CBSE Important Questions Class 11 Maths Chapter 13 – Statistics
Class 11 Maths Chapter 13 Statistics is part of the CBSE Class 11 Maths syllabus. It goes beyond the mean, median and mode studied earlier and focuses on the measures of dispersion, which tell us how spread out the data is. Students learn the range, the mean deviation about the mean and the median, the variance and the standard deviation, and the coefficient of variation used to compare two sets of data.
These CBSE important questions for Class 11 Maths Chapter 13 cover finding the standard deviation of a frequency distribution, the mean deviation about the mean, the variance of ungrouped data, the relation between variance and standard deviation, and comparing the consistency of two data sets using the coefficient of variation. Every question has a clear step-by-step solution in simple English, so students can practise on their own, parents can check the working, and teachers can use them for class tests.
Q1. Find the standard deviation for the following data.
| xᵢ | 3 | 8 | 13 | 18 | 23 |
|---|---|---|---|---|---|
| fᵢ | 7 | 10 | 15 | 10 | 6 |
Answer: Prepare the table of fᵢxᵢ and fᵢxᵢ²:
| xᵢ | fᵢ | fᵢxᵢ | xᵢ² | fᵢxᵢ² |
|---|---|---|---|---|
| 3 | 7 | 21 | 9 | 63 |
| 8 | 10 | 80 | 64 | 640 |
| 13 | 15 | 195 | 169 | 2535 |
| 18 | 10 | 180 | 324 | 3240 |
| 23 | 6 | 138 | 529 | 3174 |
| Total | N = 48 | 614 | 9652 |
Using σ = √[Σfᵢxᵢ²/N − (Σfᵢxᵢ/N)²]:
= √[9652/48 − (614/48)²]
= √[201.08 − 163.60]
= √37.48
= 6.12 (approximately)
Q2. Write the formulae for variance and standard deviation. How are the two related?
Answer: For ungrouped data with n observations and mean x̄:
Variance σ² = Σ(xᵢ − x̄)² / n
Standard deviation σ = √[Σ(xᵢ − x̄)² / n]
For a frequency distribution:
σ² = Σfᵢ(xᵢ − x̄)² / N, where N = Σfᵢ
Relation: The standard deviation is simply the square root of the variance. Standard deviation has the same unit as the data, while variance has the square of that unit.
Q3. Find the mean deviation about the mean for the data 6, 7, 10, 12, 13, 4, 8, 12.
Answer: Mean: Sum = 6 + 7 + 10 + 12 + 13 + 4 + 8 + 12 = 72
x̄ = 72/8 = 9
Deviations |xᵢ − 9|: 3, 2, 1, 3, 4, 5, 1, 3
Sum of deviations = 22
Mean deviation = 22/8 = 2.75
Q4. Find the mean deviation about the median for the data 3, 9, 5, 3, 12, 10, 18, 4, 7, 19, 21.
Answer: Arrange in ascending order:
3, 3, 4, 5, 7, 9, 10, 12, 18, 19, 21
There are 11 observations (odd), so the median is the 6th value = 9.
Deviations |xᵢ − 9|: 6, 6, 5, 4, 2, 0, 1, 3, 9, 10, 12
Sum = 58
Mean deviation about the median = 58/11 = 5.27 (approximately)
Q5. Find the variance of the first 10 natural numbers.
Answer: The numbers are 1, 2, 3, ..., 10.
Mean: x̄ = (1 + 2 + ... + 10)/10 = 55/10 = 5.5
Σxᵢ² = 1² + 2² + ... + 10² = 385
Using σ² = Σxᵢ²/n − (x̄)²:
= 385/10 − (5.5)²
= 38.5 − 30.25
= 8.25
Q6. What is the coefficient of variation? How is it used to compare two sets of data?
Answer: The coefficient of variation (CV) measures the spread of data relative to its mean:
CV = (σ / x̄) × 100
It has no unit, so it can be used to compare the variability of two sets of data even if their units or means are different.
