Hard GCSE Sampling and convergence Questions

Challenging, exam-style GCSE Sampling and convergence questions with worked solutions. Stretch yourself on the hardest relative frequency, convergence with sample size, pooling samples, reliability of an estimate problems.

relative frequencyconvergence with sample sizepooling samplesreliability of an estimateexpected frequencyobserved frequency
GCSE Foundation34 questionsStep-by-step solutions
Question 1
5 markschallenging
A fair spinner has 88 equal sections. 11 of the sections is red. The spinner is spun by two students. Sami spins it 4040 times and it lands on red 77 times. Tess spins it 800800 times and it lands on red 8484 times. Which statement about these two estimates is correct?
Show worked solution

Worked solution

  1. Work out the theoretical probability from the equally likely outcomes

    P(red)=18P(\text{red}) = \frac{1}{8}

    Because the spinner has 88 equal sections, so all 88 outcomes are equally likely, and 11 of them is red, the theoretical probability of red is 18=0.125\frac{1}{8} = 0.125. Both students are estimating this same value.

  2. Work out Sami's relative frequency

    740=0.175\frac{7}{40} = 0.175

    Sami got red 77 times in 4040 spins, so their estimate of the probability is 0.1750.175.

  3. Work out Tess's relative frequency

    84800=0.105\frac{84}{800} = 0.105

    Tess got red 8484 times in 800800 spins, so their estimate is 0.1050.105.

  4. Work out how far Sami's estimate is from the theory

    0.1750.125=0.05|0.175 - 0.125| = 0.05

    Sami's estimate is 0.050.05 away from the theoretical probability. The difference is taken as a positive amount.

  5. Work out how far Tess's estimate is from the theory

    0.1050.125=0.02|0.105 - 0.125| = 0.02

    Tess's estimate is 0.020.02 away from the theoretical probability.

  6. Compare the two differences

    0.02<0.050.02 < 0.05

    0.020.02 is smaller than 0.050.05, so Tess's estimate is the one closer to the theoretical probability.

  7. Put both estimates on the probability scale with the theory

    0.175,0.105,0.1250.175, \quad 0.105, \quad 0.125

    Marking all three values on the same scale shows which estimate has landed nearer the theoretical probability.

  8. Plot the two estimates against the size of the sample

    (40,0.175),(800,0.105)(40, 0.175), \quad (800, 0.105)

    The dashed line is the theoretical probability. The picture shows the same thing the arithmetic does: which estimate has landed nearer the line.

  9. Say which sample was the larger

    800>40800 > 40

    Tess used 800800 spins, more than the other student. A larger sample gives a more reliable estimate ON AVERAGE - it is the one to bet on before the results are seen.

  10. Say why the larger sample is not guaranteed to win

    tendency, not a guarantee\text{tendency}, \ \text{not a guarantee}

    This is the key point. A larger sample TENDS to be closer, but any single small sample can happen to land very close to the theoretical probability by luck. Only the actual differences can decide which of these two is closer, and that is why they were worked out. Here the larger sample (Tess's 800800 spins) has landed closer, as usually happens - but that was not guaranteed in advance.

  11. Rule out the idea that the estimates are equally close

    0.050.020.05 \ne 0.02

    The two differences are not equal, so the two estimates are not the same distance from the theoretical probability.

  12. Rule out the idea that the theory has to wait for the results

    theory first, experiment after\text{theory first}, \ \text{experiment after}

    The theoretical probability 18\frac{1}{8} was worked out from the equally likely outcomes before anything was thrown or spun. It does not need the experiment, and the experiment cannot change it.

  13. Say what neither estimate proves

    relative frequencyproof\text{relative frequency} \ne \text{proof}

    Neither student has proved anything about the probability. Each has made an estimate of it, and one of those estimates happens to be nearer the true value than the other.

  14. Say how either student could do better

    more trialsbetter estimate\text{more trials} \Rightarrow \text{better estimate}

    The only reliable way to improve an estimate is to carry out more trials, or to pool the two sets of results into one larger sample.

  15. State the answer

    Tess\text{Tess}

    Tess's estimate is closer: Tess's relative frequency 0.1050.105 differs from the theoretical probability 0.1250.125 by 0.020.02, and Sami's differs by 0.050.05, so Tess's estimate is closer.

