Important Questions Class 11 Maths Chapter 13 Statistics
Statistics studies data through measures that describe central value, spread, variability and interpretation.
Important Questions Class 11 Maths Chapter 13 help students practise range, mean deviation, variance and standard deviation for 2026-27 exams.
Class 11 Maths Chapter 13 moves beyond mean, median and mode to show how data values are spread around a central point. Two batsmen may have the same mean score, but one may be more consistent because his scores are less scattered. This is why Statistics focuses on measures of dispersion such as range, mean deviation, variance and standard deviation. Students should practise ungrouped data, discrete frequency distribution, continuous frequency distribution and shortcut method problems carefully. These class 11 maths statistics important questions follow the 80-mark Class 11 Maths paper pattern and use simple copy-friendly formulas.
Key Takeaways
- Range: Range is the difference between the maximum value and the minimum value.
- Mean Deviation: Mean deviation uses absolute deviations from mean or median.
- Variance: Variance is the mean of the squares of deviations from mean.
- Standard Deviation: Standard deviation is the positive square root of variance.
Important Questions Class 11 Maths Chapter 13 Structure 2026-27
| Section | Question Type | Marks and Word Limit |
| Section A | MCQs and assertion-reason | 20 marks, 1 mark each |
| Section B | Very Short Answer | 10 marks, 2 marks each |
| Section C | Short Answer | 18 marks, 3 marks each |
| Section D | Long Answer | 20 marks, 5 marks each |
| Section E | Case Study-Based | 12 marks, 4 marks each |
Section A: MCQs from Important Questions Class 11 Maths Chapter 13
Section A tests formulas, definitions and quick numerical checks. Statistics class 11 important questions in this section usually cover range, dispersion, mean deviation, variance and standard deviation.
Q1. A measure that shows how data values are scattered is called:
- Measure of central tendency
b. Measure of dispersion
c. Median
d. Mode
Answer: b. Measure of dispersion
A measure of dispersion describes the spread or variability in data.
Q2. Which of the following is a measure of dispersion?
- Mean
b. Median
c. Range
d. Mode
Answer: c. Range
Range class 11 statistics questions usually test the simplest measure of dispersion.
Q3. Range is calculated as:
- Minimum value - Maximum value
b. Maximum value - Minimum value
c. Maximum value + Minimum value
d. Mean - Median
Answer: b. Maximum value - Minimum value
Range = Maximum value - Minimum value.
Q4. For the data 4, 9, 11, 15, 20, the range is:
- 12
b. 14
c. 16
d. 20
Answer: c. 16
Range = 20 - 4 = 16.
Q5. Mean deviation is based on:
- Absolute values of deviations
b. Only positive observations
c. Only maximum value
d. Only minimum value
Answer: a. Absolute values of deviations
Mean deviation uses distances from a central value.
Q6. Mean deviation can commonly be calculated about:
- Mean and median
b. Only mode
c. Only range
d. Only maximum value
Answer: a. Mean and median
Mean deviation about mean class 11 and mean deviation about median class 11 are commonly used.
Q7. The sum of deviations from mean is:
- Always 1
b. Always 0
c. Always negative
d. Always equal to range
Answer: b. Always 0
This is why absolute values or squares of deviations are used.
Q8. Variance is the:
- Mean of observations
b. Median of observations
c. Mean of squares of deviations from mean
d. Difference between first and last value
Answer: c. Mean of squares of deviations from mean
Variance class 11 maths questions often ask this definition.
Q9. Standard deviation is the:
- Square of variance
b. Positive square root of variance
c. Sum of all observations
d. Difference between mean and median
Answer: b. Positive square root of variance
Standard deviation class 11 uses the same unit as the observations.
Q10. If variance is 49, then standard deviation is:
- 7
b. 14
c. 24.5
d. 49
Answer: a. 7
Standard deviation = √49 = 7.
Q11. If every observation is multiplied by 2, variance becomes:
- Same as before
b. 2 times the old variance
c. 4 times the old variance
d. Half of the old variance
Answer: c. 4 times the old variance
If observations are multiplied by k, variance becomes k² times the old variance.
