Advanced Statistics & Probability

A histogram shows the ages, in years, of members at a gym. The bar for the class 40–60 has a frequency density of 0.6. What is the frequency for this class?

• A12 members✓
• B36 members
• C33 members
• D0.6 members

Answer: Frequency equals frequency density multiplied by class width, and this class spans 60 minus 40, which is 20 years, so 0.6 times 20 gives a frequency of 12 members. Using the upper class boundary of 60 as the width instead of 20 gives 36, dividing the width by the density instead of multiplying gives 33, and quoting the frequency density itself ignores that it must be scaled by the class width.

In a histogram showing the salaries of 150 employees, the class £20,000–£50,000 contains 45 employees. What is the frequency density for this class?

• A45 employees per £1,000
• B1.5 employees per £1,000 of salary✓
• C0.90 employees per £1,000
• D0.67 employees per £1,000

Answer: Frequency density equals frequency divided by class width; here the class width is £50,000 minus £20,000, which is £30,000, or 30 thousand pounds, so 45 divided by 30 gives 1.5 employees per £1,000. Quoting 45 gives the frequency itself rather than the density, dividing by the upper boundary of 50 instead of the true width of 30 gives 0.9, and inverting the division gives 0.67.

A histogram has a class 0–10 with frequency density 3 and a class 10–40 with frequency density 1.2. What is the frequency of the 10–40 class, and is it larger or smaller than the frequency of the 0–10 class?

• A36, which is smaller than the 0–10 class's frequency of 30
• B36, which is larger than the 0–10 class's frequency of 30✓
• C1.2, which is smaller than the 0–10 class's frequency of 30
• D12, smaller than the 0–10 class's frequency

Answer: Frequency equals frequency density multiplied by class width, and the 10–40 class spans 30 years, giving 1.2 times 30, which is 36, and this exceeds the 0–10 class's frequency of 3 times 10, which is 30. Using the wrong width of 10 instead of 30 gives 12, wrongly judging 36 to be smaller than 30 reverses the comparison, and quoting the frequency density of 1.2 ignores that it must be scaled by the class width.

A histogram uses equal class widths of 5 for all bars. The bar for the class 15–20 has a frequency density of 4.4. How many people are in this class?

• A4.4 people
• B11 people
• C22 people✓
• D44 people

Answer: With equal class widths, frequency equals frequency density multiplied by the class width, so 4.4 times 5 gives 22 people. Quoting 4.4 gives the frequency density rather than the frequency, using a width of 10 instead of 5 doubles the answer to 44, and using a width of 2.5 instead of 5 halves it to 11.

A histogram for the heights of 100 plants has three classes: 0–20 cm with frequency density 1.5, 20–30 cm with frequency density 4, and 30–50 cm with an unknown frequency density. The first two classes have frequencies of 30 and 40. What frequency density is needed for the 30–50 cm class so the total frequency is 100?

• A30 plants per cm
• B0.67 plants per cm
• C3 plants per cm
• D1.5 plants per cm✓

Answer: The missing frequency is 100 minus 30 minus 40, which is 30 plants, and the class width is 50 minus 30, which is 20 cm, so the frequency density is 30 divided by 20, which is 1.5 plants per cm. Quoting 30 gives the frequency rather than the density, inverting the division gives 0.67, and using a width of 10 instead of 20 gives 3.

A histogram has a class 10–30 with frequency density 2.5. Assuming the people in the class are spread evenly, estimate how many are in the sub-interval 10–18.

• A8 people
• B3.2 people
• C20 people✓
• D50 people

Answer: Assuming a uniform spread, the estimated frequency in a sub-interval equals the frequency density multiplied by the width of that sub-interval, and 18 minus 10 is 8, so 2.5 times 8 gives 20 people. Quoting 8 gives only the sub-interval width, using the full class width of 20 instead of 8 gives 50, and dividing instead of multiplying gives 3.2.

A cumulative frequency graph for the marks of 60 students shows the cumulative frequency reaching 30 at a mark of 54. What is the median mark?

• A15 marks
• B30 marks
• C54 marks✓
• D60 marks

Answer: The median is the mark at the cumulative frequency of n divided by 2, which is 60 divided by 2, giving 30, and the graph shows this cumulative frequency is reached at a mark of 54. Quoting 30 confuses the cumulative frequency value with the mark itself, quoting 60 uses the total number of students rather than the halfway point, and quoting 15 mistakenly uses a quarter of the students instead of half.

On a cumulative frequency graph for the heights of 200 trees, the lower quartile is read at 12 m and the upper quartile at 19 m. What is the inter-quartile range?

