OCR MEI S1 2013 June — Question 1 6 marks

Exam BoardOCR MEI
ModuleS1 (Statistics 1)
Year2013
SessionJune
Marks6
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicMeasures of Location and Spread
TypeForward transformation: find new statistics
DifficultyModerate -0.8 This is a straightforward application of standard formulas and transformation rules. Part (i) requires direct substitution into mean and standard deviation formulas (routine recall), while part (ii) applies the well-known linear transformation properties (mean scales and shifts, SD only scales). Both are textbook exercises with no problem-solving or insight required.
Spec2.02g Calculate mean and standard deviation5.02c Linear coding: effects on mean and variance
1 The weights, \(x\) grams, of 100 potatoes are summarised as follows. $$n = 100 \quad \sum x = 24940 \quad \sum x ^ { 2 } = 6240780$$
  1. Calculate the mean and standard deviation of \(x\).
  2. The weights, \(y\) grams, of the potatoes after they have been peeled are given by the formula \(y = 0.9 x - 15\). Deduce the mean and standard deviation of the weights of the potatoes after they have been peeled.
Part (i)
Answer:
\[\text{Mean} = \frac{24940}{100} = 249.4\text{g or }249\text{g}\]
\[S_{xx} = 6240780 - \frac{24940^2}{100} = 20744\]
\[s = \sqrt{\frac{20744}{99}} = \sqrt{209.53} = 14.4751 = 14.5\text{g}\]
AnswerMarks Guidance
Marks: B1M1 A1
Guidance:
- B1: Ignore units. CAO. NB 249.40 gets B0 for over-specification
- M1: For \(S_{xx}\). M1 for \(6240780 - 100 \times \text{their mean}^2\) BUT NOTE M0 if their \(S_{xx} < 0\)
- A1: CAO ignore units. For \(s^2\) of 210 (or better) allow M1A0 with or without working. For RMSD of 14.4 (or better) allow M1A0 provided working seen. For RMSD² of 207 (or better) allow M1A0 provided working seen. Allow 14.48 but NOT 14.47
Part (ii)
Answer:
\[\text{New mean} = (0.9 \times 249.4) - 15 = 209.5\text{g}\]
\[\text{New sd} = 0.9 \times 14.48 = 13.03\text{g}\]
AnswerMarks Guidance
Marks: B1M1 A1
Guidance:
- B1: FT their mean provided answer is positive
- M1: FT their sd
- A1: FT Allow 13.0 to 13.1
- Overall: If candidate 'starts again' only award marks for CAO. Allow 209. Or for \(0.9^2 \times 14.5^2\). Deduct at most 1 mark overall in whole question for over-specification of Mean and 1 mark overall for SD
## Part (i)
**Answer:** 
$$\text{Mean} = \frac{24940}{100} = 249.4\text{g or }249\text{g}$$

$$S_{xx} = 6240780 - \frac{24940^2}{100} = 20744$$

$$s = \sqrt{\frac{20744}{99}} = \sqrt{209.53} = 14.4751 = 14.5\text{g}$$

**Marks:** B1 | M1 | A1 | [3]

**Guidance:** 
- B1: Ignore units. CAO. NB 249.40 gets B0 for over-specification
- M1: For $S_{xx}$. M1 for $6240780 - 100 \times \text{their mean}^2$ BUT NOTE M0 if their $S_{xx} < 0$
- A1: CAO ignore units. For $s^2$ of 210 (or better) allow M1A0 with or without working. For RMSD of 14.4 (or better) allow M1A0 provided working seen. For RMSD² of 207 (or better) allow M1A0 provided working seen. Allow 14.48 but NOT 14.47

## Part (ii)
**Answer:**
$$\text{New mean} = (0.9 \times 249.4) - 15 = 209.5\text{g}$$

$$\text{New sd} = 0.9 \times 14.48 = 13.03\text{g}$$

**Marks:** B1 | M1 | A1 | [3]

**Guidance:**
- B1: FT their mean provided answer is positive
- M1: FT their sd
- A1: FT Allow 13.0 to 13.1
- Overall: If candidate 'starts again' only award marks for CAO. Allow 209. Or for $0.9^2 \times 14.5^2$. Deduct at most 1 mark overall in whole question for over-specification of Mean and 1 mark overall for SD

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1 The weights, $x$ grams, of 100 potatoes are summarised as follows.

$$n = 100 \quad \sum x = 24940 \quad \sum x ^ { 2 } = 6240780$$

(i) Calculate the mean and standard deviation of $x$.\\
(ii) The weights, $y$ grams, of the potatoes after they have been peeled are given by the formula $y = 0.9 x - 15$. Deduce the mean and standard deviation of the weights of the potatoes after they have been peeled.

\hfill \mbox{\textit{OCR MEI S1 2013 Q1 [6]}}
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