Q12. If every observation is increased by 5, variance:
- Increases by 5
b. Decreases by 5
c. Becomes 25 times
d. Remains unchanged
Answer: d. Remains unchanged
Adding or subtracting a constant does not change variance.
Q13. For continuous frequency distribution, each class is represented by its:
- Lower limit
b. Upper limit
c. Mid-point
d. Frequency only
Answer: c. Mid-point
Mid-points are used for grouped-data calculations.
Q14. Mean deviation about median can be unreliable when:
- Variability is very high
b. All values are equal
c. Range is zero
d. Data has no observation
Answer: a. Variability is very high
NCERT notes that mean deviation about median may be less reliable in highly variable series.
Q15. Assertion: Range gives only a rough idea of dispersion.
Reason: Range depends only on maximum and minimum values.
- Both Assertion and Reason are true, and Reason explains Assertion
b. Both are true, but Reason does not explain Assertion
c. Assertion is true, Reason is false
d. Assertion is false, Reason is true
Answer: a. Both Assertion and Reason are true, and Reason explains Assertion
Range ignores all middle observations.
Q16. Assertion: Standard deviation is preferred over variance for interpretation.
Reason: Standard deviation is expressed in the same unit as the observations.
- Both Assertion and Reason are true, and Reason explains Assertion
b. Both are true, but Reason does not explain Assertion
c. Assertion is true, Reason is false
d. Assertion is false, Reason is true
Answer: a. Both Assertion and Reason are true, and Reason explains Assertion
Variance uses squared units, while standard deviation returns to original units.
Q17. The formula for range is:
- Range = Mean - Median
b. Range = Maximum value - Minimum value
c. Range = Sum of values / Number of values
d. Range = Variance × Standard deviation
Answer: b. Range = Maximum value - Minimum value
This is the direct formula for range.
Q18. Which symbol is commonly used for standard deviation?
- μ
b. σ
c. Σ
d. π
Answer: b. σ
Standard deviation is usually denoted by σ.
Q19. Which symbol is commonly used for variance?
- σ²
b. √σ
c. M
d. x
Answer: a. σ²
Variance is denoted by sigma square.
Q20. If all observations are equal, standard deviation is:
- 0
b. 1
c. Equal to mean
d. Equal to range plus mean
Answer: a. 0
There is no dispersion when all values are equal.
Section B: Very Short Answer Questions from Class 11 Maths Statistics Important Questions
Section B has short numerical questions. Write the formula first, substitute values clearly and keep each calculation step visible.
Q21. Find the range of 12, 18, 7, 25, 30 and 10.
Range = Maximum value - Minimum value.
Maximum value = 30.
Minimum value = 7.
Range = 30 - 7 = 23.
Q22. What is mean deviation?
Mean deviation is the mean of the absolute values of deviations from a central value.
Mean deviation can be calculated about mean or median.
Formula:
Mean deviation about a = Sum of absolute deviations from a / Number of observations.
Q23. Find the mean of 6, 7, 10, 12, 13, 4, 8 and 12.
Mean = Sum of observations / Number of observations.
Sum = 6 + 7 + 10 + 12 + 13 + 4 + 8 + 12 = 72.
Number of observations = 8.
Mean = 72 / 8 = 9.
Q24. Find the median of 3, 9, 5, 12, 10, 4 and 7.
Arrange the data in ascending order:
3, 4, 5, 7, 9, 10, 12.
Number of observations = 7.
Median = 4th observation.
Median = 7.
Q25. If variance is 33, find the standard deviation.
Standard deviation = √Variance.
Standard deviation = √33.
Standard deviation = 5.74 approximately.
Section C: Short Answer Questions from Statistics Class 11 Important Questions
Section C has 3-mark questions. These usually test range, mean deviation, variance and standard deviation for small data sets.
Q26. Find the range and comment on dispersion for scores 30, 91, 0, 64, 42, 80, 30, 5, 117, 71.
Maximum value = 117.
