Set A - Question 3

Question Details

• A. A data analyst is studying the factors affecting employee productivity in a technology company. The analyst collects the following data from 4 employees over a specific period:

  Employee	Training Hours	Break Hours	Productivity Score
  1	5	2	60
  2	10	2	68
  3	15	1	78
  4	20	1	88

  Determine which independent variable - Training Hours or Break Hours - has a stronger relationship with the Productivity Score. Support your answer using an appropriate correlation measure. (10 marks)

• B. A company has 8 employees and wants to select 3 employees to attend a training program. The order in which the employees are selected does not matter. How many different groups of 3 employees can be selected? (5 marks)

• C. The daily water consumption of households in a city is normally distributed with a mean of 180 litres and a variance of 400 litres square. What is the probability that a randomly selected household consumes more than 215 litres in a day? Assume given P(Z ≤ 1.75) = 0.9599. (5 marks)

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Model Answer

A. Correlation Analysis: Training Hours vs. Break Hours on Productivity (10 Marks)

1. Appropriate Measure of Correlation (1 Mark)

The appropriate metric is Pearson's Product-Moment Correlation Coefficient ($r$), which evaluates the strength and direction of the linear relationship between continuous/ratio scale quantitative variables: $r = \frac{\sum (X_i - \bar{X})(Y_i - \bar{Y})}{\sqrt{\sum (X_i - \bar{X})^2 \sum (Y_i - \bar{Y})^2}}$

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2. Variable Definitions & Sample Means (2 Marks)

Let:

• $X_1$: Training Hours
• $X_2$: Break Hours
• $Y$: Productivity Score
• Sample size: $n = 4$

Calculating Means:

$\bar{X}_1 = \frac{5 + 10 + 15 + 20}{4} = \frac{50}{4} = \mathbf{12.5}$ $\bar{X}_2 = \frac{2 + 2 + 1 + 1}{4} = \frac{6}{4} = \mathbf{1.5}$ $\bar{Y} = \frac{60 + 68 + 78 + 88}{4} = \frac{294}{4} = \mathbf{73.5}$

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3. Step-by-Step Computation Table (3 Marks)

Emp	$X_1$	$X_2$	$Y$	$(X_1 - \bar{X}_1)$	$(X_1 - \bar{X}_1)^2$	$(X_2 - \bar{X}_2)$	$(X_2 - \bar{X}_2)^2$	$(Y - \bar{Y})$	$(Y - \bar{Y})^2$	$(X_1 - \bar{X}_1)(Y - \bar{Y})$	$(X_2 - \bar{X}_2)(Y - \bar{Y})$
1	5	2	60	$-7.5$	$56.25$	$+0.5$	$0.25$	$-13.5$	$182.25$	$+101.25$	$-6.75$
2	10	2	68	$-2.5$	$6.25$	$+0.5$	$0.25$	$-5.5$	$30.25$	$+13.75$	$-2.75$
3	15	1	78	$+2.5$	$6.25$	$-0.5$	$0.25$	$+4.5$	$20.25$	$+11.25$	$-2.25$
4	20	1	88	$+7.5$	$56.25$	$-0.5$	$0.25$	$+14.5$	$210.25$	$+108.75$	$-7.25$
$\sum$	$50$	$6$	$294$	$0.0$	$125.0$	$0.0$	$1.0$	$0.0$	$443.0$	$+235.0$	$-19.0$

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4. Correlation 1: Training Hours ($X_1$) vs. Productivity Score ($Y$) (2 Marks)

$r(X_1, Y) = \frac{\sum (X_1 - \bar{X}_1)(Y - \bar{Y})}{\sqrt{\sum (X_1 - \bar{X}_1)^2 \cdot \sum (Y - \bar{Y})^2}}$ $r(X_1, Y) = \frac{235.0}{\sqrt{125.0 \times 443.0}} = \frac{235.0}{\sqrt{55375.0}} = \frac{235.0}{235.3189} \approx \mathbf{+0.9986}$

• Interpretation: There is an exceptionally strong, nearly perfect positive linear correlation between Training Hours and Productivity Score.

