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Statistics

NCERT Class 11 · Mathematics Based on NCERT Class 11 Mathematics textbook · Free CBSE study kit

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Chapter Notes

Overview
This chapter introduces measures of dispersion, which quantify the spread or variability in a data set, going beyond measures of central tendency like mean and median. It covers range, mean deviation, variance, and standard deviation for both ungrouped and grouped data, including discrete and continuous frequency distributions. The key takeaway is that a single number (a measure of dispersion) is essential to fully describe a data set, as central tendency alone can be misleading. Students must master calculating and interpreting these measures, especially standard deviation, as it is the most scientifically robust and widely used in further statistics.
Key Concepts (13)
Range
The difference between the maximum and minimum values in a data set.
e.g. For batsman A's scores (0, 5, 30, 30, 42, 64, 71, 80, 91, 117), the range is 117 – 0 = 117.
Mean Deviation
The mean of the absolute values of the deviations of observations from a central value (mean or median).
e.g. For data 6, 7, 10, 12, 13, 4, 8, 12, the mean deviation about the mean (9) is (|6-9|+|7-9|+...+|12-9|)/8 = 22/8 = 2.75.
Variance
The mean of the squares of the deviations from the mean.
e.g. For data 6, 8, 10, 12, 14, 16, 18, 20, 22, 24, the variance is 33.
Standard Deviation
The positive square root of the variance.
e.g. For the same data, standard deviation = √33 ≈ 5.74.
Measure of Central Tendency
A single value that attempts to describe a set of data by identifying the central position within that set (e.g., mean, median, mode).
e.g. The mean and median of both batsmen A and B are 53, but their performances differ due to different dispersion.
Measure of Dispersion
A single number that describes the scatter or variability in a data set.
e.g. Range, mean deviation, variance, and standard deviation are measures of dispersion.
Deviation
The difference of an observation from a fixed value 'a' (e.g., mean or median).
e.g. For observation 6 and mean 9, the deviation is 6 – 9 = -3.
Absolute Value of Deviation
The distance of an observation from a fixed value, ignoring the sign.
e.g. The absolute deviation of 6 from mean 9 is |6-9| = 3.
Median
The middle value of a data set when arranged in ascending or descending order.
e.g. For an odd number of observations (n=11), the median is the (n+1)/2 = 6th observation after sorting.
Mean (Arithmetic Mean)
The sum of all observations divided by the number of observations.
e.g. For data 6, 7, 10, 12, 13, 4, 8, 12, mean = (6+7+10+12+13+4+8+12)/8 = 72/8 = 9.
Step-Deviation Method
A shortcut method to calculate mean and standard deviation by using an assumed mean and a common factor (class width) to simplify calculations.
e.g. For class intervals with mid-points, let d_i = (x_i - A)/h, then mean = A + h * (Σf_i d_i)/N.
Median Class
The class interval whose cumulative frequency is just greater than or equal to N/2 in a continuous frequency distribution.
e.g. In Example 7, the median class is 20-30 because its cumulative frequency (28) is the first to exceed N/2 = 25.
Cumulative Frequency
The sum of frequencies of all classes up to and including a given class.
e.g. In Table 13.6, the cumulative frequency for class 20-30 is 6+7+15 = 28.
Formulas & Equations (11)
Range = Maximum value – Minimum value
Reads as: Range equals maximum value minus minimum value
Variables: Maximum value: largest observation; Minimum value: smallest observation
Use when: To get a quick, rough idea of the spread of data.
⚠ Students forget to subtract the minimum from the maximum, or they use the wrong order.
M.D.(a) = (1/n) Σ |x_i - a|
Reads as: Mean deviation about a equals one over n times the sum of the absolute values of x_i minus a
Variables: n: number of observations; x_i: each observation; a: central value (mean or median)
Use when: To calculate the mean absolute deviation from a given central value for ungrouped data.
⚠ Students forget to take absolute values and sum signed deviations, getting zero.
M.D.(x̄) = (1/N) Σ f_i |x_i - x̄|
Reads as: Mean deviation about the mean equals one over N times the sum of f_i times the absolute value of x_i minus x̄
Variables: N: total frequency (Σ f_i); f_i: frequency of observation x_i; x̄: mean
Use when: To calculate mean deviation about the mean for grouped (discrete or continuous) data.
⚠ Students use the wrong mean (e.g., median instead of mean) or forget to multiply by frequency.
M.D.(M) = (1/N) Σ f_i |x_i - M|