Minimum value = 0.
Range = Maximum value - Minimum value.
Range = 117 - 0 = 117.
The range is large, so the scores are highly scattered.
Q27. Find the mean deviation about mean for 6, 7, 10, 12, 13, 4, 8 and 12.
Mean = 9.
Deviations from mean:
6 - 9 = -3
7 - 9 = -2
10 - 9 = 1
12 - 9 = 3
13 - 9 = 4
4 - 9 = -5
8 - 9 = -1
12 - 9 = 3
Absolute deviations:
3, 2, 1, 3, 4, 5, 1, 3.
Sum of absolute deviations = 22.
Mean deviation about mean = 22 / 8 = 2.75.
Q28. Find the mean deviation about median for 3, 3, 4, 5, 7, 9, 10, 12, 18, 19 and 21.
Number of observations = 11.
Median = 6th observation = 9.
Absolute deviations from median:
6, 6, 5, 4, 2, 0, 1, 3, 9, 10, 12.
Sum of absolute deviations = 58.
Mean deviation about median = 58 / 11 = 5.27 approximately.
Q29. Find the variance of 6, 8, 10, 12, 14, 16, 18, 20, 22 and 24.
Mean = 15.
Deviations from mean:
-9, -7, -5, -3, -1, 1, 3, 5, 7, 9.
Squares of deviations:
81, 49, 25, 9, 1, 1, 9, 25, 49, 81.
Sum of squares = 330.
Variance = 330 / 10 = 33.
Q30. Find the standard deviation for the data in Q29.
Variance = 33.
Standard deviation = √33.
Standard deviation = 5.74 approximately.
Q31. Why are measures of central tendency not enough to describe data?
Measures of central tendency show where data values are centred.
They do not show how values are scattered around the centre. Two data sets can have the same mean and median but different spread. Measures of dispersion are therefore needed.
Range, mean deviation, variance and standard deviation help measure this spread.
Section D: Long Answer Questions from Important Questions Class 11 Maths Chapter 13
Section D has 5-mark questions. These questions usually need complete tables, formulas and final interpretation.
Q32. Find the mean deviation about mean for the following data.
| xi | 2 | 5 | 6 | 8 | 10 | 12 |
| fi | 2 | 8 | 10 | 7 | 8 | 5 |
First, find N.
N = 2 + 8 + 10 + 7 + 8 + 5 = 40.
Find Σfixi:
2×2 = 4
8×5 = 40
10×6 = 60
7×8 = 56
8×10 = 80
5×12 = 60
Σfixi = 300.
Mean = Σfixi / N = 300 / 40 = 7.5.
Now find |xi - mean| and fi|xi - mean|.
For xi = 2, |2 - 7.5| = 5.5, fi|xi - mean| = 2×5.5 = 11.
For xi = 5, |5 - 7.5| = 2.5, fi|xi - mean| = 8×2.5 = 20.
For xi = 6, |6 - 7.5| = 1.5, fi|xi - mean| = 10×1.5 = 15.
For xi = 8, |8 - 7.5| = 0.5, fi|xi - mean| = 7×0.5 = 3.5.
For xi = 10, |10 - 7.5| = 2.5, fi|xi - mean| = 8×2.5 = 20.
For xi = 12, |12 - 7.5| = 4.5, fi|xi - mean| = 5×4.5 = 22.5.
Σfi|xi - mean| = 92.
Mean deviation about mean = 92 / 40 = 2.3.
Q33. Find the mean deviation about median for the following data.
| xi | 3 | 6 | 9 | 12 | 13 | 15 | 21 | 22 |
| fi | 3 | 4 | 5 | 2 | 4 | 5 | 4 | 3 |
First, find N.
N = 3 + 4 + 5 + 2 + 4 + 5 + 4 + 3 = 30.
Cumulative frequencies:
3, 7, 12, 14, 18, 23, 27, 30.
N / 2 = 30 / 2 = 15.
The cumulative frequency just greater than 15 is 18.
So, median = 13.
Now find |xi - median| and fi|xi - median|.
