Averages & measures of spread: IGCSE Mathematics questions with answers
Free Averages & measures of spread questions for IGCSE Mathematics (0580), part of Statistics. Each question has the answer and the marking points an examiner would look for.
The mean mass of 21 boys is 36 kg. The mean mass of 10 girls is 38 kg. One more student joins the group. The mean mass of all 32 students is now 37 kg. Find the mass of the student who joined.
Here are the marks, out of 20, of 7 students in a quiz. 15, 8, 20, 2, 5, 10, 20
(a)
Write down the mode.
[1]
(b)
Find the median.
[1]
(c)
Calculate the mean.
[2]
(d)
Find the range.
[1]
Show answers and mark scheme
Part
Answer and marking points
Marks
(a)
20
20 B1
1
(b)
10
10 B1
1
(c)
11.4
80 ÷ 7 M1 11.4 A1
2
(d)
18
18 B1
1
Question 4Structured · 4 marksA* challenge
Five positive whole numbers have • a mode of 3 • a median of 5 • a mean of 6 • a range of 9. Find the five numbers.
[4]
Show answers and mark scheme
Part
Answer and marking points
Marks
3, 3, 5, 7, 12
3, 3, 5 seen B1 largest = 3 + 9 = 12 B1 total = 5 × 6 = 30 M1 3, 3, 5, 7, 12 A1
4
Question 5Structured · 13 marksA* challenge
Seven positive integers have a mean of 10 and a median of 9. The list has exactly one mode, which is 12.
(a)
Find the greatest possible range of the seven integers.
[4]
(b)
Find the least possible range.
[3]
(c)
Prove that only one list of seven integers has the range in part (b).
[3]
(d)
An eighth positive integer, k, is added to the list in part (b). The mean and the median of the eight integers are now equal. Find all the possible values of k.
[3]
Show answers and mark scheme
Part
Answer and marking points
Marks
(a)
30 (1, 2, 3, 9, 12, 12, 31)
total 70, fourth value 9, 12 at least twice above the median B1 the three values below 9 must be different (a repeat would tie with 12) M1 smallest values 1, 2, 3, largest 70 − 6 − 9 − 24 = 31 M1 30 A1
4
(b)
4 (8, 8, 9, 9, 12, 12, 12)
if 12 occurs only twice, the values below 9 are distinct, so the range is at least 10 (e.g. smallest values total at most 6 + 7 + 8 = 21, largest at least 70 − 21 − 9 − 24 = 16) M1 so 12 occurs three times and the three smallest values total 25 M1 4 A1
3
(c)
smallest is 8, so the other two total 17 with each ≤ 9: 8 and 9 only
a + b + c = 25 with a ≤ b ≤ c ≤ 9 and the largest value 12 forces a = 8 for range 4 M1 b + c = 17 with b, c ≤ 9 gives only b = 8, c = 9 M1 (a = 7 would need 9, 9 and three 9s ties the mode) so 8, 8, 9, 9, 12, 12, 12 is the only list A1
3
(d)
k = 2 or k = 14
mean = 70 + k8; median 9 if k ≤ 9, 9 + k2 if 9 ≤ k ≤ 12, 10.5 if k ≥ 12 M1 solves each case: 70 + k8 = 9 gives k = 2; 70 + k8 = 9 + k2 gives k = 343 (rejected); 70 + k8 = 10.5 gives k = 14 M1 k = 2 or k = 14 A1
3
Question 6Structured · 3 marksA* challenge
The mean mass of 22 boys is 30 kg. The mean mass of 7 girls is 33 kg. One more student joins the group. The mean mass of all 30 students is now 31.7 kg. Find the mass of the student who joined.
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