Rule: The set with the smaller CV is more consistent (less variable), and the one with the larger CV is more variable.
Q7. The mean and standard deviation of two sets of data are: Set A — mean 50, SD 5; Set B — mean 80, SD 6. Which set is more consistent?
Answer: Compare the coefficients of variation.
Set A: CV = (5/50) × 100 = 10%
Set B: CV = (6/80) × 100 = 7.5%
Since Set B has the smaller CV, Set B is more consistent.
Note that Set B has the larger standard deviation, but because its mean is also much larger, its relative variability is smaller.
Q8. Find the standard deviation of the data 6, 8, 10, 12, 14, 16, 18, 20, 22, 24.
Answer: Mean: Sum = 150, so x̄ = 150/10 = 15
Deviations (xᵢ − 15): −9, −7, −5, −3, −1, 1, 3, 5, 7, 9
Squares: 81, 49, 25, 9, 1, 1, 9, 25, 49, 81
Sum of squares = 330
Variance = 330/10 = 33
Standard deviation = √33 = 5.74 (approximately)
Q9. Name the measures of dispersion. Which one is considered the best and why?
Answer: The main measures of dispersion are:
1. Range = highest value − lowest value
2. Mean deviation (about the mean or the median)
3. Variance
4. Standard deviation
The standard deviation is considered the best measure because it uses every observation, it is based on squared deviations so the signs do not cancel out, it has the same unit as the data, and it is suitable for further mathematical and statistical work.
Q10. If the variance of a data set is 36, find its standard deviation. If each observation is multiplied by 3, what happens to the standard deviation?
Answer: Standard deviation = √variance = √36 = 6
If each observation is multiplied by 3: The standard deviation also gets multiplied by 3.
New SD = 3 × 6 = 18
(And the new variance becomes 3² × 36 = 324. Note that adding a constant to every observation does not change the standard deviation at all.)
Q11. Find the mean and variance of the data 2, 4, 5, 6, 8, 17.
Answer: Mean: Sum = 2 + 4 + 5 + 6 + 8 + 17 = 42
x̄ = 42/6 = 7
Deviations (xᵢ − 7): −5, −3, −2, −1, 1, 10
Squares: 25, 9, 4, 1, 1, 100
Sum of squares = 140
Variance = 140/6 = 23.33 (approximately)
Q12. The mean of 5 observations is 4.4 and their variance is 8.24. If three of the observations are 1, 2 and 6, find the other two.
Answer: Let the other two be x and y.
From the mean: (1 + 2 + 6 + x + y)/5 = 4.4
9 + x + y = 22, so x + y = 13 ...(i)
From the variance: Σxᵢ²/5 − (4.4)² = 8.24
Σxᵢ²/5 = 8.24 + 19.36 = 27.6
Σxᵢ² = 138
So 1 + 4 + 36 + x² + y² = 138, giving x² + y² = 97 ...(ii)
From (i), (x + y)² = 169, so x² + y² + 2xy = 169
97 + 2xy = 169, so xy = 36
Now x and y are the roots of t² − 13t + 36 = 0, that is (t − 9)(t − 4) = 0.
So the other two observations are 9 and 4.
Related Study Material on Extramarks
- CBSE Important Questions Class 11 Maths Chapter 1
- CBSE Important Questions Class 11 Maths Chapter 2
- CBSE Important Questions Class 11 Maths Chapter 3
- CBSE Important Questions Class 11 Maths Chapter 4
- CBSE Important Questions Class 11 Maths Chapter 5
- CBSE Important Questions Class 11 Maths Chapter 6
- CBSE Important Questions Class 11 Maths Chapter 7
- CBSE Important Questions Class 11 Maths Chapter 8
- CBSE Important Questions Class 11 Maths Chapter 9
- CBSE Important Questions Class 11 Maths Chapter 10
- CBSE Important Questions Class 11 Maths Chapter 11
- CBSE Important Questions Class 11 Maths Chapter 12
- CBSE Important Questions Class 11 Maths Chapter 14
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.