For xi = 3, |3 - 13| = 10, fi|xi - median| = 3×10 = 30.
For xi = 6, |6 - 13| = 7, fi|xi - median| = 4×7 = 28.
For xi = 9, |9 - 13| = 4, fi|xi - median| = 5×4 = 20.
For xi = 12, |12 - 13| = 1, fi|xi - median| = 2×1 = 2.
For xi = 13, |13 - 13| = 0, fi|xi - median| = 4×0 = 0.
For xi = 15, |15 - 13| = 2, fi|xi - median| = 5×2 = 10.
For xi = 21, |21 - 13| = 8, fi|xi - median| = 4×8 = 32.
For xi = 22, |22 - 13| = 9, fi|xi - median| = 3×9 = 27.
Σfi|xi - median| = 149.
Mean deviation about median = 149 / 30 = 4.97 approximately.
Q34. Find the variance and standard deviation for the following data.
| xi | 4 | 8 | 11 | 17 | 20 | 24 | 32 |
| fi | 3 | 5 | 9 | 5 | 4 | 3 | 1 |
First, find N.
N = 3 + 5 + 9 + 5 + 4 + 3 + 1 = 30.
Find Σfixi:
3×4 = 12
5×8 = 40
9×11 = 99
5×17 = 85
4×20 = 80
3×24 = 72
1×32 = 32
Σfixi = 420.
Mean = 420 / 30 = 14.
Now find fi(xi - mean)².
For xi = 4, xi - mean = -10, square = 100, fi square = 3×100 = 300.
For xi = 8, xi - mean = -6, square = 36, fi square = 5×36 = 180.
For xi = 11, xi - mean = -3, square = 9, fi square = 9×9 = 81.
For xi = 17, xi - mean = 3, square = 9, fi square = 5×9 = 45.
For xi = 20, xi - mean = 6, square = 36, fi square = 4×36 = 144.
For xi = 24, xi - mean = 10, square = 100, fi square = 3×100 = 300.
For xi = 32, xi - mean = 18, square = 324, fi square = 1×324 = 324.
Σfi(xi - mean)² = 1374.
Variance = 1374 / 30 = 45.8.
Standard deviation = √45.8 = 6.77 approximately.
Q35. Find the mean, variance and standard deviation for the following continuous distribution.
| Class | 30-40 | 40-50 | 50-60 | 60-70 | 70-80 | 80-90 | 90-100 |
| Frequency | 3 | 7 | 12 | 15 | 8 | 3 | 2 |
Mid-points are:
35, 45, 55, 65, 75, 85, 95.
N = 3 + 7 + 12 + 15 + 8 + 3 + 2 = 50.
Find Σfixi:
3×35 = 105
7×45 = 315
12×55 = 660
15×65 = 975
8×75 = 600
3×85 = 255
2×95 = 190
Σfixi = 3100.
Mean = 3100 / 50 = 62.
Now find fi(xi - mean)².
For xi = 35, xi - mean = -27, square = 729, fi square = 3×729 = 2187.
For xi = 45, xi - mean = -17, square = 289, fi square = 7×289 = 2023.
For xi = 55, xi - mean = -7, square = 49, fi square = 12×49 = 588.
For xi = 65, xi - mean = 3, square = 9, fi square = 15×9 = 135.
For xi = 75, xi - mean = 13, square = 169, fi square = 8×169 = 1352.
For xi = 85, xi - mean = 23, square = 529, fi square = 3×529 = 1587.
For xi = 95, xi - mean = 33, square = 1089, fi square = 2×1089 = 2178.
Σfi(xi - mean)² = 10050.
Variance = 10050 / 50 = 201.
Standard deviation = √201 = 14.18 approximately.
Section E: Case Study-Based Questions from Class 11 Maths Chapter 13 Statistics
Section E usually checks whether students can connect data with consistency, spread and interpretation. Identify the measure first, then calculate step by step.
Q36. Case Study: Two batsmen with the same average
Two batsmen scored runs in their last ten matches.