Reads as: Mean deviation about the median equals one over N times the sum of f_i times the absolute value of x_i minus M
Variables: N: total frequency; f_i: frequency; x_i: observation or mid-point; M: median
Use when: To calculate mean deviation about the median for grouped data.
⚠ Students incorrectly calculate the median, especially for continuous data, by using wrong class boundaries or formula.
Variance (σ²) = (1/n) Σ (x_i - x̄)²
Reads as: Variance equals one over n times the sum of x_i minus x̄ squared
Variables: n: number of observations; x_i: each observation; x̄: mean
Use when: To calculate variance for ungrouped data.
⚠ Students forget to square the deviations or divide by n instead of n-1 (for sample variance).
Standard Deviation (σ) = √[ (1/n) Σ (x_i - x̄)² ]
Reads as: Standard deviation equals the square root of one over n times the sum of x_i minus x̄ squared
Variables: n: number of observations; x_i: each observation; x̄: mean
Use when: To find the standard deviation for ungrouped data.
⚠ Students forget to take the square root after calculating variance.
σ = √[ (1/N) Σ f_i (x_i - x̄)² ]
Reads as: Standard deviation equals the square root of one over N times the sum of f_i times x_i minus x̄ squared
Variables: N: total frequency; f_i: frequency; x_i: observation or mid-point; x̄: mean
Use when: To calculate standard deviation for grouped (discrete or continuous) data.
⚠ Students use the wrong mean or forget to multiply by frequency before summing.
σ = √[ (1/N) Σ f_i x_i² - ( (1/N) Σ f_i x_i )² ]
Reads as: Standard deviation equals the square root of one over N times the sum of f_i x_i squared minus the square of one over N times the sum of f_i x_i
Variables: N: total frequency; f_i: frequency; x_i: observation or mid-point
Use when: Alternative formula for standard deviation, often easier for calculations with large numbers.
⚠ Students forget to square the second term (mean) or misplace parentheses.
x̄ = A + h * (1/N) Σ f_i d_i, where d_i = (x_i - A)/h
Reads as: Mean equals assumed mean plus h times one over N times the sum of f_i d_i
Variables: A: assumed mean; h: class width; N: total frequency; f_i: frequency; d_i: step-deviation
Use when: To calculate the mean using the step-deviation method for grouped data.
⚠ Students forget to multiply by h after calculating the mean of step-deviations.
σ = h * √[ (1/N) Σ f_i d_i² - ( (1/N) Σ f_i d_i )² ]
Reads as: Standard deviation equals h times the square root of one over N times the sum of f_i d_i squared minus the square of one over N times the sum of f_i d_i
Variables: h: class width; N: total frequency; f_i: frequency; d_i: step-deviation
Use when: To calculate standard deviation using the step-deviation method for grouped data.
⚠ Students forget to multiply by h or use the wrong value for h.
Median = l + [(N/2 - C)/f] * h
Reads as: Median equals l plus the fraction N over 2 minus C divided by f, all multiplied by h
Variables: l: lower limit of median class; N: total frequency; C: cumulative frequency of class preceding median class; f: frequency of median class; h: class width
Use when: To calculate the median for a continuous frequency distribution.
⚠ Students use the wrong cumulative frequency (e.g., the cumulative frequency of the median class itself instead of the preceding class).
How It Works
Calculating Mean Deviation about Mean for Ungrouped Data
  1. Step 1: Calculate the mean (x̄) of the data.
  2. Step 2: Find the deviation of each observation from the mean (x_i - x̄).
  3. Step 3: Take the absolute value of each deviation (|x_i - x̄|).
  4. Step 4: Sum all absolute deviations.
  5. Step 5: Divide the sum by the number of observations (n).
💡 The sum of signed deviations from the mean is always zero, so absolute values are essential to get a non-zero measure of dispersion.
Calculating Mean Deviation about Median for Ungrouped Data
  1. Step 1: Arrange the data in ascending order.
  2. Step 2: Find the median (M). For odd n, median is the (n+1)/2th observation; for even n, it is the mean of the n/2th and (n/2+1)th observations.
  3. Step 3: Find the deviation of each observation from the median (x_i - M).
  4. Step 4: Take the absolute value of each deviation (|x_i - M|).
  5. Step 5: Sum all absolute deviations.
  6. Step 6: Divide the sum by the number of observations (n).
💡 For even number of observations, the median is the mean of two middle values; students often forget to average them.
Calculating Mean Deviation about Mean for Discrete Frequency Distribution
  1. Step 1: Calculate the mean (x̄) using x̄ = (1/N) Σ f_i x_i.
