CAIE S1 2003 November — Question 1 4 marks

Exam BoardCAIE
ModuleS1 (Statistics 1)
Year2003
SessionNovember
Marks4
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicMeasures of Location and Spread
TypeCalculate statistics from discrete frequency table
DifficultyEasy -1.8 This is a straightforward calculation requiring only basic formulas for mean and variance from a simple two-value frequency distribution. No problem-solving or conceptual understanding is needed—just direct substitution into standard formulas with minimal arithmetic.
Spec2.02f Measures of average and spread2.02g Calculate mean and standard deviation

1 A computer can generate random numbers which are either 0 or 2 . On a particular occasion, it generates a set of numbers which consists of 23 zeros and 17 twos. Find the mean and variance of this set of 40 numbers.

Part (i):
AnswerMarks Guidance
\(x\)0 2
freq23 17
OR
\(P(0) = \frac{23}{40}\), \(P(2) = \frac{17}{40}\)
Mean \(= \frac{34}{40} = 0.850\)
AnswerMarks Guidance
Variance \(= \frac{(4 \times 17)}{40} - (0.85)^2 = 0.978\) (exact answer 0.9775) (\(\frac{391}{400}\))M1 For reasonable attempt at the mean using freqs or probs but not using prob=0.5
A1For correct mean
M1For correct variance formula
A1ftFor correct answer
4 marks total
Part (ii):
frequencies: 3, 7, 6, 3, 1
scaled frequencies: 3, 7, 3, 1.5, 0.5
AnswerMarks Guidance
or 0.006, 0.014, 0.006, 0.003, 0.001M1 For frequencies and attempt at scaling, accept cw/freq but not cw \(\times\) freq, not cw/mid point
Histogram with:
- Correct heights from scaled frequencies
- Correct widths of bars, uniform horiz scale, no halves or gaps or less-than-or-equal tos
AnswerMarks Guidance
- Both axes labelled, fd and area or m². Not class widthA1 For correct heights from their scaled frequencies seen on the graph
B1For correct widths of bars, uniform horiz scale, no halves or gaps or less-than-or-equal tos
B1Both axes labelled, fd and area or m². Not class width
4 marks total
**Part (i):**
$x$ | 0 | 2
freq | 23 | 17

OR

$P(0) = \frac{23}{40}$, $P(2) = \frac{17}{40}$

Mean $= \frac{34}{40} = 0.850$

Variance $= \frac{(4 \times 17)}{40} - (0.85)^2 = 0.978$ (exact answer 0.9775) ($\frac{391}{400}$) | M1 | For reasonable attempt at the mean using freqs or probs but not using prob=0.5
A1 | For correct mean
M1 | For correct variance formula
A1ft | For correct answer
**4 marks total**

**Part (ii):**
frequencies: 3, 7, 6, 3, 1

scaled frequencies: 3, 7, 3, 1.5, 0.5
or 0.006, 0.014, 0.006, 0.003, 0.001 | M1 | For frequencies and attempt at scaling, accept cw/freq but not cw $\times$ freq, not cw/mid point

**Histogram with:**
- Correct heights from scaled frequencies
- Correct widths of bars, uniform horiz scale, no halves or gaps or less-than-or-equal tos
- Both axes labelled, fd and area or m². Not class width | A1 | For correct heights from their scaled frequencies seen on the graph
B1 | For correct widths of bars, uniform horiz scale, no halves or gaps or less-than-or-equal tos
B1 | Both axes labelled, fd and area or m². Not class width
**4 marks total**

---
1 A computer can generate random numbers which are either 0 or 2 . On a particular occasion, it generates a set of numbers which consists of 23 zeros and 17 twos. Find the mean and variance of this set of 40 numbers.

\hfill \mbox{\textit{CAIE S1 2003 Q1 [4]}}