- 2072 Chaitra · 6 marks
Corrected mean and SD after item error
Short answer
Corrected sums: (i) omitting 8: Σx=192, Σx²=2016; mean=10.105, σ≈1.997. (ii) replacing with 12: Σx=204, Σx²=2160; mean=10.2, σ≈1.990.
What are the differences between measures of central tendency and measures of dispersion? The mean and standard deviation of 20 items is found to be 10 and 2 respectively. At the time of checking it was found that one item 8 was incorrect. Calculate the mean and standard deviation if: (i) the wrong item is omitted (ii) it is replaced by 12.
Central tendency vs dispersion
| Basis | Measures of central tendency | Measures of dispersion |
|---|---|---|
| Purpose | Give the central or typical value | Show how far values are spread about the centre |
| Examples | Mean, median, mode | Range, quartile deviation, mean deviation, standard deviation |
| Answers | "Where is the middle?" | "How scattered is the data?" |
| Units | Same units as data | Absolute measures in the same units; relative measures have none |
| Use | Summarise a data set by one value | Judge reliability of the average and compare consistency |
Numerical
Given n = 20, x̄ = 10, σ = 2. The wrong item is 8. The SD is taken with divisor n (σ² = Σx²/n − x̄²).
Σx = 20 × 10 = 200
Σx² = 20 (4 + 100) = 2080
(i) Wrong item omitted (n = 19)
Σx = 200 − 8 = 192
Σx² = 2080 − 64 = 2016
Mean = 192/19 = 10.105
σ² = 2016/19 − (10.105)² = 106.105 − 102.116 = 3.989
σ = 1.997
(ii) Replaced by 12 (n = 20)
Σx = 200 − 8 + 12 = 204
Σx² = 2080 − 64 + 144 = 2160
Mean = 204/20 = 10.2
σ² = 2160/20 − (10.2)² = 108 − 104.04 = 3.96
σ = 1.990
Answer: (i) mean ≈ 10.11, SD ≈ 2.00; (ii) mean = 10.2, SD ≈ 1.99.
✓ Answer checked by Claude Sonnet 5.5
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