Question 1Statistics
TMUA Statistics — Questions and Methods
These are SummitPapers original questions, not official past paper questions. Official TMUA questions sorted by topic are on TMUA past papers by topic.
- M6Statistics
- 18 questions10 on Paper 1 · 8 on Paper 2
- 8 free solutionsThe rest show the correct letter only
Covers: means after one value is swapped between two sets, mean, median and range together, when the mean equals the median, the mean of the squares and the square of the mean, how adding, removing or changing a value moves the mean and median, quartiles and the interquartile range, combining two groups.
What this topic tests
These statistics questions are about the mean, the median, the mode, the range and the interquartile range. They are not probability questions. Counting, conditional probability, and expectation sit under probability, which has its own page.
The questions use the mean, the median, the mode, the range and the quartiles, including two groups combined into one. A change in a mean is a change in the sum. The median is a term in the ordered list. One question compares the mean of the squares with the square of the mean. Probability is a different topic.
How it is assessed
Statistics is Section 1, so both papers can test it. This set has questions on Paper 1 and on Paper 2.
Each question has five options. A full paper is 20 questions in 75 minutes, and a calculator is not allowed. UAT-UK does not publish a fixed number of statistics questions per paper, so this note does not invent one.
This note is written for SummitPapers. The official list of what can be examined is the specification, together with the Notes on Mathematics and, for Paper 2, the Notes on Logic and Proof.
Key methods
A change in the mean is a change in the sum
If a set of n numbers has its mean increased by d, the sum has increased by nd. When one value a is swapped for a value b, the sum changes by b − a. Two such changes, on two sets, can fix the ratio of the set sizes even when the sizes were not given.
The combined mean is weighted
The mean of all the data is the total of the sums divided by the total of the sizes. It is the average of the two means only when the sets have the same size.
The median is a term in the ordered list
For an odd count, the median is the middle term after the numbers are ordered. A triple with median 6 and sum 27 can be written p, 6, 21 − p, with p less than 6 when the integers are distinct. The range is then limited by how large the other end must be.
The mean of the squares, minus the square of the mean
Adding the same number to every term changes the mean of the squares and the square of the mean by the same amount, so the difference does not change. A run of consecutive integers can be shifted to 0, 1, …, k − 1 before that difference is calculated.
Common mistakes
Averaging the two means
The unweighted average of 10 and 22 is 16. That is the combined mean only if the sets are the same size. A swap that moves one mean by 2 and the other by 4 says the sizes are not equal.
Saying the sizes cannot be found
The sizes themselves may stay unknown. The combined mean needs only their ratio, and the two changes in the mean can give that ratio.
Repeating a value the question forbids
If the integers must be distinct, a triple such as 6, 6, 15 is not allowed, even when it makes the range look smaller.
Necessary is not the same as sufficient
An arithmetic progression has its mean equal to its median, so that equality is necessary for the progression. A list can have mean equal to median without being an arithmetic progression, so the equality is not sufficient.
Worked example
This question is also in the list below, with the solution folded. It is opened here so the method is on the page before the other questions.
One set of data has mean 10 and another has mean 22. One value from the first set is exchanged with one value from the second set. The mean of the first set then rises from 10 to 12, and the mean of the second set falls from 22 to 18. What is the mean of all the data combined?
Answer: B. 14
Worked solution. Let the sets have n and m members. Exchanging a value a from the first set with a value b from the second changes the first sum by b − a and the second sum by a − b. The mean changes then give b − a = 2n and b − a = 4m, so n = 2m. The combined sum is 10n + 22m = 42m and the combined size is 3m, so the mean is 14.
Why the other options look right. A assumes the sizes of the two sets are needed and are not given, but the two mean changes fix the ratio of the sizes as 2:1, which is all the combined mean needs. C is 15, the unweighted average of the new means 12 and 18. D is 16, the unweighted average of the original means 10 and 22. E is 18, from reversing the size ratio and taking n:m = 1:2, giving (10 + 44)/3.
Paper 1 questions
Paper 1 is Applications of Mathematical Knowledge. Calculators are not allowed.
Q1. One set of data has mean 10 and another has mean 22. One value from the first set is exchanged with one value from the second set. The mean of the first set then rises from 10 to 12, and the mean of the second set falls from 22 to 18. What is the mean of all the data combined?