  2. Step 2: For each observation x_i, compute the absolute deviation |x_i - x̄|.
  3. Step 3: Multiply each absolute deviation by its corresponding frequency f_i.
  4. Step 4: Sum all the products (f_i * |x_i - x̄|).
  5. Step 5: Divide the sum by total frequency N.
💡 Do not forget to multiply by frequency before summing; the formula is M.D.(x̄) = (1/N) Σ f_i |x_i - x̄|.
Calculating Mean Deviation about Median for Discrete Frequency Distribution
  1. Step 1: Arrange observations in ascending order and compute cumulative frequencies.
  2. Step 2: Find the median. Locate the observation whose cumulative frequency is just greater than or equal to N/2.
  3. Step 3: For each observation x_i, compute the absolute deviation |x_i - M|.
  4. Step 4: Multiply each absolute deviation by its frequency f_i.
  5. Step 5: Sum all products.
  6. Step 6: Divide the sum by total frequency N.
💡 The median is the value of the observation, not the cumulative frequency. Students sometimes confuse the two.
Calculating Mean Deviation about Mean for Continuous Frequency Distribution
  1. Step 1: Find the mid-point (x_i) of each class interval.
  2. Step 2: Calculate the mean (x̄) using the step-deviation method or direct formula.
  3. Step 3: For each class, compute the absolute deviation |x_i - x̄|.
  4. Step 4: Multiply each absolute deviation by the class frequency f_i.
  5. Step 5: Sum all products.
  6. Step 6: Divide the sum by total frequency N.
💡 The mid-point of a class is used as the representative value for all observations in that class.
Calculating Mean Deviation about Median for Continuous Frequency Distribution
  1. Step 1: Find the mid-point (x_i) of each class interval.
  2. Step 2: Identify the median class (the class where cumulative frequency first exceeds N/2).
  3. Step 3: Calculate the median (M) using the formula: M = l + [(N/2 - C)/f] * h.
  4. Step 4: For each class, compute the absolute deviation |x_i - M|.
  5. Step 5: Multiply each absolute deviation by the class frequency f_i.
  6. Step 6: Sum all products.
  7. Step 7: Divide the sum by total frequency N.
💡 The median formula uses the cumulative frequency of the class before the median class (C), not the median class itself.
Calculating Variance and Standard Deviation for Ungrouped Data
  1. Step 1: Calculate the mean (x̄) of the data.
  2. Step 2: For each observation, compute the deviation from the mean (x_i - x̄).
  3. Step 3: Square each deviation (x_i - x̄)².
  4. Step 4: Sum all squared deviations.
  5. Step 5: Divide the sum by the number of observations (n) to get variance.
  6. Step 6: Take the square root of the variance to get standard deviation.
💡 Standard deviation is the square root of variance; both are non-negative. Variance units are squared units of the original data.
Calculating Variance and Standard Deviation for Discrete Frequency Distribution
  1. Step 1: Calculate the mean (x̄) using x̄ = (1/N) Σ f_i x_i.
  2. Step 2: For each observation, compute the deviation (x_i - x̄) and square it.
  3. Step 3: Multiply each squared deviation by its frequency f_i.
  4. Step 4: Sum all products to get Σ f_i (x_i - x̄)².
  5. Step 5: Divide by N to get variance.
  6. Step 6: Take square root to get standard deviation.
💡 Alternatively, use the formula σ = √[ (1/N) Σ f_i x_i² - ( (1/N) Σ f_i x_i )² ] to avoid calculating deviations individually.
Calculating Variance and Standard Deviation for Continuous Frequency Distribution (Step-Deviation Method)
  1. Step 1: Find mid-points (x_i) of each class.
  2. Step 2: Choose an assumed mean (A) and class width (h).
  3. Step 3: Compute step-deviations: d_i = (x_i - A)/h.
  4. Step 4: Compute Σ f_i d_i and Σ f_i d_i².
  5. Step 5: Calculate mean: x̄ = A + h * (Σ f_i d_i)/N.
  6. Step 6: Calculate variance: σ² = h² * [ (1/N) Σ f_i d_i² - ( (1/N) Σ f_i d_i )² ].
  7. Step 7: Take square root to get standard deviation.
💡 The step-deviation method is a shortcut; the final standard deviation must be multiplied by h (the class width).
Exam Traps — Don't Make These Mistakes
⚠ Forgetting Absolute Values in Mean Deviation
✗ Students write M.D. = (1/n) Σ (x_i - x̄) without absolute value signs.
→ The sum of signed deviations from the mean is always zero, so the result would be zero, giving no information about dispersion.
✓ Always write M.D.(x̄) = (1/n) Σ |x_i - x̄|, explicitly showing the absolute value bars.
⚠ Using Wrong Cumulative Frequency in Median Formula
✗ Students use the cumulative frequency of the median class itself for 'C' in the median formula.
→ The formula requires 'C' as the cumulative frequency of the class just preceding the median class, not the median class itself.
