Edexcel IAL S1 October 2022 question paper and mark scheme (WST01/01)

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1. Data and Summary Statistics · 9 marks

   Two different groups of students took a memory test. Group A consists of \(20\) students. Their mean score is \(15\) and the standard deviation is \(3\). Group B consists of \(30\) students. Their mean score is \(20\) and the standard deviation is \(4\). The two groups are combined into a single group of \(50\) students.

   1. Find the combined mean score.

   2. By calculating the total sum of squares (\(\sum x^2\)) for each group separately, find the overall standard deviation of the combined group.

   Worked answer

   1. Sum A \(= 20 \times 15 = 300\). Sum B \(= 30 \times 20 = 600\). M1
      Total Sum \(= 300 + 600 = 900\). Total students \(= 50\).
      Combined mean \(= \frac{900}{50} = 18\). A1

   2. For Group A: \(\sigma_A^2 = 3^2 = 9\).
      \(9 = \frac{\sum x_A^2}{20} - 15^2 \implies \frac{\sum x_A^2}{20} = 9 + 225 = 234 \implies \sum x_A^2 = 4680\). M1A1
      For Group B: \(\sigma_B^2 = 4^2 = 16\).
      \(16 = \frac{\sum x_B^2}{30} - 20^2 \implies \frac{\sum x_B^2}{30} = 16 + 400 = 416 \implies \sum x_B^2 = 12480\). M1A1
      Total sum of squares \(\sum x^2 = 4680 + 12480 = 17160\). A1
      Combined Variance \(= \frac{17160}{50} - 18^2 = 343.2 - 324 = 19.2\). M1
      Combined SD \(= \sqrt{19.2} = 4.3817... \implies 4.38\) (3sf). A1

2. Representation of Data · 5 marks

   A histogram is drawn. The class $0 \le x < 20$ has frequency $10$, and the class $20 \le x < 30$ has frequency $30$. If the height of the bar for the first class is $2\text{ cm}$, find the height of the bar for the second class, assuming the width of the bars is proportional to the class width.

   Worked answer

   1. M1 for stating the class widths: $20$ and $10$.
   2. M1 for establishing the width ratio: if class width $20$ has a certain drawn width, class width $10$ has half that drawn width. Let drawn width of first bar be $2w$, then second bar is $w$.
   3. M1 for using Area $\propto$ Frequency for the first bar: Area $= 2w \times 2 = 4w$. So $4w$ represents frequency $10$, meaning $w$ represents frequency $2.5$.
   4. M1 for applying to the second bar: Area $= w \times h$. Frequency $= 30$. So $w \times h$ represents $30$. Since $w$ represents $2.5$, Area must be $12w$.
   5. A1 for solving for height: $w \times h = 12w \implies h = 12\text{ cm}$.

3. Discrete Random Variables · 4 marks

   The discrete random variable $X$ takes the values $1, 2,$ and $3$ with probabilities $a, b,$ and $0.3$ respectively. Given that $E(X) = 1.9$, find the values of $a$ and $b$.

   Worked answer

   1. M1 for forming the sum of probabilities equation: $a + b + 0.3 = 1 \implies a + b = 0.7$.
   2. M1 for forming the expected value equation: $1(a) + 2(b) + 3(0.3) = 1.9 \implies a + 2b = 1.0$.
   3. M1 for solving the simultaneous equations (e.g., subtracting the first from the second: $b = 1.0 - 0.7$).
   4. A1 for concluding $b = 0.3$ and $a = 0.4$.

More on these topics: Data and Summary Statistics · Representation of Data · Discrete Random Variables

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• Data and Summary Statistics
• Representation of Data
• Probability
• Correlation and Regression
• Discrete Random Variables
• Normal Distribution
• Measures of Location and Spread

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Formula booklet

Mathematical Formulae and Statistical Tables (International AS/A Level Mathematics) · Pearson Edexcel · For the 2018 IAL specification (current)