Master Stats.

Add all values: 2 + 5 + 3 + 8 + 4 + 3 + 6 = 31
Divide by the number of values (7):
Mean = 31 ÷ 7 ≈ 4.43 hours Why not C (4)? — 4 is the median, not the mean. Sorted: 2, 3, 3, 4, 5, 6, 8 → middle value = 4.
Always distinguish: mean = arithmetic average, median = positional middle.

Median: Sorted: 42, 43, 45, 47, 210 → middle = $45K
Mean: (42+43+45+47+210) ÷ 5 = 387 ÷ 5 = $77.4K

The outlier ($210K) pulls the mean toward it but does NOT move the median. This is why:
→ Skewed-right data: mean > median
→ The median is called a resistant measure — it doesn't care about extreme values.

The key rule: mean < median → left skew (negatively skewed).

Here mean (69) < median (78), so the tail pulls to the left. This means a few students scored very low, dragging the mean down below the median.

Memory trick: The mean "chases the tail."
→ Tail on right → mean is pulled right → mean > median → right skew
→ Tail on left → mean is pulled left → mean < median → left skew

IQR = Q3 − Q1 = 35 − 20 = 15 Common mistakes:
→ A: That's the range (Max − Min), not IQR.
→ D: That's Median − Q1, which has no standard name.

The IQR tells us the spread of the middle 50% of the data. A larger IQR means the data is more spread out in the middle.

A — Actually true! The box (Q1 to Q3) always contains the middle 50%. ✓ But wait — option C is the best and most insightful statement.

B — Wrong! Range = Max − Min = 95 − 30 = 65, not 25. (25 is the IQR = 75 − 50.)

C ✓ — The median (65) is closer to Q3 (75) than to Q1 (50). This means the upper half of the box is compressed → left skew (the longer whisker is on the left).

D — The median is the 50th percentile. 50% of students scored above 65, not 75%.

1450 = Q3, which is the 75th percentile. This means the student scored at or above approximately 75% of all admitted students.

A — Wrong. Only above Q3 doesn't mean above all — 25% of students scored higher.
B/C — Wrong. 50% describes the median (Q2 = 1350), not Q3.
D ✓ — Correct. Q3 = 75th percentile means 75% of the data falls at or below this value.

The key question is: Are these all the individuals you care about, or just a subset?

Here, the 8 students ARE the entire group of interest → use population SD (σ), divide by N.

B/D — Wrong. Use n−1 only when you are drawing conclusions about a larger population from a subset.
C — Wrong. They give different (close but not equal) results.

Memory trick: Population = P = σ (sigma) | Sample = S = s

Population variance: σ² = (sum of squared deviations) ÷ N = 18 ÷ 4 = 4.5
σ = √4.5 ≈ 2.12 A — Wrong. σ² is the variance; σ requires a square root.
C — Wrong. Dividing by n−1 = 3 gives the sample standard deviation (s ≈ 2.45), not σ.
D — Wrong. 4.5 is the variance (σ²), not the standard deviation (σ).

Always remember: Standard Deviation = √(Variance)

Standard deviation measures spread/variability, not performance level.

Class A (SD = 3): most scores likely between 72 and 78. Consistent!
Class B (SD = 15): scores likely range from ~45 to ~105. Very spread out.

B — Wrong. Higher SD doesn't mean higher scores; the means are equal.
C — Misleading. Same mean ≠ same performance. The spread tells a very different story.
D — Wrong. SD has no relationship to the number of students (sample size = n).

First, count standard deviations from the mean:
70 − 64 = 6 inches = 2σ below the mean
76 − 70 = 6 inches = 2σ above the mean
So the range 64 to 76 = μ ± 2σ → 95% of data Quick check: 70 ± 1×3 = [67, 73] → 68% | 70 ± 2×3 = [64, 76] → 95% ✓ | 70 ± 3×3 = [61, 79] → 99.7%

130 − 100 = 30 = 2σ (since σ = 15)
Within 2σ (between 70 and 130): 95% of the population.
That leaves 5% outside the range — split equally in both tails (symmetry!):
Above 130 = 5% ÷ 2 = 2.5% A — 5% is in both tails combined, not just the upper one.
C — Wrong. 16% applies to 1σ above mean (score = 115), not 2σ.
D — Wrong. 3σ from mean = 100 ± 45 = [55, 145], not 130.

z = (x − μ) / σ = (85 − 75) / 5 = 10 / 5 = 2.0 This means the student scored 2 standard deviations above the mean.