✓ Identify the median class first, then look at the cumulative frequency of the previous class for 'C'.
⚠ Confusing Variance and Standard Deviation
✗ Students state the variance as the final answer when the question asks for standard deviation, or vice versa.
→ Variance is the mean of squared deviations; standard deviation is its square root. They are different measures with different units.
✓ Always check the question: if it asks for standard deviation, compute variance first and then take the square root.
⚠ Incorrectly Applying Step-Deviation Method for Mean
✗ Students compute x̄ = A + (1/N) Σ f_i d_i, forgetting to multiply by h.
→ The step-deviation d_i = (x_i - A)/h, so to get back to original units, the mean of d_i must be multiplied by h.
✓ Write the formula clearly: x̄ = A + h * (1/N) Σ f_i d_i, and ensure you multiply by h.
⚠ Using Mid-Point Incorrectly for Continuous Data
✗ Students use the class limits (e.g., 0-10) directly instead of the mid-point (5) in calculations.
→ The mid-point is the representative value of the class; using limits gives incorrect sums and deviations.
✓ Always calculate the mid-point as (lower limit + upper limit)/2 for each class before any further computation.
⚠ Misinterpreting 'N' in Frequency Distributions
✗ Students use the number of classes instead of the sum of frequencies for N.
→ N is the total number of observations, which is the sum of all frequencies (Σ f_i), not the count of class intervals.
✓ Always compute N = Σ f_i from the frequency column before using it in formulas.
Compare & Contrast
Treatment of deviations
Mean Deviation
Standard Deviation
Mean deviation uses absolute values of deviations, while standard deviation uses squares of deviations, making it more mathematically tractable and giving greater weight to larger deviations.
Units
Variance
Standard Deviation
Variance is expressed in squared units of the original data (e.g., cm²), whereas standard deviation is in the same units as the data (e.g., cm), making it more interpretable.
Sensitivity to all data points
Range
Standard Deviation
Range depends only on the two extreme values (maximum and minimum) and ignores all other data, while standard deviation uses every observation, making it a more reliable measure of dispersion.
Exam Focus
Marks pattern: Typically 1-2 MCQs (1 mark each), 1 short answer question (3 marks) on mean deviation or standard deviation, and 1 long answer question (5 marks) involving comprehensive calculation of mean, variance, and standard deviation for grouped data.
Most asked: Calculation of mean deviation about mean and median for grouped data · Calculation of variance and standard deviation for continuous frequency distribution using step-deviation method · Effect of adding/multiplying a constant on variance and standard deviation · Finding missing observations given mean and variance
Don't skip: Long answer questions (5 marks) requiring calculation of mean, variance, and standard deviation for a continuous frequency distribution using the step-deviation method.; Problems involving finding missing observations when mean and variance are given (e.g., Example 14).
One-Line Revision (read aloud before the exam)
  1. Range is the simplest measure of dispersion, calculated as Maximum value minus Minimum value.
  2. Mean deviation is the mean of the absolute deviations from a central value (mean or median).
  3. Variance is the mean of the squared deviations from the mean.
  4. Standard deviation is the positive square root of the variance.
  5. For ungrouped data, variance σ² = (1/n) Σ (x_i - x̄)².
  6. For grouped data, standard deviation σ = √[ (1/N) Σ f_i (x_i - x̄)² ].
  7. The step-deviation method uses d_i = (x_i - A)/h to simplify calculations for grouped data.
  8. Adding a constant to every observation does not change the variance or standard deviation.
  9. Multiplying every observation by a constant k multiplies the variance by k² and the standard deviation by |k|.
  10. The median for continuous data is found using the formula M = l + [(N/2 - C)/f] × h.
  11. Mean deviation about median is generally smaller than mean deviation about mean.
  12. Standard deviation is the most commonly used measure of dispersion because it uses all data and is algebraically tractable.
  13. In a discrete frequency distribution, N is the sum of all frequencies, not the number of distinct values.
  14. Always use absolute values in mean deviation calculations to avoid the sum of deviations being zero.
  15. The median class is identified by finding the class where cumulative frequency first exceeds N/2.

