Question 1 OR (d)

Mean and SD of two machines' output; consistency; effect of changing every observation

9 marks

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

Question

Question 1 OR (d) · 9 marks

A toy manufacturing firm kept daily production data of two machines for a period of ten days, as listed below:

Day no.	1	2	3	4	5	6	7	8	9	10
Output of Machine A	74	68	76	63	69	83	69	78	68	70
Output of Machine B	72	66	68	72	74	75	67	74	64	68

Calculate and comment on the Mean and Standard Deviation.
Which machine makes toys more consistently?
What change will occur in the statistics if every observation of Machine A is decreased by 10, and every observation of Machine B is increased by 25? Show the revised results.

Reworded from the university paper: names and figures differ, the task is the same. The original paper belongs to the University of Delhi. See DU's question bank.

Before you write

Read off the question and its marks

18 minfor 9 marks

Read from the question

Calculate…Which…?What…?4 questions in all: answer every one.
and comment onThe comment carries marks too: say what the figures show, in two or three lines.
CalculateFormula first, then the working, then the answer with its unit (₹, %, units).

Write it in this order

Read the data; note exactly what is asked 1 min
1 Working table, means and SDs, with comment 4 marks 7 min
2 Consistency by CV 2 marks 4 min
3 Revised results after the change, with reasons 3 marks 5 min
Check: arithmetic, units (₹, %), final answer underlined 1 min

123

Try it first. About 18 minutes for 9 marks. Write it out, then open the solution.

Solution

Hover or tap a marked figure, like 72,000, to see how it is made and where else it appears.

Method. Mean x̄ = Σx / n; standard deviation σ = √[Σ(x − x̄)² / n], with deviations about the actual mean; consistency is judged by the coefficient of variation, CV = (σ / x̄) × 100.

Working table (daily output, toys; d = x − x̄)

Day	A: x	A: d	A: d²	B: x	B: d	B: d²
1	74	2.2	4.84	72	2	4
2	68	−3.8	14.44	66	−4	16
3	76	4.2	17.64	68	−2	4
4	63	−8.8	77.44	72	2	4
5	69	−2.8	7.84	74	4	16
6	83	11.2	125.44	75	5	25
7	69	−2.8	7.84	67	−3	9
8	78	6.2	38.44	74	4	16
9	68	−3.8	14.44	64	−6	36
10	70	−1.8	3.24	68	−2	4
Total	718	0	311.6	700	0	134

(i) Mean and standard deviation

Measure	Machine A	Machine B
Mean	718 / 10 = 71.8	700 / 10 = 70.0
Variance	311.6 / 10 = 31.16	134 / 10 = 13.40
Standard deviation	√31.16 = 5.58	√13.40 = 3.66
Coefficient of variation	(5.58 / 71.8) × 100 = 7.77%	(3.66 / 70.0) × 100 = 5.23%

Comment. Machine A produces slightly more on average (71.8 against 70.0 toys a day), but its output varies much more from day to day (SD 5.58 against 3.66).

(ii) Which machine is more consistent? Machine B: its CV (5.23%) is lower than A's (7.77%), so its daily output is the more uniform.

(iii) Effect of the change

Subtracting 10 from every output of A, or adding 25 to every output of B, is a change of origin. The mean moves by the same constant, while the standard deviation is unchanged; the CV changes because the mean changes.

Revised measure	Machine A (each − 10)	Machine B (each + 25)
Mean	71.8 − 10 = 61.8	70.0 + 25 = 95.0
Standard deviation	5.58 (unchanged)	3.66 (unchanged)
Coefficient of variation	(5.58 / 61.8) × 100 = 9.03%	(3.66 / 95.0) × 100 = 3.85%

After the change, A's relative variability rises (7.77% to 9.03%) and B's falls (5.23% to 3.85%); B now also has the higher average, and it remains the more consistent machine, by a wider margin.

Where the 9 marks sit

#	Component	Marks
1	Working table, means and SDs, with comment	4
2	Consistency by CV	2
3	Revised results after the change, with reasons	3
	Total	9

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