A — Wrong. z=1.0 would mean scoring 75 + 1×5 = 80, not 85.
D — Wrong. Always subtract the mean first before dividing by σ.

Calculate both z-scores:
Math z = (680 − 650) / 30 = 30/30 = 1.0 English z = (75 − 65) / 8 = 10/8 = 1.25 Sarah's English z-score (1.25) > Math z-score (1.0).
She performed relatively better in English compared to her peers.

A — NEVER compare raw scores across different tests. They're on different scales!
Key lesson: z-scores allow fair comparison across different distributions.

First compute the z-score: z = (170 − 150) / 20 = 1.0

The z-table gives P(Z < 1.0) = 0.8413 — this is the area to the LEFT (below 170g).
We want apples above 170g:
P(X > 170) = 1 − P(Z < 1.0) = 1 − 0.8413 = 0.1587 = 15.87% A — Wrong. 84.13% is the proportion below 170g — those are rejected, not accepted.
Critical tip: The z-table always gives area to the LEFT. For "greater than," always subtract from 1.

From the Empirical Rule: 68% of data is within ±1σ. That means 32% is outside, and by symmetry, 16% is above 1σ from the mean.

So "top 16%" corresponds to z = +1.0.
x = μ + z·σ = 72 + (1.0)(10) = 82 B — Top 2.5% would need z = +2, giving x = 72 + 2(10) = 92. That's too strict.
Remember: To find a cutoff value, use x = μ + z·σ (the reverse z-score formula).

z = (35 − 45) / 8 = −10/8 = −1.25 The question asks P(X < 35) = P(Z < −1.25).
The z-table directly gives us the area to the left: 0.1056 = 10.56%

B — 89.44% is P(X > 35), not P(X < 35).
D — Wrong. Do not double the value unless finding a two-tailed probability.

Step 1 — Find z-scores:
z for 80°F: (80 − 78) / 6 = 2/6 ≈ 0.33 z for 74°F: (74 − 78) / 6 = −4/6 ≈ −0.67 Step 2 — Use subtraction:
P(74 < X < 80) = P(Z < 0.33) − P(Z < −0.67) = 0.63 − 0.25 = 0.38 B — NEVER add z-table values. Always subtract (bigger minus smaller).
D — The range is NOT symmetric: 80−78 = 2, but 78−74 = 4 (different distances from mean).

Sorted data (excluding 14.8): 10.1, 10.1, 10.1, 10.2, 10.2, 10.3, 10.3, 10.4, 10.5
Q1 ≈ 10.1, Q3 ≈ 10.4, IQR = 0.3
Upper fence = Q3 + 1.5 × IQR = 10.4 + 0.45 = 10.85 mm
14.8 mm >> 10.85 mm → definitely an outlier ✓

B — Wrong. Any single data point can be an outlier.
C — The opposite is true! Outliers make the median a better (more resistant) measure.
D — Wrong. Outliers strongly inflate the standard deviation (because squaring the large deviation makes it enormous).

97.5th percentile means 97.5% of students scored below her.
From the Empirical Rule: 95% of data is within 2σ. The upper 2.5% tail begins at z = +2.
x = μ + z·σ = 500 + (2.0)(100) = 700 A — z = +1 corresponds to the 84th percentile, not 97.5th.
D — Never convert percentile directly to raw score without using μ and σ. That logic is completely wrong.

Key percentile-z links to memorize:
84th ↔ z=+1 | 97.5th ↔ z=+2 | 99.85th ↔ z=+3 | 16th ↔ z=−1

This is the classic trap: same median ≠ same distribution.

Neighborhood A: prices clustered tightly between $220K–$280K. Low SD ≈ $18K. Predictable.
Neighborhood B: includes extreme values ($420K, $890K). Much higher SD ≈ $215K. Very unpredictable.

The medians may be similar, but the risk and variability are completely different.

A — Wrong. Two distributions with identical medians can look totally different (see above).
B — Wrong. With outliers, the mean of B is inflated (~$308K), making median better, not worse.
C — Partially true but incomplete — the key issue is overall variability, not just the max.

Lesson: Always report both a measure of center AND a measure of spread. Never judge a distribution by one number alone.