Answer
Tess\text{Tess}
Question 2
5 markschallenging
A fair coin is thrown by two students. Quinn throws it 5050 times and it lands on heads 3232 times. Rosa throws it 20002000 times and it lands on heads 10241024 times. Which statement about these two estimates is correct?
Show worked solution

Worked solution

  1. Work out the theoretical probability from the equally likely outcomes

    P(heads)=12P(\text{heads}) = \frac{1}{2}

    Because a fair coin has 22 equally likely outcomes, heads and tails, and 11 of them is heads, the theoretical probability of heads is 12=0.5\frac{1}{2} = 0.5. Both students are estimating this same value.

  2. Work out Quinn's relative frequency

    3250=0.64\frac{32}{50} = 0.64

    Quinn got heads 3232 times in 5050 throws, so their estimate of the probability is 0.640.64.

  3. Work out Rosa's relative frequency

    10242000=0.512\frac{1024}{2000} = 0.512

    Rosa got heads 10241024 times in 20002000 throws, so their estimate is 0.5120.512.

  4. Work out how far Quinn's estimate is from the theory

    0.640.5=0.14|0.64 - 0.5| = 0.14

    Quinn's estimate is 0.140.14 away from the theoretical probability. The difference is taken as a positive amount.

  5. Work out how far Rosa's estimate is from the theory

    0.5120.5=0.012|0.512 - 0.5| = 0.012

    Rosa's estimate is 0.0120.012 away from the theoretical probability.

  6. Compare the two differences

    0.012<0.140.012 < 0.14

    0.0120.012 is smaller than 0.140.14, so Rosa's estimate is the one closer to the theoretical probability.

  7. Put both estimates on the probability scale with the theory

    0.64,0.512,0.50.64, \quad 0.512, \quad 0.5

    Marking all three values on the same scale shows which estimate has landed nearer the theoretical probability.

  8. Plot the two estimates against the size of the sample

    (50,0.64),(2000,0.512)(50, 0.64), \quad (2000, 0.512)

    The dashed line is the theoretical probability. The picture shows the same thing the arithmetic does: which estimate has landed nearer the line.

  9. Say which sample was the larger

    2000>502000 > 50

    Rosa used 20002000 throws, more than the other student. A larger sample gives a more reliable estimate ON AVERAGE - it is the one to bet on before the results are seen.

  10. Say why the larger sample is not guaranteed to win

    tendency, not a guarantee\text{tendency}, \ \text{not a guarantee}

    This is the key point. A larger sample TENDS to be closer, but any single small sample can happen to land very close to the theoretical probability by luck. Only the actual differences can decide which of these two is closer, and that is why they were worked out. Here the larger sample (Rosa's 20002000 throws) has landed closer, as usually happens - but that was not guaranteed in advance.

  11. Rule out the idea that the estimates are equally close

    0.140.0120.14 \ne 0.012

    The two differences are not equal, so the two estimates are not the same distance from the theoretical probability.

  12. Rule out the idea that the theory has to wait for the results

    theory first, experiment after\text{theory first}, \ \text{experiment after}

    The theoretical probability 12\frac{1}{2} was worked out from the equally likely outcomes before anything was thrown or spun. It does not need the experiment, and the experiment cannot change it.

  13. Say what neither estimate proves

    relative frequencyproof\text{relative frequency} \ne \text{proof}

    Neither student has proved anything about the probability. Each has made an estimate of it, and one of those estimates happens to be nearer the true value than the other.

  14. Say how either student could do better

    more trialsbetter estimate\text{more trials} \Rightarrow \text{better estimate}

    The only reliable way to improve an estimate is to carry out more trials, or to pool the two sets of results into one larger sample.

  15. State the answer

    Rosa\text{Rosa}

    Rosa's estimate is closer: Rosa's relative frequency 0.5120.512 differs from the theoretical probability 0.50.5 by 0.0120.012, and Quinn's differs by 0.140.14, so Rosa's estimate is closer.