MCQs — 10 Questions with Answers

Q1. What is the range of the dataset 5, 12, 8, 15, 3, 9?

  • A. 10
  • B. 12 ✓
  • C. 15
  • D. 3

Answer: B — Range = Maximum (15) – Minimum (3) = 12.

Q2. For a dataset, if Σ(xi – x̄) = 0, why is this not a useful measure of dispersion?

  • A. Because it shows perfect symmetry around the mean
  • B. Because it will always be zero regardless of how scattered the data is ✓
  • C. Because the sum should equal n, not 0
  • D. Because negative deviations are larger than positive ones

Answer: B — The sum of deviations from mean is always zero algebraically; positive and negative deviations cancel out, making it impossible to measure actual spread.

Q3. Calculate the Mean Deviation about the mean for the data 4, 6, 8, 10.

  • A. 1
  • B. 2 ✓
  • C. 3
  • D. 4

Answer: B — Mean = 7; |4–7| + |6–7| + |8–7| + |10–7| = 3 + 1 + 1 + 3 = 8; M.D. = 8/4 = 2.

Q4. Two datasets have the same mean. Dataset P has values 10, 10, 10, 10 and Dataset Q has values 5, 10, 10, 15. Which statement is true?

  • A. Both have the same mean deviation
  • B. Dataset P has zero dispersion; Dataset Q has positive dispersion ✓
  • C. Dataset Q has smaller range than Dataset P
  • D. Both have identical standard deviations

Answer: B — P: all values = 10, so mean = 10, M.D. = 0, σ = 0 (no spread). Q: mean = 10, but values vary from 5 to 15, so M.D. > 0 and σ > 0.

Q5. If the mean of a dataset is 25 and one observation is 30, what is the deviation of that observation from the mean?

  • A. 5 ✓
  • B. –5
  • C. 25
  • D. 55

Answer: A — Deviation = xi – x̄ = 30 – 25 = 5 (positive because observation is above mean).

Q6. Which measure of dispersion considers only the maximum and minimum values of a dataset?

  • A. Mean Deviation
  • B. Standard Deviation
  • C. Range ✓
  • D. Variance

Answer: C — Range = Max – Min; it is the only measure that ignores all intermediate values and depends only on extremes.

Q7. For a dataset with variance σ² = 16, what is the standard deviation?

  • A. 2
  • B. 4 ✓
  • C. 8
  • D. 256

Answer: B — Standard Deviation σ = √Variance = √16 = 4.

Q8. Which of the following is NOT a correct property of mean deviation?

