Question 1 OR (d)

Average, SD and CV of two dealers' daily sales, and the effect of a constant change

9 marks

Time to write: About 18 minutes
Topic: Measures of dispersion

Question

Question 1 OR (d) · 9 marks

Taraa Ltd., which manufactures electric scooters, recorded the daily sale of two authorized dealers over ten days of the festive season, as given below:

Dealer A	41	36	42	34	37	47	39	44	33	31
Dealer B	38	34	36	38	39	41	36	41	37	30

Calculate and comment on:

Average number of electric scooters sold by each of the dealers,
Standard Deviation and Coefficient of variation of each dealer,
What will be the change in the above statistics if the daily sales of dealer A rise by 6 scooter per day owing to promotional schemes, and the daily sales of dealer B fall by 8 per day owing to lower demand?

Reworded from the university paper; the task is the same. The original paper belongs to the University of Delhi. See DU's question bank.

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18 minfor 9 marks

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Read; underline the asks; jot your points 1 min
1 Means of the two dealers 2 marks 4 min
2 Standard deviations and CVs, with comment on consistency 4 marks 7 min
3 Revised mean, SD and CV after the change, with reasons 3 marks 5 min
Check: every ask answered, key terms underlined 1 min

123

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Solution

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Method. Mean x̄ = Σx / n; standard deviation σ = √[Σ(x − x̄)² / n], working from deviations about the actual mean; coefficient of variation CV = (σ / x̄) × 100. The dealer with the smaller CV is the more consistent.

Working table (daily sales, scooters; d = x − x̄)

Day	A: x	A: d	A: d²	B: x	B: d	B: d²
1	41	2.6	6.76	38	1	1
2	36	−2.4	5.76	34	−3	9
3	42	3.6	12.96	36	−1	1
4	34	−4.4	19.36	38	1	1
5	37	−1.4	1.96	39	2	4
6	47	8.6	73.96	41	4	16
7	39	0.6	0.36	36	−1	1
8	44	5.6	31.36	41	4	16
9	33	−5.4	29.16	37	0	0
10	31	−7.4	54.76	30	−7	49
Total	384	0	236.4	370	0	98

(i) Average sales

Dealer A: x̄ = 384 / 10 = 38.4 scooters a day.
Dealer B: x̄ = 370 / 10 = 37.0 scooters a day.

Dealer A sells about 1.4 scooters a day more on average.

(ii) Standard deviation and coefficient of variation

Dealer A: σ² = 236.4 / 10 = 23.64, σ = 4.86; CV = (4.86 / 38.4) × 100 = 12.66%.
Dealer B: σ² = 98 / 10 = 9.80, σ = 3.13; CV = (3.13 / 37.0) × 100 = 8.46%.

Comment. B's CV is lower, so B's daily sales are the more consistent (uniform); A sells more on average but its sales swing more from day to day.

(iii) Effect of the change in daily sales

Adding or subtracting a constant from every observation shifts the mean by that constant but leaves the standard deviation unchanged, since SD is not affected by a change of origin; the AM moves with the origin. The CV changes because its denominator changes.

Measure	Dealer A (+6 a day)	Dealer B (−8 a day)
Mean	38.4 + 6 = 44.4	37.0 − 8 = 29.0
Standard deviation	4.86 (unchanged)	3.13 (unchanged)
Coefficient of variation	(4.86 / 44.4) × 100 = 10.95%	(3.13 / 29.0) × 100 = 10.79%

Comment. A's relative variability falls (12.66% to 10.95%) because the same spread now sits on a larger average; B's rises (8.46% to 10.79%) because the same spread sits on a smaller average. B remains slightly the more consistent dealer, but the gap almost closes.

Where the 9 marks sit

#	Component	Marks
1	Means of the two dealers	2
2	Standard deviations and CVs, with comment on consistency	4
3	Revised mean, SD and CV after the change, with reasons	3
	Total	9

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