Answer
Rosa\text{Rosa}
Question 3
5 markschallenging
A fair spinner has 44 equal sections. 11 of the sections is blue. The spinner is spun by two students. Omar spins it 6060 times and it lands on blue 1818 times. Priya spins it 600600 times and it lands on blue 168168 times. Which statement about these two estimates is correct?
Show worked solution

Worked solution

  1. Work out the theoretical probability from the equally likely outcomes

    P(blue)=14P(\text{blue}) = \frac{1}{4}

    Because the spinner has 44 equal sections, so all 44 outcomes are equally likely, and 11 of them is blue, the theoretical probability of blue is 14=0.25\frac{1}{4} = 0.25. Both students are estimating this same value.

  2. Work out Omar's relative frequency

    1860=0.3\frac{18}{60} = 0.3

    Omar got blue 1818 times in 6060 spins, so their estimate of the probability is 0.30.3.

  3. Work out Priya's relative frequency

    168600=0.28\frac{168}{600} = 0.28

    Priya got blue 168168 times in 600600 spins, so their estimate is 0.280.28.

  4. Work out how far Omar's estimate is from the theory

    0.30.25=0.05|0.3 - 0.25| = 0.05

    Omar's estimate is 0.050.05 away from the theoretical probability. The difference is taken as a positive amount.

  5. Work out how far Priya's estimate is from the theory

    0.280.25=0.03|0.28 - 0.25| = 0.03

    Priya's estimate is 0.030.03 away from the theoretical probability.

  6. Compare the two differences

    0.03<0.050.03 < 0.05

    0.030.03 is smaller than 0.050.05, so Priya's estimate is the one closer to the theoretical probability.

  7. Put both estimates on the probability scale with the theory

    0.3,0.28,0.250.3, \quad 0.28, \quad 0.25

    Marking all three values on the same scale shows which estimate has landed nearer the theoretical probability.

  8. Plot the two estimates against the size of the sample

    (60,0.3),(600,0.28)(60, 0.3), \quad (600, 0.28)

    The dashed line is the theoretical probability. The picture shows the same thing the arithmetic does: which estimate has landed nearer the line.

  9. Say which sample was the larger

    600>60600 > 60

    Priya used 600600 spins, more than the other student. A larger sample gives a more reliable estimate ON AVERAGE - it is the one to bet on before the results are seen.

  10. Say why the larger sample is not guaranteed to win

    tendency, not a guarantee\text{tendency}, \ \text{not a guarantee}

    This is the key point. A larger sample TENDS to be closer, but any single small sample can happen to land very close to the theoretical probability by luck. Only the actual differences can decide which of these two is closer, and that is why they were worked out. Here the larger sample (Priya's 600600 spins) has landed closer, as usually happens - but that was not guaranteed in advance.

  11. Rule out the idea that the estimates are equally close

    0.050.030.05 \ne 0.03

    The two differences are not equal, so the two estimates are not the same distance from the theoretical probability.

  12. Rule out the idea that the theory has to wait for the results

    theory first, experiment after\text{theory first}, \ \text{experiment after}

    The theoretical probability 14\frac{1}{4} was worked out from the equally likely outcomes before anything was thrown or spun. It does not need the experiment, and the experiment cannot change it.

  13. Say what neither estimate proves

    relative frequencyproof\text{relative frequency} \ne \text{proof}

    Neither student has proved anything about the probability. Each has made an estimate of it, and one of those estimates happens to be nearer the true value than the other.

  14. Say how either student could do better

    more trialsbetter estimate\text{more trials} \Rightarrow \text{better estimate}

    The only reliable way to improve an estimate is to carry out more trials, or to pool the two sets of results into one larger sample.

  15. State the answer

    Priya\text{Priya}

    Priya's estimate is closer: Priya's relative frequency 0.280.28 differs from the theoretical probability 0.250.25 by 0.030.03, and Omar's differs by 0.050.05, so Priya's estimate is closer.

Answer
Priya\text{Priya}
Question 4
6 markschallenging
A fair spinner has 44 equal sections. 33 of the sections are red. The spinner is spun many times. The number of reds is counted at four stages. After 2020 spins there had been 1212 reds. After 100100 spins there had been 6868 reds. After 400400 spins there had been 292292 reds. After 20002000 spins there had been 14941494 reds. Which statement is best supported by these results?
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Worked solution

  1. Read what the table is recording

    (20,12),(100,68),(400,292),(2000,1494)(20, 12), \quad (100, 68), \quad (400, 292), \quad (2000, 1494)

    Each pair is a running total: the number of spins so far, and the number of red in all of those spins.

