IC252 — Introduction to Statistics

The z-score formula is z = (X − μ) / σ. z_Arya = 1.50 | z_Bhanu = 0.50 | z_Charu = 0.75

A higher z-score means farther above the class mean. Ranking: Arya (1.50) > Charu (0.75) > Bhanu (0.50). Despite Charu's higher raw score (91) and Bhanu's second-highest raw score (85), Arya performed the best relative to her class because her raw score was 1.5 standard deviations above her class mean — the largest relative margin.

If every score increases by a constant c, the new mean becomes μ' = μ + c and the new standard deviation is unchanged (σ' = σ). Therefore: Arya's z-score remains exactly 1.50.

A uniform shift of origin does not change relative standing — everyone moves equally.

If every score is multiplied by k = 2, the new mean is μ' = 2μ = 176 and the new standard deviation is σ' = 2σ = 8. Charu's new score is 2 × 91 = 182. z-score is unchanged: 0.75

Principle: Multiplying all data by a nonzero constant scales both the mean and standard deviation equally, leaving z-scores invariant. The z-score is dimensionless and scale-free.

Rearranging z = (X − μ) / σ: Dev's raw score = 47

Interpretation: Dev scored 23 points below the class average, placing him 2.3 standard deviations below the mean. In a roughly normal distribution, fewer than ~2% of students would score this low, indicating Dev performed in the bottom tail of his class.

Mean ≈ ₹12.44L | Median = ₹7L | Mode = ₹5L and ₹8L

Mean dropped ~46.7% | Median dropped only ~7.1%

One extreme outlier caused almost a 47% shift in the mean but barely affected the median — a vivid demonstration of the median's robustness.

The original dataset is heavily right-skewed (mean ≫ median, due to the ₹60L outlier). In such distributions, the mean is pulled far to the right by the outlier, overestimating the salary of a typical employee. The median (₹7L) accurately reflects what most employees earn.

The 8 employees have mean = 6.625. Adding one employee at 7 lakhs gives n = 9. The new mean is: New mean ≈ ₹6.667 lakhs

TRUE — always exactly zero, for any dataset.

This is why we cannot use the average of deviations as a measure of spread — we must square them first (variance) or take absolute values (mean absolute deviation).

Variable	Level	Key Reason
Department	Nominal	Named categories with no natural order; CSE ≠ > EEE
Year of Study	Ordinal	Ordered (1st < 2nd < 3rd), but gaps between years are not equal in meaning
Temp (°C)	Interval	Equal units (1°C = 1°C everywhere), but 0°C does not mean "no temperature" — zero is arbitrary
Pocket Money (₹)	Ratio	True zero exists (₹0 = no money), ratios are meaningful (₹5000 is 2.5× ₹2000)
Laptop Ownership	Nominal (Binary)	Dichotomous category; no ordering, no scale

Variable	Mode	Median	Mean
Department (Nominal)	✅ Only valid option	❌ No ordering	❌ Not meaningful
Year of Study (Ordinal)	✅	✅ Ordering exists	⚠️ Technically invalid (unequal gaps)
Temp (°C) (Interval)	✅	✅	✅
Pocket Money (Ratio)	✅	✅	✅
Laptop Ownership (Binary)	✅ (Majority category)	⚠️ Tells only majority	✅ Gives proportion

Not rigorously valid. Year of study is an ordinal variable. While the labels 1, 2, 3, 4 are numbers, the difference between "1st year" and "2nd year" is not necessarily the same in academic workload, course count, or difficulty as between "3rd year" and "4th year." The arithmetic mean assumes equal intervals — a property ordinal data does not guarantee. "Year 2.5" has no real-world meaning.

Encoding: Yes=1, No=0 → data: 1, 0, 1, 1 Mean = 0.75 = 75% of students own a laptop

The mean of a binary 0/1 variable equals the proportion of "1"s in the sample. This is the one case where the mean is meaningful for nominal data.

No. Celsius is an interval scale — it lacks a true zero. 0°C does not mean "no heat"; it is the arbitrary freezing point of water. Therefore, ratios like "25°C is 25% warmer than 20°C" are mathematically invalid.

To make meaningful temperature ratios, you must convert to Kelvin (a ratio scale with absolute zero): 25°C = 298 K, 20°C = 293 K → ratio = 298/293 ≈ 1.017, i.e., only ~1.7% warmer in absolute thermodynamic terms.

Only Ratio scales support meaningful ratio comparisons.

Q1 = 7 mm | Q2 = 25 mm | Q3 = 58 mm

140 mm is a suspected outlier. No extreme outliers.

Conclusion: The box is asymmetric (upper half larger), and the upper whisker is dramatically longer. Distribution is strongly right-skewed (positively skewed).

This makes physical sense — rainfall has a hard lower bound of 0 mm but no upper limit (cloudburst events extend the right tail).

Range increased by 60 mm | IQR increased by only 4.5 mm

The range is extremely sensitive to the new outlier. The IQR, covering only the middle 50%, is largely resistant to it.