Free · worked solution included
Answer and worked solution
Answer: B. 14
Worked solution. Let the sets have n and m members. Exchanging a value a from the first set with a value b from the second changes the first sum by b − a and the second sum by a − b. The mean changes then give b − a = 2n and b − a = 4m, so n = 2m. The combined sum is 10n + 22m = 42m and the combined size is 3m, so the mean is 14.
Why the other options look right. A assumes the sizes of the two sets are needed and are not given, but the two mean changes fix the ratio of the sizes as 2:1, which is all the combined mean needs. C is 15, the unweighted average of the new means 12 and 18. D is 16, the unweighted average of the original means 10 and 22. E is 18, from reversing the size ratio and taking n:m = 1:2, giving (10 + 44)/3.
Q2. Set P contains 5 numbers and set Q contains 15 numbers. The mean of Q is 4 more than the mean of P. One number from P is exchanged with one number from Q, and the two means become equal. By how much has the mean of P increased?
Free · worked solution included
Answer and worked solution
Answer: C. 3
Worked solution. Let the original mean of P be m, so the mean of Q is m + 4. Suppose the exchange increases the sum of P by d. The new mean of P is m + d/5, and the new mean of Q is m + 4 − d/15. Setting them equal gives d/5 + d/15 = 4, so 4d/15 = 4 and d = 15. The mean of P therefore increases by 15/5 = 3.
Why the other options look right. A uses the size of P in the numerator instead of the size of Q, giving 4 × 5/20 = 1. B assumes the two means move by equal amounts and meet halfway, a change of 4/2 = 2, ignoring that the larger set's mean moves only a third as far. D assumes the smaller set's mean must rise by the whole gap of 4, forgetting that the larger set's mean falls at the same time. E multiplies the gap by 15/5 and does not divide by the combined size 20, which gives 12.
Q3. A set of k consecutive integers, where k ≥ 2, has smallest term 11. By how much does the mean of the squares of the terms exceed the square of the mean of the terms?
Free · worked solution included
Answer and worked solution
Answer: A. (k2 − 1)/12
Worked solution. Adding the same constant c to every term increases the mean of the squares by 2cm + c2, where m is the old mean, and increases the square of the mean by the same amount, so the excess (mean of squares) − (mean)2 does not change. Shift the terms to 0, 1, ..., k − 1. Their mean is (k − 1)/2, and the mean of their squares is (k − 1)(2k − 1)/6. The excess is (k − 1)(2k − 1)/6 − (k − 1)2/4 = (k − 1)[2(2k − 1) − 3(k − 1)]/12 = (k − 1)(k + 1)/12 = (k2 − 1)/12. The smallest term 11 does not affect the answer.
Why the other options look right. B uses the sum of squares 12 + ... + k2 = k(k + 1)(2k + 1)/6 instead of 02 + ... + (k − 1)2 while keeping the mean (k − 1)/2, which gives (k + 1)(2k + 1)/6 − (k − 1)2/4 = (k2 + 12k − 1)/12. C shifts the terms to 0, 1, ..., k − 1 and gives the mean of their squares, forgetting to subtract the square of their mean. D lets the terms run from 11 to 11 + k, which is k + 1 integers, and so gets ((k + 1)2 − 1)/12 = k(k + 2)/12. E assumes that the mean of the squares equals the square of the mean.
Q4. A data set has mean 7 and range 10. Each value x in the data set is replaced by a − 2x, where a is a constant. In the new data set, the mean is equal to the range. What is the value of a?
Free · worked solution included
Answer and worked solution
Answer: D. 34
Worked solution. Replacing each x by a − 2x multiplies every value by −2 and then adds a. The mean is transformed in the same way, so the new mean is a − 2 × 7 = a − 14. Adding a constant does not change the range, and multiplying by −2 reverses the order of the values and doubles every distance between them, so the new range is 2 × 10 = 20 (a range is never negative). Setting a − 14 = 20 gives a = 34.
Why the other options look right. A takes the new range to be −2 × 10 = −20 instead of 20, so a − 14 = −20 and a = −6. B forgets to multiply the range by 2, so a − 14 = 10 and a = 24. C forgets to multiply the mean by 2, so a − 7 = 20 and a = 27. E multiplies the range by (−2)² = 4 instead of by 2, as if the range scaled like a variance, so a − 14 = 40 and a = 54.
Q5. The integers from 1 to 10, together with a real number k, form a list of eleven numbers whose mean is equal to its median. What is the sum of all the possible values of k?
