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Unit VI: Statistics and Probability — Test 1

Concepts and applications · 9 questions · 25 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

Which number cannot be a probability?

  1. 0
  2. 0.4
  3. 1
  4. 1.2
Model answer and reasoning

(d) 1.2

  1. Probabilities lie on the scale from 0 to 1 inclusive.
  2. 1.2 is larger than 1 and therefore cannot be a probability.

Question 2 · 1 mark

For one throw of a fair six-sided die, the probability of a prime number is:

  1. 1/6
  2. 1/3
  3. 1/2
  4. 2/3
Model answer and reasoning

(c) 1/2

  1. The equally likely sample space is {1,2,3,4,5,6}.
  2. Prime outcomes are 2,3,5, so probability = 3/6 = 1/2.

Question 3 · 1 mark

For data 2, 2, 3, 5, 8, the mode is:

  1. 2
  2. 3
  3. 4
  4. 8
Model answer and reasoning

(a) 2

  1. The mode is the value with the greatest frequency.
  2. 2 occurs twice; every other listed value occurs once.

Question 4 · 4 marks

For travel times 10, 15, 15, 20 and 40 minutes, compute mean, median and mode. Explain which measure best describes a typical trip if 40 minutes was an exceptional traffic delay, and suggest an investigative question.

Model answer and reasoning

Mean 20, median 15, mode 15 minutes; the median is less affected by the exceptional delay.

  1. The total is 100 minutes for 5 trips, so the mean is 20.
  2. The middle sorted value is 15, giving the median; 15 also occurs most often, giving the mode.
  3. The unusually long trip raises the mean. The median gives a useful typical value here, though the right measure depends on the purpose.
  4. An investigative question could be “How does travel time change by day or route?” Record comparable trips and organise them in a table.

Question 5 · 3 marks

A class of 20 students has average score 60; another class of 30 has average 80. Find the combined average and explain why averaging 60 and 80 directly is wrong.

Model answer and reasoning

Combined average 72.

  1. The total scores are 20 × 60 = 1200 and 30 × 80 = 2400.
  2. Combined average = (1200 + 2400)/(20 + 30) = 3600/50 = 72.
  3. The classes have different sizes; weight each average by its student count instead of treating the averages equally.

Question 6 · 4 marks

Class A has 12 students travelling by bus and 8 walking; Class B has 15 by bus and 15 walking. Describe ordinary stacked and 100% stacked bar graphs for these data. Which graph compares proportions more directly?

Model answer and reasoning

Ordinary bar totals 20 and 30; percentage bars show A 60%/40% and B 50%/50%.

  1. An ordinary stacked bar combines category counts: A = 12 + 8 = 20, B = 15 + 15 = 30.
  2. Divide each category by its own class total to form percentage shares.
  3. For A, bus share 12/20 = 60%, walking 40%; for B the shares are 50% each.
  4. 100% stacked bars have equal total height and directly compare composition; ordinary bars retain absolute totals.

Question 7 · 3 marks

A spinner lands on red 47 times in 100 trials. Estimate its probability of red. Does this prove its theoretical probability is exactly 0.47?

Model answer and reasoning

Empirical estimate 0.47; it does not prove an exact theoretical probability.

  1. Relative frequency = red outcomes/total trials = 47/100.
  2. Finite experiments fluctuate randomly, so the observed fraction need not equal an exact theoretical value.
  3. A theoretical model requires information about the spinner and outcome probabilities; more trials can help assess the estimate.

Question 8 · 4 marks

Two fair coins are tossed independently. Use a tree or table to list the sample space and find probabilities of exactly one head and at least one head.

Model answer and reasoning

Sample space {HH, HT, TH, TT}; probabilities 1/2 and 3/4.

  1. Branch H or T for the first coin, then H or T for the second.
  2. Independence gives each ordered outcome probability (1/2)(1/2) = 1/4.
  3. Exactly one head is {HT,TH}, so its probability is 2/4 = 1/2.
  4. At least one head is {HH,HT,TH}, so its probability is 3/4.
Probability tree for two independent fair coin tossesH: ½T: ½HH: ¼HT: ¼TH: ¼TT: ¼

Question 9 · 4 marks

Write a step-by-step simulation of 1000 fair die throws to estimate the probability of an even outcome. State the theoretical target and why a simulation may differ from it.

Model answer and reasoning

Choose uniform random integers 1–6; estimated probability = even count/1000; theoretical target 1/2.

  1. Initialise an even-outcome counter to zero.
  2. Repeat 1000 times: sample one of 1,2,3,4,5,6 with equal probability and increment the counter for 2,4 or 6.
  3. Divide the final counter by 1000.
  4. Three of six equally likely outcomes are even, so theoretical probability is 1/2. Random fluctuations mean a finite estimate need not be exactly 0.5.