Stats.

Add all values: 72 + 85 + 90 + 68 + 95 = 410. Divide by the number of values (n = 5): 410 ÷ 5 = 82. The mean formula is \(\bar{x} = \frac{\sum x_i}{n}\).

The $1,200K house is an outlier that drastically inflates the mean. Sorted data: 185, 195, 205, 210, 220, 340, 1200. The median is the 4th value = 210. The mean would be ≈ 365K — not representative of the typical house. When outliers exist, median is the better measure of center.

Key rule: Mean < Median → Left (negative) skew. Mean > Median → Right (positive) skew. Here mean (50) < median (65), so the distribution is left-skewed. Low outliers drag the mean down below the median. Think of it as the "tail" pointing left.

IQR = Q3 − Q1 = 54 − 25 = 29. The IQR measures the spread of the middle 50% of data. It is resistant to outliers because it ignores the top 25% and bottom 25% of values.

IQR = 16 − 8 = 8. Upper fence = Q3 + 1.5 × IQR = 16 + 12 = 28. Since 55 > 28, it IS an outlier. Lower fence = Q1 − 1.5 × IQR = 8 − 12 = −4. No values are below −4. The max value CAN be an outlier — that's a common misconception!

From the box plots: Class A box spans ≈ 55 to 85 (IQR ≈ 30). Class B box spans ≈ 70 to 85 (IQR ≈ 22). So Class B has a smaller IQR — definitely true. (A) is wrong — box plots don't show means. (B) is wrong — individual scores can't be compared between classes. (C) confuses range with IQR — range is larger for A, but IQR is the better measure of spread.

Standard deviation measures spread, not center. Same mean = same average height. But Team B (SD = 6") has much more variability — some players might be 5'8" while others are 6'8". Team A (SD = 1") is very consistent — nearly everyone is right around 6'2".

Since this is a sample, use n − 1 = 4 in the denominator. Variance = 0.528 ÷ 4 = 0.132. Standard deviation = √0.132 ≈ 0.363. Why n − 1? The sample tends to underestimate the true population spread, so we divide by a smaller number to correct upward. This is called Bessel's correction.

Adding a constant to every value shifts the distribution but does NOT change the standard deviation. Think about it: if every bolt gets 0.5mm added, each bolt is still the same distance from the (new) mean. SD measures deviation from the mean — and those deviations don't change. However, multiplying by a constant DOES change the SD (new SD = old SD × constant).

70 = 100 − 2(15), and 130 = 100 + 2(15). So 70 and 130 are exactly 2 standard deviations from the mean. By the empirical rule, 95% of data falls within ±2σ of the mean in a normal distribution.

76 inches = 70 + 2(3), so it's 2 standard deviations above the mean. The 95% rule means 5% falls outside ±2σ. By symmetry, half of that (2.5%) is in each tail. So 2.5% of men are taller than 76 inches. (Note: options B and C are both 2.5% here — the correct reasoning leads to 2.5%.)

9 oz = 8 + 2(0.5), which is exactly 2 SDs above the mean. By the empirical rule, 95% of cups fall within ±2σ, so 5% fall outside. Since we only care about the upper tail (overflow), it's 5% ÷ 2 = 2.5%. Out of 1,000 cups: 1,000 × 0.025 = 25 cups.

z = (750 − 520) / 115 = 230 / 115 = 2.0. This means her score is exactly 2 standard deviations above the mean. From the empirical rule, about 97.5% of test-takers scored below her.

Alex: z = (88 − 80) / 10 = 0.8. Jordan: z = (72 − 60) / 8 = 1.5. Jordan's z-score (1.5) is higher than Alex's (0.8), meaning Jordan scored further above the class average. Z-scores allow fair comparison across different scales — raw scores alone are misleading.

The z-table gives the area to the LEFT of z. If the area left of z = 1.28 is 0.8997 (89.97%), then the area to the right (above) is 1 − 0.8997 = 0.1003 ≈ 10.03%. Always remember: total area under the curve = 1 (= 100%).

To find x from z: rearrange the z-score formula. x = μ + z · σ = 500 + (1.28)(100) = 500 + 128 = 628. This is the "reverse z-score" problem — common on exams! Always check: does the answer make sense? 628 is above the mean (500), which makes sense for the 90th percentile.

Dataset X has SD = 0 — every value IS the mean, so every deviation is 0. Dataset Y has values spread away from 5: deviations are −2, −1, 0, +1, +2. Therefore Y has a larger SD. Key insight: SD = 0 means ALL values are identical. SD can never be negative.

Population = the entire group of interest (all 800 students). Sample = the subset selected (40 students). Parameter = a number that describes the population (usually unknown). Statistic = a number that describes the sample (calculated from data). So the sample average is a statistic, and the true school-wide average is a parameter. Memory trick: Population → Parameter · Sample → Statistic

z = (15.2 − 16.0) / 0.4 = −0.8 / 0.4 = −2.0. The probability of being underfilled = area left of z = −2.0 = 0.0228 = 2.28%. Expected underfilled boxes = 5,000 × 0.0228 = 114 boxes. Don't confuse 0.0228 with 0.228 — the decimal placement matters!

Skew check: Mean (74) < Median (80) → left skew ✓. Outlier check: IQR = Q3 − Q1 = 85 − 65 = 20. Lower fence = Q1 − 1.5(IQR) = 65 − 30 = 35. Since 30 < 35, the value 30 IS an outlier ✓. Both claims are correct! The minimum value CAN be an outlier — this is a very commonly missed point on exams.