Free · correct letter only
Answer
Answer: E. 16.5
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Q6. Six years ago, the members of a committee had a mean age of 40. The membership did not change until today, when one member, aged 58, left and was replaced by a new member aged 34. The mean age of the committee is now 44. How many members does the committee have?
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Answer
Answer: D. 12
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Q7. Seven positive integers have median 6 and mean 6, and their only mode is 9. What is the largest possible range of the seven integers?
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Answer
Answer: C. 11
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Q8. In a test, the students in group X had mean mark 50, the students in group Y had mean mark 55, and the mean mark of all the students in X and Y together was 52. Five more students then took the test and each scored 62, after which the mean mark of all the students who had taken the test was 54. How many students are in group X?
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Answer
Answer: B. 12
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Q9. For a list of seven numbers arranged in increasing order, the lower quartile is the 2nd number, the median is the 4th number and the upper quartile is the 6th number. The seven numbers 3, 7, 10, 16, 21, 22 and x, where x is a real number, have interquartile range (upper quartile minus lower quartile) equal to their median. What is the value of x?
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Answer
Answer: B. 14
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Q10. A list contains n numbers, where n ≥ 3. If the largest number is removed, the mean of the remaining numbers is 3 less than the mean of the list. If instead the smallest number is removed, the mean of the remaining numbers is 2 more than the mean of the list. The range of the list is 60. What is n?
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Answer
Answer: B. 13
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Paper 2 questions
Paper 2 is Mathematical Reasoning. It can test this same topic. Argument, proof, and identifying errors are the topics that appear on Paper 2 only.
Q1. A set of six distinct integers is split into two sets of three. The first set has mean 9 and median 6. The second set has mean 15 and median 12. What is the smallest possible range of the six integers?
Free · worked solution included
Answer and worked solution
Answer: D. 17
Worked solution. The first triple sums to 27 with median 6, so it is p, 6, 21 − p with p ≤ 5 (the integers are distinct). The second sums to 45 with median 12, so it is c, 12, 33 − c with c ≤ 11, and its largest value 33 − c is at least 22. The smallest of the six integers is at most p ≤ 5 and the largest is at least 33 − c ≥ 22, so the range is at least 22 − 5 = 17. Taking p = 5 and c = 11 gives 5, 6, 16 and 11, 12, 22, which are six distinct integers with range 22 − 5 = 17. The smallest possible range is 17.
Why the other options look right. A is 11, the range of the single triple 5, 6, 16 (or of 11, 12, 22), not of all six integers. B is 15, from allowing repeated values: 6, 6, 15 and 12, 12, 21 give 21 − 6 = 15, but the integers must be distinct. C is 16, from letting only the first triple repeat its median: 6, 6, 15 with 11, 12, 22 gives 22 − 6 = 16. E takes the second triple's smallest value as 9 instead of the largest allowed value 11, so its largest value is 33 − 9 = 24 and the range is 24 − 5 = 19.
Q2. A list contains n real numbers, where n can be any integer with n ≥ 2. Let P be the statement that the mean of the list equals its median, and let Q be the statement that the numbers, when arranged in increasing order, form an arithmetic progression. Which description is correct for all such lists?
Free · worked solution included
Answer and worked solution
Answer: B. P is necessary but not sufficient for Q
Worked solution. Suppose Q holds, so the ordered list is a, a + d, ..., a + (n − 1)d. Its mean is a + (n − 1)d/2. If n is odd, the median is the middle term a + ((n − 1)/2)d. If n is even, the median is the average of a + (n/2 − 1)d and a + (n/2)d, which is again a + (n − 1)d/2. So Q forces P, that is, P is necessary for Q. P is not sufficient: the list 0, 1, 1, 2 has mean 1 and median 1, but its differences 1, 0, 1 are not constant. So the answer is P is necessary but not sufficient for Q.
Why the other options look right. A tests only lists of three numbers a ≤ b ≤ c, where mean = median means a + b + c = 3b, so b − a = c − b and the list is a progression; with four numbers, 0, 1, 1, 2 breaks this. C reverses the two directions: it is Q that forces P, not P that forces Q. D rejects both directions, thinking that a progression need not have its mean equal to its median; but for a, a + d, ..., a + (n − 1)d both equal a + (n − 1)d/2, and 0, 1, 1, 2 shows only that P does not force Q. E checks only odd lengths, where the median is the middle term, and doubts the even case because the median 4 of 1, 3, 5, 7 is not one of the entries; the median is still the average of the two middle terms, which equals the mean 4.