• A19 m
• B7 m✓
• C12 m
• D31 m

Answer: The inter-quartile range equals the upper quartile minus the lower quartile, so 19 minus 12 gives 7 m. Adding the two quartiles instead of subtracting gives 31, and quoting either quartile alone gives 19 or 12 rather than the spread between them.

A cumulative frequency graph for the waiting times of 150 patients shows that 120 patients waited less than 25 minutes. Estimate how many patients waited 25 minutes or more.

• A120 patients
• B30 patients✓
• C270 patients
• D70 patients

Answer: The total number of patients is 150, and 120 of them waited less than 25 minutes, so 150 minus 120 gives 30 who waited 25 minutes or more. Quoting 120 repeats the given figure instead of subtracting it, adding instead of subtracting gives 270, and misreading the graph's cumulative frequency as 80 instead of 120 gives 70 rather than the correct 30.

A cumulative frequency graph plots the journey times of 80 commuters. At what cumulative frequency should the graph be read to estimate the upper quartile?

• A20
• B80
• C40
• D60✓

Answer: The upper quartile is estimated at three-quarters of the way through the data, so the cumulative frequency to read from is 3 times 80 divided by 4, which is 60. Using 40 instead finds the median position, using 20 finds the lower quartile position, and using 80 refers to the entire sample rather than three-quarters of it.

A cumulative frequency graph for exam scores gives a lower quartile of 42 and an upper quartile of 68. What is the interquartile range?

• A13
• B110
• C55
• D26✓

Answer: The interquartile range is the upper quartile minus the lower quartile: 68 minus 42 is 26. Halving it gives 13, which is not a measure used at GCSE; averaging the two quartiles instead of subtracting them gives 55, and simply adding the quartiles gives 110.

A box plot for the times of 50 sprinters shows the box drawn from 11.2 seconds to 12.6 seconds, with the median line at 11.8 seconds. What is the inter-quartile range?

• A1.4 seconds✓
• B0.60 seconds
• C12.6 seconds
• D23.8 seconds

Answer: The box of a box plot spans from the lower quartile to the upper quartile, so the inter-quartile range is 12.6 minus 11.2, which gives 1.4 seconds. Adding the two ends of the box instead of subtracting gives 23.8, finding the gap between the median and the lower quartile only gives 0.6, and quoting the upper quartile alone gives 12.6.

Box plot A for one class's test scores has an inter-quartile range of 8 marks, and box plot B for another class has an inter-quartile range of 14 marks. Which class shows more consistent, less varied scores in the middle 50% of results?

• AClass B, because its inter-quartile range of 14 is smaller than class A's 8
• BClass A, because its inter-quartile range of 8 is larger than class B's 14
• CClass B, because a larger inter-quartile range means more consistent scores
• DClass A, because its inter-quartile range of 8 is smaller than class B's 14✓

Answer: A smaller inter-quartile range means the middle 50% of results are more tightly clustered, so class A, with an inter-quartile range of 8 marks, is more consistent than class B's 14 marks. Reversing the comparison, wrongly stating 8 is larger than 14, or claiming a bigger inter-quartile range shows more consistency are all incorrect, since consistency corresponds to a smaller spread.

A box plot has a minimum of 5, lower quartile 20, median 21, upper quartile 24, and maximum 45. What does the position of the median within the box, together with the whisker lengths, suggest about the distribution?

• AThe distribution is positively skewed, with a longer tail towards the higher values✓
• BThe distribution is positively skewed, with a longer tail towards the lower values
• CThe distribution is negatively skewed, with a longer tail towards the lower values
• DThe distribution is symmetrical, since the box plot shows the data is evenly spread

Answer: The median sits closer to the lower quartile than the upper quartile, and the upper whisker of 45 minus 24, which is 21, is longer than the lower whisker of 20 minus 5, which is 15, both signs that the data stretches out towards higher values, giving positive skew. Describing the skew as negative reverses the direction, calling the distribution symmetrical ignores this clear asymmetry, and pairing positive skew with a tail towards lower values contradicts what positive skew means.

A dataset of 11 ordered values is 3, 5, 6, 8, 9, 11, 12, 14, 15, 18, 20. Using the median of the lower five values as the lower quartile and the median of the upper five values as the upper quartile, what is the inter-quartile range?

• A15
• B9✓
• C17
• D6

Answer: With the data in order, the median is the 6th value, 11, so the lower quartile is the median of the first five values, 6, and the upper quartile is the median of the last five values, 15, giving an inter-quartile range of 15 minus 6, which is 9. Subtracting the minimum from the maximum instead gives the range, 17, and quoting either quartile alone gives 6 or 15 rather than their difference.