The claim is partially incorrect. The IQR describes the spread of only the central 50% of observations. It effectively ignores the bottom 25% and top 25% — the very extreme events that matter most to a meteorologist studying floods or droughts. In this dataset, IQR = 51 mm but the range is 137 mm; the IQR tells us nothing about the extreme rainfall of 80 or 140 mm. For applications where tails matter (risk, flood planning), range, 90th/99th percentiles, or full distributional analysis are needed.

s_A ≈ 0.141 mm | s_B ≈ 0.772 mm | Both have X̄ = 10.0 mm

Both machines have the same mean (10.0 mm), but Machine A has a standard deviation of only 0.141 mm vs. Machine B's 0.772 mm — more than 5× larger. Machine A is far more consistent. Despite producing identical average diameters, Machine B's bearings vary wildly, making them unsuitable for precision manufacturing.

When we compute a sample standard deviation, we first estimate the population mean μ using the sample mean X̄. This estimation "uses up" one piece of information — once we fix X̄ from n data points, only (n−1) of the deviations (Xᵢ − X̄) are free to vary (the last one is determined by the constraint that they sum to zero). Hence we have only n−1 degrees of freedom.

Dividing by n instead of n−1 produces a biased estimator that systematically underestimates the population variance σ². Dividing by (n−1) corrects this bias, making s² an unbiased estimator of σ².

n−1 ensures s² is an unbiased estimator of σ²

Here the target happens to equal the mean, so both sums are equal. This illustrates the least squares property: the sample mean minimises the sum of squared deviations. No other value, when subtracted from each data point and squared, produces a smaller sum. This is a defining optimality property of the mean. Σ(Xᵢ − c)² is minimised when c = X̄

Adding a constant to every value shifts the mean but does not change the spread. Standard deviation measures how data points deviate from the mean; if all values shift by the same amount, each deviation (Xᵢ + 0.5) − (X̄ + 0.5) = Xᵢ − X̄ is unchanged. Standard deviations remain unchanged: s_A ≈ 0.141 mm, s_B ≈ 0.772 mm

CV: Bond = 25% | Stock = 33.3% | Crypto = 360%

In this case both rankings agree. However, CV provides additional insight: the Crypto Token is not merely 18× riskier than the Bond in absolute terms — per unit of expected return, it carries 360/25 = 14.4× more relative risk. The CV properly normalises risk per unit of reward.

The analyst compares standard deviations across assets with very different scales of return. Crypto's std dev of 18% is large, but its mean return is also 5% — vs. Bond's mean of only 0.40%. A fair comparison must normalise by the mean (CV). The Bond's CV = 25% means for every 1% of expected return, there is 0.25% of risk. Crypto's CV = 360% means for every 1% of expected return, there is 3.6% of risk. The Bond carries far less risk per unit of expected return. The analyst's comparison is like comparing the weight fluctuation of a mouse (±5g) to an elephant (±10kg) and declaring the elephant is "heavier in every way" — context matters.

Standard deviation = 1.5%

CV is problematic when:

• Mean ≈ 0: If the mean return is close to zero or negative (e.g., an asset losing money on average), CV becomes meaningless or infinite — dividing by a near-zero denominator produces absurd numbers.
• Data can take negative values: CV assumes the mean is positive and meaningful as a denominator. For temperatures in Celsius (which can be negative), CV has no coherent interpretation.
• Comparing variables of different sign conventions (e.g., gains vs. losses).

Since Mean (5.50) > Median (3.75), the distribution is right-skewed (positively skewed). The 24-hour outlier pulls the mean rightward while barely affecting the median.

24.0 hours is confirmed as a suspected outlier.

Box plot description: The box spans from 2.25 to 5.5 hours (IQR = 3.25 hrs). The median line at 3.75 hours sits closer to Q1 than to Q3 (distance from Q1: 1.5 hrs; distance to Q3: 1.75 hrs), indicating slight asymmetry even within the box. The right whisker would extend to the fence at ~10.375 hrs (since the next-largest value is 8, which is inside the fence), while the left whisker extends to 1.5 hrs. The value 24.0 would be plotted as a separate dot (outlier). The right whisker is considerably longer.

• (i) Mean: Drops significantly. The outlier 24.0 contributed (24−5.5) = 18.5 hours above the mean. New mean = (66−24)/11 = 42/11 ≈ 3.82 hrs — a large decrease.
• (ii) Median: Shifts slightly. With 11 observations, the new median is the 6th value = 3.5 hrs (was 3.75). Small change — the median is robust.
• (iii) Standard Deviation: Decreases substantially. The 24-hour value was 18.5 hours above the mean; removing it dramatically reduces the sum of squared deviations.
• (iv) Skewness: Decreases (distribution becomes more symmetric). The right tail shrinks; mean and median move closer together.

Right-skewed: Income/wealth. The lower bound is ₹0 (you can't earn negative income), but there is no upper limit — a few billionaires extend the right tail to enormous values. Most people earn modest amounts, creating a long right tail.

Left-skewed: Test scores on a very easy exam. The upper bound is 100% — scores cannot exceed this. Most students score near the top (90–100%), while a few who struggle extend the left tail toward lower marks. The ceiling compresses the right side and forces the tail leftward.