Q3. Five real numbers have mean 2, median 4 and range 12. What is the smallest possible value of the largest of the five numbers?
Free · worked solution included
Answer and worked solution
Answer: B. 14/3
Worked solution. Write the numbers in order as a ≤ b ≤ 4 ≤ d ≤ e, where 4 is the median. The range gives a = e − 12, and the mean gives a + b + 4 + d + e = 10. Substituting for a gives b + d = 18 − 2e. Since b ≤ 4 and d ≤ e, b + d ≤ 4 + e, so 18 − 2e ≤ 4 + e and e ≥ 14/3. Equality needs b = 4 and d = e = 14/3, with a = 14/3 − 12 = −22/3. The list −22/3, 4, 4, 14/3, 14/3 has sum 10, median 4 and range 12, so the smallest possible value of the largest number is 14/3.
Why the other options look right. A makes the largest number equal to the median 4 without checking the sum: then a = −8 and b + d would have to be 10, but b ≤ 4 and d ≤ 4. C makes the second number equal to the smallest and the fourth equal to the largest, so 2a + 4 + 2e = 10 with e − a = 12, which gives e = 15/2; that list is possible, but its largest number is not the smallest possible. D places the numbers symmetrically about the mean 2, so the largest is 2 + 12/2 = 8. E is 26/3, the largest possible value of the largest number, reached by the list −10/3, −10/3, 4, 4, 26/3.
Q4. A list of nine positive integers has median 8. One of the numbers in the list is increased by 9, and the median of the new list is 9. Which of the following must be true? I: The original list contains the number 9. II: The number that was increased was at most 8 before it was increased. III: The number that was increased was 8.
Free · worked solution included
Answer and worked solution
Answer: A. I and II only
Worked solution. Write the original list in increasing order as a1 ≤ a2 ≤ … ≤ a9, so a5 = 8. If the number increased is one of a6, …, a9, then a1, …, a5 are still the five smallest numbers and the median is still 8, not 9. So the number increased is one of a1, …, a5, and it was at most 8: II is true. After the change, the four other numbers from a1, …, a5 are all at most 8, and the fifth smallest number of the new list is the smaller of a6 and the increased number. The increased number is at least 1 + 9 = 10, so it cannot be 9; hence the new median 9 must be a6, and the original list contains 9: I is true. III is false: in 1, 1, 1, 1, 8, 9, 9, 9, 11, increasing the first 1 to 10 gives 1, 1, 1, 8, 9, 9, 9, 10, 11, whose median is 9, although the number increased was 1, not 8. The answer is I and II only.
Why the other options look right. B rejects I, thinking the increased number itself could become the new median 9; it would have to have been 0, which is not a positive integer. C rejects II, thinking any number could be increased; increasing one of the four largest leaves the five smallest, and so the median 8, unchanged. D also accepts III, but increasing a 1 to 10 in 1, 1, 1, 1, 8, 9, 9, 9, 11 gives median 9 without changing the 8. E makes two errors: it accepts III, which the same list refutes, and it rejects I, thinking the increased number could be the new median 9, although it is at least 1 + 9 = 10.
Q5. A list contains at least one number. A new number, equal to the median of the list, is added to the list. Which of the following must be true? I: The mean of the new list equals the mean of the original list. II: The median of the new list equals the median of the original list. III: The range of the new list equals the range of the original list.
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Answer
Answer: E. II and III only
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Q6. Consider the claim: “If the mean of list P is greater than the mean of list Q, and the mean of list R is greater than the mean of list S, then the mean of the list formed by combining P and R is greater than the mean of the list formed by combining Q and S.” Which of the following is a counterexample to the claim?
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Answer
Answer: C. P: 7; Q: 6, 6, 6; R: 2, 2, 2; S: 1
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Q7. X and Y are two lists of numbers, and Z is the list formed by combining all the numbers in X and all the numbers in Y. Which of the following must be true? I: The mean of Z is between the mean of X and the mean of Y, inclusive. II: If X and Y contain the same number of values, then the median of Z is the mean of the median of X and the median of Y. III: If X and Y each have exactly one mode, then every mode of Z is the mode of X or the mode of Y.
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Answer
Answer: A. I only
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Q8. Five positive integers have range 6 and exactly one mode. Which of the following could be their mean, median and mode, in that order?
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Answer
Answer: A. mean 5, median 4, mode 8
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