RowQ
The Vault
RowQ
The Vault
CBSE Class 11 Maths · 10 questions · 24 marks
Averages alone can hide a lot: two classes can have identical mean marks while one is far more consistent than the other. This chapter introduces measures of dispersion — mean deviation, variance, and standard deviation — that quantify exactly how spread out data is around its centre. You will compute these for both raw and grouped data, and use the coefficient of variation to compare the consistency of two different datasets.
The range of the data set 12, 7, 25, 18, 3, 30 is:
Answer
The range is 27. Range = maximum - minimum. Here maximum = 30 and minimum = 3. Range = 30 - 3 = 27.
If each observation in a data set is increased by 5, the standard deviation:
Answer
The standard deviation remains unchanged. Adding a constant shifts every value by the same amount, so the spread of the data around its (also shifted) mean stays exactly the same. Standard deviation is unaffected by such a shift.
For the data 4, 8, 6, 10, 2 (mean = 6), the mean deviation about the mean is:
Answer
The mean deviation is 2.4. Deviations from mean 6: |4-6|=2, |8-6|=2, |6-6|=0, |10-6|=4, |2-6|=4. Sum = 2+2+0+4+4=12. MD = 12/5 = 2.4.
Two batsmen A and B have equal mean scores, but A's standard deviation is 4 and B's is 7 (same mean). Which batsman is more consistent?
Answer
Batsman A is more consistent. A lower standard deviation (relative to the same mean, hence lower CV) means the scores are more tightly clustered around the mean, i.e., more consistent performance. Since A's SD (4) is less than B's SD (7), A is more consistent.
Assertion (A): If every observation of a data set is multiplied by 3, the variance becomes 9 times the original variance. Reason (R): Multiplying each observation by a constant k multiplies the standard deviation by |k|.
Answer
Both A and R are true and R is the correct explanation of A. If SD is multiplied by 3 (from R), then variance (SD squared) is multiplied by 3² = 9, confirming Assertion A is true. Reason R gives the scaling rule for standard deviation under multiplication, and squaring that rule is exactly how Assertion A's variance-scaling follows, so R correctly explains A.
Find the mean and range of the data: 15, 22, 18, 9, 26.
Answer
Mean = (15+22+18+9+26)/5 = 90/5 = 18. Range = maximum - minimum = 26 - 9 = 17. So the mean is 18 and the range is 17.
Find the variance of the data: 2, 4, 6, 8, 10.
Answer
Mean x̄ = (2+4+6+8+10)/5 = 30/5 = 6. Deviations: (2-6)=-4, (4-6)=-2, (6-6)=0, (8-6)=2, (10-6)=4. Squares: 16, 4, 0, 4, 16. Sum = 40. Variance = 40/5 = 8.
Calculate the mean, variance and standard deviation for the following frequency data: x: 2, 4, 6, 8; f: 3, 5, 4, 2.
Answer
Step 1 — Total frequency: Σf = 3+5+4+2 = 14. Step 2 — Compute Σfx: 2×3=6, 4×5=20, 6×4=24, 8×2=16. Sum = 6+20+24+16 = 66. Step 3 — Mean x̄ = Σfx/Σf = 66/14 = 33/7 ≈ 4.714. Step 4 — Compute Σfx²: 2²×3=12, 4²×5=80, 6²×4=144, 8²×2=128. Sum = 12+80+144+128 = 364. Step 5 — Variance σ² = Σfx²/Σf - (x̄)² = 364/14 - (33/7)² = 26 - 1089/49. Convert 26 to 49ths: 26 = 1274/49. So σ² = 1274/49 - 1089/49 = 185/49 ≈ 3.776. Step 6 — Standard deviation σ = √(185/49) ≈ √3.776 ≈ 1.943. So mean ≈ 4.71, variance ≈ 3.78, and SD ≈ 1.94.
The mean and standard deviation of 20 observations were found to be 10 and 2 respectively. On checking, it was found that one observation, recorded as 8, was actually 12. Find the correct mean and correct standard deviation.
Answer
Step 1 — Incorrect Σx = mean × n = 10 × 20 = 200. Step 2 — Correct Σx = 200 - 8 + 12 = 204. Step 3 — Correct mean = 204/20 = 10.2. Step 4 — Incorrect variance = 4 (since SD=2), so incorrect Σx² /n - (mean)² = 4, giving incorrect Σx²/20 = 4 + 100 = 104, so incorrect Σx² = 2080. Step 5 — Correct Σx² = 2080 - 8² + 12² = 2080 - 64 + 144 = 2160. Step 6 — Correct variance = correct Σx²/n - (correct mean)² = 2160/20 - (10.2)² = 108 - 104.04 = 3.96. Step 7 — Correct SD = √3.96 ≈ 1.99. So the correct mean is 10.2 and the correct standard deviation is approximately 1.99.
Read the following and answer the questions that follow: The marks (out of 10) of 5 students in a surprise quiz are: 6, 8, 4, 10, 2. (a) Find the mean of the marks. (b) Find the deviations of each mark from the mean. (c) Find the mean deviation about the mean. (d) Find the variance of the marks.
Answer
(a) Mean = (6+8+4+10+2)/5 = 30/5 = 6. (b) Deviations: 6-6=0, 8-6=2, 4-6=-2, 10-6=4, 2-6=-4. (c) Absolute deviations: 0,2,2,4,4, sum=12. MD = 12/5 = 2.4. (d) Squared deviations: 0,4,4,16,16, sum=40. Variance = 40/5 = 8.
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