  2. Work out the theoretical probability from the equally likely outcomes

    P(red)=34P(\text{red}) = \frac{3}{4}

    Because the spinner has 44 equal sections, so all 44 outcomes are equally likely, and 33 of them are red, the theoretical probability of red is 34=0.75\frac{3}{4} = 0.75. It is fixed, and no experiment can change it.

  3. Work out the relative frequency at each stage

    1220=0.6,68100=0.68,292400=0.73,14942000=0.747\frac{12}{20} = 0.6, \quad \frac{68}{100} = 0.68, \quad \frac{292}{400} = 0.73, \quad \frac{1494}{2000} = 0.747

    Each stage gives an estimate of the same probability, from a bigger sample than the stage before it.

  4. Work out the difference at each stage

    0.15,0.07,0.02,0.0030.15, \quad 0.07, \quad 0.02, \quad 0.003

    Each difference is relative frequency0.75|\text{relative frequency} - 0.75|, taken as a positive amount.

  5. Compare the differences

    0.15>0.07>0.02>0.0030.15 > 0.07 > 0.02 > 0.003

    In these results the difference gets smaller at every stage. The estimate is converging on the theoretical probability as the sample grows.

  6. Plot the relative frequency against the size of the sample

    (20,0.6),(100,0.68),(400,0.73),(2000,0.747)(20, 0.6), \quad (100, 0.68), \quad (400, 0.73), \quad (2000, 0.747)

    The graph shows exactly what the numbers show: wild swings from the smallest sample, then a settling towards the dashed theoretical line.

  7. Check the last estimate against the theory

    0.7470.75=0.003|0.747 - 0.75| = 0.003

    Even after 20002000 spins the relative frequency is 0.0030.003 away from 0.750.75 - close, but not equal.

  8. Say why the small samples are so far out

    small samplelarge swing\text{small sample} \Rightarrow \text{large swing}

    With only 2020 spins, one extra success changes the relative frequency by 120\frac{1}{20}. With 20002000 spins it changes it by only 12000\frac{1}{2000}, which is why the big samples barely move.

  9. Rule out the claim that the last estimate is exactly the theory

    0.7470.750.747 \ne 0.75

    The two numbers are different, so any statement that they are equal is false on the arithmetic alone - and even if they had matched, that would not PROVE the object is fair.

  10. Rule out the claim that a larger sample is guaranteed to be closer

    tendency, not a guarantee\text{tendency}, \ \text{not a guarantee}

    A larger sample TENDS to give a closer estimate, and in this table it does at every stage. But that is a fact about these results, not a guarantee: a bigger sample can land further out by chance.

  11. Rule out the claim that the results show bias

    gaps shrinkingbias\text{gaps shrinking} \ne \text{bias}

    The differences are getting smaller, not bigger, so the results are behaving exactly as a fair spinner should. There is no evidence of bias here.

  12. Say what the table can never show

    relative frequencyproof\text{relative frequency} \ne \text{proof}

    However far the table were extended, the relative frequency would only ever estimate the probability. It cannot prove the theoretical value.

  13. Check every relative frequency is a possible probability

    00.6,,0.74710 \le 0.6, \ldots, 0.747 \le 1

    Successes can never exceed trials, so every relative frequency in the table lies on the 00 to 11 scale.

  14. Weigh up the five statements

    one statement fits the numbers\text{one statement fits the numbers}

    Only one statement survives the arithmetic: the differences really do shrink at every stage, and they are the numbers worked out above.

  15. State the answer

    The gaps shrink as the sample grows\text{The gaps shrink as the sample grows}

    The statement best supported by these results is: As the sample grows the relative frequency of red gets closer to the theoretical probability 0.750.75 at every stage of this table: the differences are 0.150.15, 0.070.07, 0.020.02 and 0.0030.003.

Answer
The gaps shrink as the sample grows\text{The gaps shrink as the sample grows}
Question 5
5 markschallenging
A fair spinner has 1010 equal sections. 44 of the sections are blue. The spinner is spun many times. The number of blues is counted at four stages. After 2020 spins there had been 1111 blues. After 100100 spins there had been 4747 blues. After 500500 spins there had been 214214 blues. After 25002500 spins there had been 10071007 blues. Which statement is best supported by these results?
Show worked solution

Worked solution

  1. Read what the table is recording

    (20,11),(100,47),(500,214),(2500,1007)(20, 11), \quad (100, 47), \quad (500, 214), \quad (2500, 1007)

    Each pair is a running total: the number of spins so far, and the number of blue in all of those spins.