A box plot for annual salaries, in £1,000s, at a company shows a minimum of 18, lower quartile 24, median 30, upper quartile 42, and maximum 95. What is the range of salaries?

• A£77,000✓
• B£95,000
• C£18,000
• D£65,000

Answer: The range is the maximum value minus the minimum value, so £95,000 minus £18,000 gives £77,000. Quoting the minimum or maximum alone gives £18,000 or £95,000, and subtracting the median from the maximum instead of the minimum gives £65,000 rather than the true range.

A dataset of 12 ordered values is 4, 7, 9, 9, 12, 14, 16, 18, 19, 21, 24, 27. Using the 3rd value as the lower quartile and the 9th value as the upper quartile, what is the inter-quartile range?

• A9
• B23
• C10✓
• D19

Answer: The lower quartile is the 3rd value, 9, and the upper quartile is the 9th value, 19, so the inter-quartile range is 19 minus 9, which is 10. Subtracting the smallest value from the largest gives the range, 23, not the inter-quartile range, and quoting either quartile on its own gives 9 or 19 rather than their difference.

A two-way table shows that of 200 students, 90 are boys, of whom 40 study French. Of the 110 girls, 70 study French. A student is picked at random and is known to study French. What is the probability that the student is a boy?

• A4/9
• B9/20
• C4/11✓
• D1/5

Answer: Given that the student studies French, the sample space is restricted to the 40 plus 70, which is 110 French students, of whom 40 are boys, so the probability is 40 out of 110, which simplifies to 4/11. Finding 90 out of 200 gives the overall probability of being a boy regardless of French, finding 40 out of 90 gives the reverse conditional probability of studying French given the student is a boy, and finding 40 out of 200 gives the unconditional probability of being a boy and studying French without restricting to French students.

A two-way table for 300 customers shows 180 bought a coffee, of whom 45 also bought a pastry. Of the 120 who did not buy a coffee, 20 bought a pastry. A customer who bought a pastry is chosen at random. What is the probability that they also bought a coffee?

• A1/4
• B3/5
• C3/20
• D9/13✓

Answer: Given that the customer bought a pastry, the sample space is restricted to the 45 plus 20, which is 65 pastry buyers, of whom 45 also bought a coffee, so the probability is 45 out of 65, which simplifies to 9/13. Finding 180 out of 300 gives the overall probability of buying a coffee regardless of pastry, finding 45 out of 180 gives the reverse conditional probability of buying a pastry given the customer bought a coffee, and finding 45 out of 300 gives the unconditional probability of buying both without restricting to pastry buyers.

A bag contains 5 red and 3 blue counters. Two counters are drawn without replacement. Given that the first counter drawn is red, what is the probability that the second counter is also red?

• A5/7
• B5/8
• C1/2
• D4/7✓

Answer: After one red counter is removed without replacement, 4 red and 3 blue counters remain out of 7, so the probability the second is also red is 4/7. Using 5/8 ignores that the first counter was removed at all, using 5/7 reduces the total but not the red count, and using 1/2, which is 4 out of the original 8 counters, reduces the red count but not the total.

A factory tests components from two machines. Machine A makes 60% of components and 5% of its components are defective; machine B makes 40% of components and 8% of its components are defective. A component is selected at random and found to be defective. What is the probability it came from machine B?

• A15/31
• B16/31✓
• C2/25
• D2/5

Answer: Machine A contributes 0.6 times 0.05, which is 0.03, to the probability of a defective component, and machine B contributes 0.4 times 0.08, which is 0.032, so the total probability of a defective component is 0.062, and the probability it came from machine B is 0.032 divided by 0.062, which simplifies to 16/31. Finding 0.03 divided by 0.062, which is 15/31, gives the probability it came from machine A instead, quoting 2/5 gives the overall proportion made by machine B regardless of being defective, and quoting 2/25 gives machine B's defect rate rather than the reverse conditional probability required.

In a Venn diagram for 50 students, 22 study Spanish, 18 study German, 8 study both, and the rest study neither. A student who studies Spanish is chosen at random. What is the probability that they also study German?

• A4/9
• B4/11✓
• C4/25
• D7/11

Answer: Given that a student studies Spanish, the sample space is restricted to the 22 Spanish students, of whom 8 also study German, so the probability is 8 out of 22, which simplifies to 4/11. Finding 8 out of 18, which is 4/9, gives the reverse conditional probability of studying Spanish given German, finding 8 out of 50, which is 4/25, gives the unconditional probability of studying both without restricting to Spanish students, and finding 14 out of 22, which is 7/11, gives the probability of studying Spanish only rather than also studying German.