B; its long right tail marks upper bands.

  4. Use the Rule for neither; standard deviation blocks normality.

C5 · Flex extension · FLEX

Symmetry splits the outside area into equal tails.

Only after the approximately normal condition is established may central coverage be split.

Outside 2 SDs \(100\%-95\%=5\%\)

Approximately 2.5% in each tail.

Outside 3 SDs \(100\%-99.7\%=0.3\%\)

Approximately 0.15% in each tail.

These are approximate normal-tail statements—not exact probabilities for arbitrary shapes.

Reusable decision chain

Make the claim only after five checks.

A calculation is defensible only when its comparison and model conditions travel with it.

  1. 1 Comparison group

    Name the distribution the value belongs to.

  2. 2 Position measure

    Boundary/rank question or standardized distance?

  3. 3 Signed distance

    Compute and interpret z with a positive SD.

  4. 4 Shape condition

    Read the tail; decide whether bell-shaped evidence exists.

  5. 5 Bounded claim

    State only the percentile, distance, or approximation the evidence permits.

Comparison group → position measure → signed distance → shape condition → bounded claim.

Independent transfer · commit first

Audit two production lines.

Write one recommendation for each line before the model appears.

Write: Which line supports an Empirical Rule estimate? What can the supplied summaries establish—and what can they not?

Line A \(\bar{x}=8.0\text{ mm},\;s=0.3\text{ mm}\)

Bell-shaped and symmetric; specifications 7.1–8.9 mm.

Line B \(\bar{x}=12.0\text{ mm},\;s=1.2\text{ mm}\)

Sharp peak at 10 mm, long right tail; specifications 10–14 mm.

For each line: choose a method, compute what is justified, and name the evidence still needed.

Independent transfer · model

A defensible recommendation separates evidence from action.

Compare your reasoning, not just your numerical answer.

  1. Line A · standardize
    \(\frac{7.1-8.0}{0.3}=-3,\quad\frac{8.9-8.0}{0.3}=+3\)

    The approximately normal model estimates ≈99.7% inside and ≈0.3% outside combined.

  2. Line B · stop the model
    Long right tail

    The summaries do not support a normal estimate; request direct counts or raw data.

  3. Operational boundary
    Evidence triggers review

    Neither an unusual value nor an approximation automatically authorizes process action.

Use the method the shape justifies, then make the smallest honest claim.

Anonymous check

Which report is statistically defensible?

A measurement has z = −1.5 in a clearly right-skewed distribution. No model, table, software, or raw-data counts are supplied.

  1. It is 1.5 raw units below the median; no percentile claim is available.

  2. It is 1.5 SDs below the mean; about 68% must fall below it.

  3. It is 1.5 SDs below the mean; right skew blocks a normal-percentile claim.

  4. It is at the 6.7th percentile; right skew confirms that rank.

Anonymous exit reflection · confidence, not mastery

How confident are you with each idea?

Use the exit diagnosis as evidence first. Then rate whether you could explain each statement—not merely recognize it.

C1

I can compute percentile, quartile, or decile boundaries and distinguish them from rank.

C2

I can compute a z-score with the correct centre and positive SD.

C3

I can interpret sign and standardized distance without inventing a percentile.

C4

I can identify skew from the long tail and use centre order as supporting evidence.

C5

I can check shape before using the approximate 68–95–99.7 Rule.

0 Not yet 1 Very unsure 2 Beginning 3 Mostly understand 4 Confident 5 I could explain it

Synthesis · enrichment bridge

78 never spoke for itself.

The comparison distribution turned one raw value into opposite relative positions.

North \(z=+2\)

Two SDs above its mean.

South \(z=-2\)

Two SDs below its mean.

The rule Check shape first

Approximate coverage needs bell-shaped evidence.

The claim Say only what follows

Distance, percentile, and probability are different statements.

Name the group. Locate the value. Read the shape. Bound the claim.