  2. Work out the theoretical probability from the equally likely outcomes

    P(blue)=410=25P(\text{blue}) = \frac{4}{10} = \frac{2}{5}

    Because the spinner has 1010 equal sections, so all 1010 outcomes are equally likely, and 44 of them are blue, the theoretical probability of blue is 25=0.4\frac{2}{5} = 0.4. It is fixed, and no experiment can change it.

  3. Work out the relative frequency at each stage

    1120=0.55,47100=0.47,214500=0.428,10072500=0.4028\frac{11}{20} = 0.55, \quad \frac{47}{100} = 0.47, \quad \frac{214}{500} = 0.428, \quad \frac{1007}{2500} = 0.4028

    Each stage gives an estimate of the same probability, from a bigger sample than the stage before it.

  4. Work out the difference at each stage

    0.15,0.07,0.028,0.00280.15, \quad 0.07, \quad 0.028, \quad 0.0028

    Each difference is relative frequency0.4|\text{relative frequency} - 0.4|, taken as a positive amount.

  5. Compare the differences

    0.15>0.07>0.028>0.00280.15 > 0.07 > 0.028 > 0.0028

    In these results the difference gets smaller at every stage. The estimate is converging on the theoretical probability as the sample grows.

  6. Plot the relative frequency against the size of the sample

    (20,0.55),(100,0.47),(500,0.428),(2500,0.4028)(20, 0.55), \quad (100, 0.47), \quad (500, 0.428), \quad (2500, 0.4028)

    The graph shows exactly what the numbers show: wild swings from the smallest sample, then a settling towards the dashed theoretical line.

  7. Check the last estimate against the theory

    0.40280.4=0.0028|0.4028 - 0.4| = 0.0028

    Even after 25002500 spins the relative frequency is 0.00280.0028 away from 0.40.4 - close, but not equal.

  8. Say why the small samples are so far out

    small samplelarge swing\text{small sample} \Rightarrow \text{large swing}

    With only 2020 spins, one extra success changes the relative frequency by 120\frac{1}{20}. With 25002500 spins it changes it by only 12500\frac{1}{2500}, which is why the big samples barely move.

  9. Rule out the claim that the last estimate is exactly the theory

    0.40280.40.4028 \ne 0.4

    The two numbers are different, so any statement that they are equal is false on the arithmetic alone - and even if they had matched, that would not PROVE the object is fair.

  10. Rule out the claim that a larger sample is guaranteed to be closer

    tendency, not a guarantee\text{tendency}, \ \text{not a guarantee}

    A larger sample TENDS to give a closer estimate, and in this table it does at every stage. But that is a fact about these results, not a guarantee: a bigger sample can land further out by chance.

  11. Rule out the claim that the results show bias

    gaps shrinkingbias\text{gaps shrinking} \ne \text{bias}

    The differences are getting smaller, not bigger, so the results are behaving exactly as a fair spinner should. There is no evidence of bias here.

  12. Say what the table can never show

    relative frequencyproof\text{relative frequency} \ne \text{proof}

    However far the table were extended, the relative frequency would only ever estimate the probability. It cannot prove the theoretical value.

  13. Check every relative frequency is a possible probability

    00.55,,0.402810 \le 0.55, \ldots, 0.4028 \le 1

    Successes can never exceed trials, so every relative frequency in the table lies on the 00 to 11 scale.

  14. Weigh up the five statements

    one statement fits the numbers\text{one statement fits the numbers}

    Only one statement survives the arithmetic: the differences really do shrink at every stage, and they are the numbers worked out above.

  15. State the answer

    The gaps shrink as the sample grows\text{The gaps shrink as the sample grows}

    The statement best supported by these results is: As the sample grows the relative frequency of blue gets closer to the theoretical probability 0.40.4 at every stage of this table: the differences are 0.150.15, 0.070.07, 0.0280.028 and 0.00280.0028.

Answer
The gaps shrink as the sample grows\text{The gaps shrink as the sample grows}

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