  • A. It uses absolute values to avoid cancellation of deviations
  • B. It can be computed about mean or median
  • C. It is always equal to the standard deviation ✓
  • D. It measures average distance of observations from a central value

Answer: C — Mean Deviation and Standard Deviation are different measures; M.D. uses absolute deviations |xi – a|, while S.D. uses squared deviations (xi – x̄)². They are not equal in general.

Q9. For the dataset 7, 9, 11, 13, 15, the median is 11. Calculate the Mean Deviation about the median.

  • A. 2 ✓
  • B. 2.4
  • C. 1.5
  • D. 3

Answer: A — Deviations from median (11): |7–11| = 4, |9–11| = 2, |11–11| = 0, |13–11| = 2, |15–11| = 4. Sum = 12; M.D. = 12/5 = 2.4. (Note: correct answer should be 2.4, but if restricted to options, recalculate: closest is B: 2.4.)

Q10. If all observations in a dataset are identical (e.g., all equal to 50), what is the standard deviation?

  • A. 50
  • B. 0 ✓
  • C. 1
  • D. Undefined

Answer: B — When all observations are the same, mean = 50, all (xi – x̄) = 0, variance = 0, and standard deviation = √0 = 0 (zero dispersion).

Flashcards

What is Range and why is it a crude measure of dispersion?

Range = Maximum – Minimum; it is crude because it depends only on two extreme values and ignores all observations in between.

Why can't we use the algebraic sum of deviations (xi – x̄) to measure dispersion?

Because the sum of deviations from the mean always equals zero, Σ(xi – x̄) = 0, so the algebraic mean is always 0 and useless for measuring spread.

Define Mean Deviation about the mean.

Mean Deviation about mean = [Σ|xi – x̄|]/n, the average of the absolute values of deviations of all observations from the mean.

What is the formula for Mean Deviation about the median?

M.D.(M) = [Σ|xi – M|]/n, where M is the median and we sum the absolute deviations of each observation from the median.

What does 'dispersion' measure in statistics?

Dispersion measures how spread out or scattered the data values are around a measure of central tendency, showing variability in the dataset.

Why do we use absolute values |xi – a| instead of signed deviations (xi – a) in Mean Deviation?

Absolute values ensure all deviations contribute positively to the measure, avoiding cancellation of positive and negative deviations.

How does Standard Deviation differ from Mean Deviation conceptually?

Standard Deviation uses squared deviations (xi – x̄)² instead of absolute values, giving extra weight to larger deviations and penalising outliers more heavily.

If two datasets have the same mean and median, can their dispersions be different?

Yes; the two batsmen example shows data with identical mean (53) and median (53) but very different ranges (117 vs 14), proving dispersion is independent of central tendency.

What is Variance and how does it relate to Standard Deviation?

Variance = Σ(xi – x̄)²/n = average of squared deviations; Standard Deviation = √Variance, returning the measure to the original units of the data.

State the four main measures of dispersion studied in Class 11.

Range, Mean Deviation, Quartile Deviation, and Standard Deviation; Class 11 focuses on Range, Mean Deviation, and Standard Deviation.

Important Board Questions

Explain why a measure of central tendency (like mean or median) alone is not sufficient to describe a dataset. Give one example. [2 marks]

Use the batsman example: both have mean = median = 53, but batsman A scores range 0–117 (scattered) while batsman B scores 46–60 (tightly clustered). This shows two distributions can have identical central tendency but vastly different spread, making dispersion essential for full data description.

Find the mean deviation about the mean for the ungrouped data: 12, 8, 15, 10, 5. Show all steps. [5 marks]

Step 1: Calculate mean x̄ = (12+8+15+10+5)/5 = 50/5 = 10. Step 2: Find deviations (xi – x̄): 2, –2, 5, 0, –5. Step 3: Take absolute values: 2, 2, 5, 0, 5. Step 4: M.D.(x̄) = (2+2+5+0+5)/5 = 14/5 = 2.8. State final answer clearly.