Batsman A: 30, 91, 0, 64, 42, 80, 30, 5, 117, 71
Batsman B: 53, 46, 48, 50, 53, 53, 58, 60, 57, 52
Both have mean score 53 and median score 53.
Q36(a). Can both batsmen be called equally consistent?
No, both batsmen cannot be called equally consistent.
Their averages are same, but their spread is different.
Q36(b). Find the range for Batsman A.
Maximum score = 117.
Minimum score = 0.
Range = 117 - 0 = 117.
Q36(c). Find the range for Batsman B.
Maximum score = 60.
Minimum score = 46.
Range = 60 - 46 = 14.
Q36(d). Which batsman is more consistent?
Batsman B is more consistent.
His scores have a smaller range and are closer to the mean.
Q37. Case Study: Heights of Students
The heights of 5 students are:
150 cm, 152 cm, 155 cm, 158 cm and 160 cm.
Q37(a). Find the mean height.
Mean = (150 + 152 + 155 + 158 + 160) / 5.
Mean = 775 / 5 = 155 cm.
Q37(b). Find the range.
Maximum height = 160 cm.
Minimum height = 150 cm.
Range = 160 - 150 = 10 cm.
Q37(c). Find the absolute deviations from mean.
Mean = 155.
Absolute deviations:
|150 - 155| = 5
|152 - 155| = 3
|155 - 155| = 0
|158 - 155| = 3
|160 - 155| = 5
Q37(d). Find the mean deviation about mean.
Sum of absolute deviations = 5 + 3 + 0 + 3 + 5 = 16.
Mean deviation about mean = 16 / 5 = 3.2 cm.
Q38. Case Study: Marks in a Class Test
A teacher records the marks of 6 students:
4, 6, 8, 10, 12, 14.
Q38(a). Find the mean.
Mean = (4 + 6 + 8 + 10 + 12 + 14) / 6.
Mean = 54 / 6 = 9.
Q38(b). Find the squared deviations from mean.
Mean = 9.
Squared deviations:
(4 - 9)² = 25
(6 - 9)² = 9
(8 - 9)² = 1
(10 - 9)² = 1
(12 - 9)² = 9
(14 - 9)² = 25
Q38(c). Find the variance.
Sum of squared deviations = 25 + 9 + 1 + 1 + 9 + 25 = 70.
Variance = 70 / 6 = 11.67 approximately.
Q38(d). Find the standard deviation.
Standard deviation = √11.67.
Standard deviation = 3.42 approximately.
Formula-Based Revision for Important Questions Class 11 Maths Chapter 13
Important questions class 11 maths chapter 13 should be revised through formulas first, then solved examples. Keep formulas simple and copy-friendly.
Range Class 11 Statistics
Range = Maximum value - Minimum value.
Range gives a rough idea of spread because it uses only two values.
Mean Deviation Class 11
Mean deviation about a value is the average of absolute deviations from that value.
Mean deviation about mean = Sum of |xi - mean| / n.
Mean deviation about median = Sum of |xi - median| / n.
For frequency data:
Mean deviation about mean = Σfi|xi - mean| / Σfi.
Mean deviation about median = Σfi|xi - median| / Σfi.
Variance Class 11 Maths
Variance is the mean of squared deviations from mean.
For ungrouped data:
Variance = Σ(xi - mean)² / n.
For frequency data:
Variance = Σfi(xi - mean)² / Σfi.
Standard Deviation Class 11
Standard deviation is the positive square root of variance.
Standard deviation = √Variance.
For ungrouped data:
Standard deviation = √[Σ(xi - mean)² / n].
For frequency data:
Standard deviation = √[Σfi(xi - mean)² / Σfi].
Standard Deviation Shortcut Method Class 11
Use the shortcut method when values are large or class intervals are equal.
Let A = assumed mean.
Let h = class width.
Let yi = (xi - A) / h.
Mean = A + h × [Σfiyi / Σfi].
Variance = h² × {[Σfiyi² / Σfi] - [Σfiyi / Σfi]²}.
Standard deviation = √Variance.