~8,200 below | ~1,800 at or above

Priya (347, P82) and Sita (350, P83) differ by only 1 percentile despite a 3-mark difference. This means only about 0.01 × 10,000 = 100 students scored between 347 and 350. The distribution is sparse in this region. Contrast this with regions where many students cluster (e.g., a jump from P50 to P55 might cover just 2 marks in a dense middle region). Percentile rank changes fast where students are densely packed and slowly where they are sparse.

At least 50 days had "at least extreme" rainfall (≥ P95).

The 50th percentile (median) is the value at which half the data falls above and half below. The mean is the arithmetic average. They coincide only in symmetric distributions.

Example: Salaries: ₹20K, ₹25K, ₹30K, ₹35K, ₹500K (n=5). Median = ₹30K (50th percentile). Mean = ₹122K. These are drastically different. A student at the 50th percentile earned exactly ₹30K, not ₹122K.

In right-skewed data: mean > median. Scores at P50 ≠ mean.

Yes — they are all the same value, by definition: D5 = Q2 = P50 = Median — all three refer to the same quantity.

Adding a constant c shifts the location measures by c but doesn't affect spread. Mean = 55 | Median = 51 | s = 8

Multiplying by k scales both location and spread by |k|. Mean = 120 | Median = 108 | s = 24 | s² = 576

For Y = aX + b: new mean = aX̄ + b, new std dev = |a|×s, new variance = a²×s². Mean=70 | Median=62 | s=16 | s²=256 | CV≈22.9%

The general rule: if Y = aX + b, then Median_Y = a × Median_X + b. This holds because:

• Multiplying by a (positive) preserves the order of observations
• Adding b shifts all values equally
• The middle value of the transformed data is the transformation of the middle value of the original data

Therefore: Median_Y = 2×36 − 10 = 62. ✓ Consistent with the rule.

This is the linear transformation Y = (1/s)X − (X̄/s), i.e., a = 1/s, b = −X̄/s. Mean of Z = 0 | Std Dev of Z = 1 | Variance = 1

Importance: Standardisation converts any variable into a dimensionless z-score, enabling comparison across variables measured in different units and scales. It is the foundation of z-tests, normal tables, and principal component analysis. All z-scores share the same scale, making cross-variable and cross-study comparisons rigorous.

Server A:

• Mean (3.2) ≈ Median (3.1) → nearly symmetric, slight right skew
• Q2−Q1 = 3.1−2.9 = 0.2 s; Q3−Q2 = 3.5−3.1 = 0.4 s; Upper whisker: 4.5−3.5 = 1.0 s; Lower whisker: 2.9−2.0 = 0.9 s → mild right skew

Server B:

• Mean (3.2) >> Median (2.1) → strong right skew
• Q2−Q1 = 2.1−1.2 = 0.9 s; Q3−Q2 = 4.0−2.1 = 1.9 s (upper much larger); Upper whisker: 18.0−4.0 = 14.0 s; Lower whisker: 1.2−0.5 = 0.7 s → extreme right skew

Both methods agree: A ≈ symmetric; B is heavily right-skewed.

Server A's IQR is 4.7× smaller (much tighter middle 50%), and its CV is 7× lower. Both confirm Server A is dramatically more consistent per unit of mean response time.

Server A: 1 mild outlier | Server B: 1 extreme outlier (18.0s)

Choose Server A, based on every relevant dimension:

1. Std Dev: A = 0.4s vs. B = 2.8s. Server B's variability is 7× larger.
2. IQR: A = 0.6s vs. B = 2.8s. Even the middle 50% of B's responses are highly variable.
3. Maximum: A's worst case is 4.5s. B's worst case is 18.0s — 4× worse. For real-time apps, worst-case latency matters enormously.
4. Skewness: B's strong right skew means frequent spikes (the mean is pulled to 3.2s by rare 18s responses). A is symmetric — predictable.
5. CV: A = 12.5% (excellent consistency) vs. B = 87.5% (highly erratic per unit of mean).

Server B sometimes responds faster (mode = 1.5s) but is wildly inconsistent. A real-time application requires predictable performance, not occasionally fast performance. Server A wins comprehensively.

The manager's claim relies solely on the mean — a single statistic that collapses all distributional information into one number. Here is every piece of evidence against his claim:

1. Median: A = 3.1s, B = 2.1s. The typical (middle) response is 1 second faster on B, yet B's mean equals A's due to extreme tail events. The servers are not equivalent for the typical user.
2. Standard deviation: 0.4s vs. 2.8s — a factor of 7 difference in spread.
3. Maximum: 4.5s vs. 18.0s — Server B occasionally fails catastrophically.
4. Skewness: A is symmetric; B is right-skewed. Identical means can coexist with radically different shapes.
5. IQR: 0.6s vs. 2.8s — the middle 50% of users experience vastly different service on each server.
6. Outliers: B has an extreme outlier at 18s inflating its mean to match A's. Remove outliers and B's mean would be < A's.

The manager's claim is statistically unfounded on 6 independent dimensions.