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Unit VI: Statistics and Probability — Test 2

Reasoning, proof and modelling · 9 questions · 26 practice marks · 40 minutes.
MCQs are scored automatically; written answers use step-wise self/teacher review.

Section A
Question 1 (1 Mark)
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Full question paper and step-wise model answers

Practice marks are not official examination weightage. Written answers require self- or teacher assessment.

Question 1 · 1 mark

Two equally large groups have means 12 and 18. Their combined mean is:

  1. 6
  2. 12
  3. 15
  4. 30
Model answer and reasoning

(c) 15

  1. The weights are equal.
  2. The weighted mean reduces to (12 + 18)/2 = 15.

Question 2 · 1 mark

An impossible event has probability:

  1. −1
  2. 0
  3. 1/2
  4. 1
Model answer and reasoning

(b) 0

  1. An impossible event contains no possible outcomes.
  2. Its probability is 0. A certain event has probability 1.

Question 3 · 1 mark

A 100% stacked bar graph always makes each complete bar represent:

  1. The same absolute count
  2. 100% of its own group
  3. Only its largest category
  4. The arithmetic mean
Model answer and reasoning

(b) 100% of its own group

  1. Each group’s category frequencies are converted to percentages of that group.
  2. The complete bar is 100%, even when the group sizes differ.

Question 4 · 4 marks

A survey records daily reading times 5, 10, 10, 15, 20, 60 minutes. Find mean, median and mode. Explain one sampling issue to consider before using this survey to represent the whole school.

Model answer and reasoning

Mean 20 min; median 12.5 min; mode 10 min.

  1. The sum is 120, so the mean is 120/6 = 20.
  2. There are six sorted values; average the third and fourth: (10 + 15)/2 = 12.5.
  3. 10 occurs most frequently, so it is the mode.
  4. A small or selected group may not represent all ages/classes. Use a fair sampling plan, a clear question and a consistent measurement period.

Question 5 · 3 marks

Rice costing ₹40/kg and ₹60/kg is mixed in quantities 3 kg and 2 kg. Find the average cost per kilogram and verify it by total cost.

Model answer and reasoning

₹48/kg.

  1. Total cost is 3 × 40 + 2 × 60 = ₹240.
  2. Total mass is 3 + 2 = 5 kg.
  3. The weighted mean cost is 240/5 = ₹48/kg, not the unweighted average ₹50/kg.

Question 6 · 4 marks

A bag contains 3 red and 2 blue balls. Two balls are drawn without replacement. Draw a probability tree and find the probability that the colours differ.

Model answer and reasoning

3/5.

  1. First branches have probabilities R = 3/5 and B = 2/5.
  2. After R, the next blue probability is 2/4. After B, the next red probability is 3/4.
  3. P(RB) = (3/5)(2/4) = 3/10 and P(BR) = (2/5)(3/4) = 3/10.
  4. The paths are disjoint, so add them: 3/10 + 3/10 = 3/5. Without replacement the second probabilities change.

Question 7 · 4 marks

School X reports 30 bus, 10 bicycle and 10 walking journeys; School Y reports 36 bus, 12 bicycle and 12 walking journeys. Compare their stacked and 100% stacked bars and explain what can and cannot be concluded.

Model answer and reasoning

X total 50, Y total 60; both have proportions 60% bus, 20% bicycle, 20% walking.

  1. Ordinary stacked bars show different totals: 50 and 60.
  2. For X divide each count by 50; for Y divide by 60.
  3. Both percentage bars have segments 60%, 20% and 20%, hence identical composition.
  4. Equal composition does not imply equal absolute numbers. Nor does this observational table alone establish the cause of transport choices.

Question 8 · 4 marks

A player wins 18 of 30 games and then 12 of 20 games. Find the combined empirical win probability. Can we conclude the next 10 games must contain exactly 6 wins?

Model answer and reasoning

Estimate 30/50 = 0.6; no exact future count is guaranteed.

  1. Add successes: 18 + 12 = 30. Add trials: 30 + 20 = 50.
  2. Combined relative frequency is 30/50 = 0.6.
  3. This is an empirical estimate under comparable playing conditions, not a guaranteed theoretical probability.
  4. Even if a suitable model had probability 0.6, future random outcomes need not match 6 out of 10 exactly.

Question 9 · 4 marks

A learner lists sums 2,3,…,12 for two fair dice and assumes all sums are equally likely. Correct the model using a table and describe a simulation for estimating P(sum = 7).

Model answer and reasoning

Use 36 equally likely ordered pairs; P(sum = 7) = 6/36 = 1/6.

  1. Make a 6-by-6 table: rows identify the first die and columns the second, giving 36 ordered pairs.
  2. Sum 7 occurs at (1,6), (2,5), (3,4), (4,3), (5,2), (6,1), so its probability is 1/6.
  3. Sums are not equally likely: sum 2 has only one ordered pair, while sum 7 has six.
  4. Simulate independent uniform throws of each die repeatedly, count sums of 7, and divide by total trials. Compare this relative frequency with 1/6.