Chapter-Wise Revision for Statistics Class 11 Important Questions
Statistics class 11 important questions should be revised in four parts: range, mean deviation, variance and standard deviation.
Start with range because it is the simplest measure of dispersion. It gives a quick idea of spread but ignores middle observations.
Next, revise mean deviation class 11. Practise mean deviation about mean class 11 and mean deviation about median class 11 for ungrouped and grouped data.
Then revise variance class 11 maths. Focus on why squared deviations are used and how variance measures scatter around mean.
Finally, revise standard deviation class 11. Practise direct method and shortcut method for continuous frequency distributions.
Important Questions Class 11 Maths Chapter-Wise
| Chapter No. | Chapter Name |
| Chapter 1 | Sets |
| Chapter 2 | Relations and Functions |
| Chapter 3 | Trigonometric Functions |
| Chapter 4 | Complex Numbers and Quadratic Equations |
| Chapter 5 | Linear Inequalities |
| Chapter 6 | Permutations and Combinations |
| Chapter 7 | Binomial Theorem |
| Chapter 8 | Sequences and Series |
| Chapter 9 | Straight Lines |
| Chapter 10 | Conic Sections |
| Chapter 11 | Introduction to Three Dimensional Geometry |
| Chapter 12 | Limits and Derivatives |
| Chapter 13 | Statistics |
| Chapter 14 | Probability |
Q.1 If f(x) =xx, x00, x=0 , the limx0f(x) equals
Marks:1
Ans
LHL = limx0-f(x)=limx0-xx=limx0x-x=-1 RHL = limx0+f(x)=limx0+xx=limx0xx=1 limx0-f(x) limx0+f(x)So limit does not exist.
Q.2 The value of 
is
Marks:1
Ans
Q.3 Find the derivative of :(a)3x4x2 (b)(5y42y2+1)(y1)
Marks:4
Ans
(a) f(x) =3x4x23df(x)dx=3x4x2=34‹2x=38x(b)f(y) =5y42y2+1y1 Using the product rule,
df(y)dy =ddy5y42y2+1(y1) =5y42y2+1(y1)+5y42y2+1(y1) =20y34y(y1)+5y42y2+1(1)
=20y420y34y2+4y+5y42y2+1 =25y420y36y2+4y+1
Q.4 If f(x)={mx2+n,x<0nx+m,0x1nx3+m,x>1For what values of m and n dolimx0f (x) and limx1f(x) both exist.
Marks:4
Ans
It is given that fx=mx2+n,x<0nx+m,0x1nx3+m,x>1limx 0f(x)and limx1f(x) both exist.limx 0f(x)=limx0+f(x)and limx1f(x)=limx1+f(x)limh0f(0h)
=limh 0f (0+h)andlimh0f(1h) =limh0f(1+ h)limh0m(h)2+n =limh0{nh + m}andlimh0{n(1h) + m} =limh0n (1 + h)3+ mn=m and n+m=n+m Hence limx 0f
(x) and limx1f (x) both exist for n=m.
Q.5 Differentiate with respect to x:ax2+bx+cx.
Marks:4
Ans
ddxax2+bx+cx =ddxax2x+bxx+cx
=ddxax32+bx12+cx12 =ddxax32+ddxbx12+ddxcx12
=addxx32+bddxx12+cddxx12 =a32x12+b12x12+c12x32 =3a2x12+b2x12c2x32.
FAQs (Frequently Asked Questions)
The most important questions cover range, mean deviation about mean, mean deviation about median, variance, standard deviation and shortcut method. Grouped-data questions are especially important for calculation practice.
Variance is the mean of squared deviations from mean. Standard deviation is the positive square root of variance and is easier to interpret because it uses the same unit as the data.
Range is called a rough measure because it depends only on maximum and minimum values. It ignores all other observations in the data set.
First find the mean or median. Then find absolute deviations from that value. Add those absolute deviations and divide by the number of observations or total frequency.
Yes, standard deviation shortcut method is important when values are large or data is grouped. It reduces calculation work by using assumed mean and step deviations.