2 The table summarises the diameters, \(d\) millimetres, of a random sample of 60 new cricket balls to be used in junior cricket.
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Question 2:
Part (a):
Answer Marks
Guidance
Answer Marks
Guidance
Midpoints: \(65.5, 66.5, 67.5, 68.5, 69.5, 70.5, 71.5\) B1
At least one correct midpoint seen
\(\bar{d} = \dfrac{5(65.5)+9(66.5)+12(67.5)+15(68.5)+10(69.5)+7(70.5)+2(71.5)}{60}\) M1
Correct method for mean
\(= \dfrac{4115}{60} = 68.583...\) A1
Awrt \(68.6\) mm
Variance \(= \dfrac{\sum fd^2}{\sum f} - \bar{d}^2\) M1
Correct method for variance
\(= \dfrac{282110.5}{60} - 68.583...^2 = 4701.84... - 4703.69... \) Variance \(\approx 1.87\) (mm²) A1
Awrt \(1.87\)
Part (b):
Answer Marks
Guidance
Answer Marks
Guidance
New mean \(= \dfrac{68.583...}{25.4} \approx 2.70\) inches B1
Awrt \(2.70\)
New variance \(= \dfrac{1.87...}{25.4^2} \approx 0.00290\) inches² B1
Awrt \(0.00290\); divide by \(25.4^2\)
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# Question 2:
## Part (a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Midpoints: $65.5, 66.5, 67.5, 68.5, 69.5, 70.5, 71.5$ | B1 | At least one correct midpoint seen |
| $\bar{d} = \dfrac{5(65.5)+9(66.5)+12(67.5)+15(68.5)+10(69.5)+7(70.5)+2(71.5)}{60}$ | M1 | Correct method for mean |
| $= \dfrac{4115}{60} = 68.583...$ | A1 | Awrt $68.6$ mm |
| Variance $= \dfrac{\sum fd^2}{\sum f} - \bar{d}^2$ | M1 | Correct method for variance |
| $= \dfrac{282110.5}{60} - 68.583...^2 = 4701.84... - 4703.69... $ | | |
| Variance $\approx 1.87$ (mm²) | A1 | Awrt $1.87$ |
## Part (b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| New mean $= \dfrac{68.583...}{25.4} \approx 2.70$ inches | B1 | Awrt $2.70$ |
| New variance $= \dfrac{1.87...}{25.4^2} \approx 0.00290$ inches² | B1 | Awrt $0.00290$; divide by $25.4^2$ |
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2 The table summarises the diameters, $d$ millimetres, of a random sample of 60 new cricket balls to be used in junior cricket.
\hfill \mbox{\textit{AQA S1 2015 Q2 [6]}}