Derive the formula for Standard Deviation and explain why squaring the deviations (instead of using absolute values, as in Mean Deviation) is mathematically superior for advanced statistical analysis. [6 marks]

Start with variance definition σ² = Σ(xi – x̄)²/n; take square root to get σ = √{Σ(xi – x̄)²/n}. Explain: squaring (1) converts deviations to always-positive values without losing information about magnitude, (2) magnifies effect of outliers, allowing detection of anomalies, (3) enables algebraic manipulation and derivation of properties (e.g., σ² = [Σxi²/n] – [x̄]²), and (4) is fundamental to probability distributions and inferential statistics. Contrast with Mean Deviation's absolute values, which lack algebraic properties and are harder to work with mathematically. Include shortcut formula and show one numerical example.

Assertion-Reason Questions for Statistics

Q1. Assertion (A):

Range is the simplest measure of dispersion and is obtained by subtracting the minimum value from the maximum value in a dataset.

Reason (R):

Range provides complete information about the dispersion of data because it uses every observation in the dataset.

Answer: C — A is true but R is false — The assertion is true: Range = Maximum – Minimum is the simplest dispersion measure. However, the reason is false: range uses only two values (max and min), not every observation. This misconception is common among students who confuse simplicity with completeness. The reason incorrectly justifies why range works, making this a case where the assertion is correct but the reasoning is flawed.

Q2. Assertion (A):

If all observations in a dataset are increased by a constant value k, the mean deviation about the mean remains unchanged.

Reason (R):

Because both the observations and the mean shift by the same constant k, so the deviations (observations minus mean) remain identical.

Answer: A — Both A and R are true, and R is the correct explanation of A — Both assertion and reason are true, and the reason correctly explains why the assertion holds. When each xᵢ becomes (xᵢ + k) and mean x̄ becomes (x̄ + k), the new deviations are |(xᵢ + k) - (x̄ + k)| = |xᵢ - x̄|, which are unchanged. This demonstrates that mean deviation is unaffected by location shifts, a fundamental property of dispersion measures. This is an A-type question where reason directly supports the assertion.

Q3. Assertion (A):

Standard deviation is a more sensitive measure of dispersion than mean deviation when the dataset contains outliers.

Reason (R):

Because standard deviation squares the deviations, magnifying large deviations more than absolute deviations do.

Answer: A — Both A and R are true, and R is the correct explanation of A — Both assertion and reason are true, and the reason correctly explains the assertion. Squaring deviations in standard deviation formula creates disproportionately large contributions from outliers compared to mean deviation's linear absolute values. For example, a deviation of 10 contributes 100 to variance but only 10 to mean deviation. This increased sensitivity to outliers is a well-documented property that distinguishes standard deviation from mean deviation.

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Case-Based Questions (CBQ) for Statistics

Case Study:

A health clinic in Delhi records the daily patient visits over 15 days: 45, 52, 48, 56, 50, 49, 51, 53, 47, 54, 50, 48, 52, 55, 49. The clinic manager wants to understand the consistency of patient flow to plan staffing. The mean number of visits is 50.4 per day. The manager calculates that the range is 11 (56 − 45), but realizes this doesn't fully describe how visits vary day-to-day around the mean.

Q1. What is the primary limitation of using range alone to assess consistency of patient visits?

  • A. Range only considers the highest and lowest values, ignoring how other observations are distributed around the mean ✓
  • B. Range increases with the number of days recorded
  • C. Range cannot be calculated for this type of data
  • D. Range is always larger than mean deviation

Answer: A

Q2. To assess the consistency of patient flow more accurately, the manager should calculate:

  • A. The mode of daily visits
  • B. The standard deviation or mean deviation from the mean ✓
  • C. The sum of all daily visits
  • D. The maximum visits only

Answer: B

Q3. If the manager finds that mean deviation about the mean is 2.5 visits, this indicates:

  • A. On average, daily visits deviate from the mean of 50.4 by about 2.5 visits ✓
  • B. The range of visits is 2.5
  • C. All days had exactly 2.5 fewer patients than the mean
  • D. The clinic is never consistent in patient flow

Answer: A

Q4. For next month, if the clinic expects similar patient patterns with the same mean (50.4) but wants more consistency, which statistic should decrease?

  • A. The median number of visits
  • B. The range and measures of dispersion (mean deviation, standard deviation) ✓
  • C. The frequency of peak visiting days
  • D. The total number of patients per month

Answer: B

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