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5. Correlation 2: Break Hours ($X_2$) vs. Productivity Score ($Y$) (1 Mark)

$r(X_2, Y) = \frac{\sum (X_2 - \bar{X}_2)(Y - \bar{Y})}{\sqrt{\sum (X_2 - \bar{X}_2)^2 \cdot \sum (Y - \bar{Y})^2}}$ $r(X_2, Y) = \frac{-19.0}{\sqrt{1.0 \times 443.0}} = \frac{-19.0}{\sqrt{443.0}} = \frac{-19.0}{21.0476} \approx \mathbf{-0.9027}$

• Interpretation: There is a strong negative linear correlation between Break Hours and Productivity Score.

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6. Determination of Stronger Relationship (1 Mark)

The strength of a relationship is evaluated by comparing the absolute values of the correlation coefficients ($\lvert r \rvert$):

Independent Variable	Correlation ($r$)	Absolute Strength ($\lvert r \rvert$)	Coefficient of Determination ($R^2$)	Rank
Training Hours ($X_1$)	$+0.9986$	$0.9986$	$(0.9986)^2 \approx \mathbf{99.73\%}$	1 (Stronger)
Break Hours ($X_2$)	$-0.9027$	$0.9027$	$(-0.9027)^2 \approx \mathbf{81.49\%}$	2

• Final Decision: Training Hours has a stronger relationship with the Productivity Score than Break Hours because $\lvert r(X_1, Y) \rvert = 0.9986 > \lvert r(X_2, Y) \rvert = 0.9027$.
• Statistical Rationale: Training Hours accounts for $99.73\%$ of the variance in productivity, compared to $81.49\%$ accounted for by Break Hours.

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B. Combinatorics: Employee Selection (5 Marks)

1. Concept Identification (1 Mark)

Because the order of employee selection does not matter, the problem represents an unordered combination: $\binom{n}{r} = {}^nC_r$

2. Formula (1 Mark)

$\binom{n}{r} = \frac{n!}{r! \cdot (n - r)!}$

3. Step-by-Step Calculation (3 Marks)

• Total available employees: $n = 8$
• Target group size: $r = 3$

$\binom{8}{3} = \frac{8!}{3! \cdot (8 - 3)!} = \frac{8!}{3! \cdot 5!}$ $\binom{8}{3} = \frac{8 \times 7 \times 6 \times 5!}{(3 \times 2 \times 1) \times 5!}$ $\binom{8}{3} = \frac{8 \times 7 \times 6}{6} = 8 \times 7 = \mathbf{56}$

• Result: 56 different groups of 3 employees can be selected.

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C. Normal Distribution: Household Water Consumption (5 Marks)

1. Given Parameters (1 Mark)

Let $X$ denote the daily household water consumption in litres:

• Distribution: $X \sim \mathcal{N}(\mu, \sigma^2)$
• Population Mean: $\mu = 180 \text{ litres}$
• Population Variance: $\sigma^2 = 400 \text{ litres}^2$
• Population Standard Deviation: $\sigma = \sqrt{400} = \mathbf{20 \text{ litres}}$
• Consumption Threshold: $x = 215 \text{ litres}$
• Given Standard Normal Cumulative Probability: $P(Z \le 1.75) = 0.9599$

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2. Standardization ($Z$-Score Transformation) (2 Marks)

Converting the threshold value $x = 215$ into the standard normal variable $Z \sim \mathcal{N}(0, 1)$: $Z = \frac{X - \mu}{\sigma}$ $Z = \frac{215 - 180}{20} = \frac{35}{20} = \mathbf{1.75}$

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3. Probability Calculation (2 Marks)

We seek the probability of consuming more than 215 litres: $P(X > 215) = P(Z > 1.75)$

Applying the complement rule for cumulative probabilities: $P(Z > 1.75) = 1 - P(Z \le 1.75)$

Substituting the given value $P(Z \le 1.75) = 0.9599$: $P(X > 215) = 1 - 0.9599 = \mathbf{0.0401}$

• Final Result: The probability that a randomly selected household consumes more than 215 litres in a day is $0.0401$ (or $4.01\%$).