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KS3 · Statistics

KS3.M-S9 Comparing Distributions

AQA KS3.M-S9 · Calculators not allowed · about 75 minutes

Name: _______________________________ Date: ____ / ____ / ______

Answer ALL questions. Show all your working.

Write down the mean of the five test scores: 6, 9, 8, 7, 10.

(2)

State the median of the list: 12, 7, 9, 14, 11.

(1)

Give the mode of these shoe sizes: 4, 6, 5, 6, 7, 6, 5.

(1)

Calculate the range of these weights in kg: 48, 52, 45, 50, 47.

(2)

A frequency table shows times (minutes) taken by pupils to run 1 km.
Times: 4, 5, 6 with frequencies 3, 4, 3 respectively. Work out the mean time.

(2)

Class A has mean score 72 from 10 pupils. Class B has mean 68 from 10 pupils. Which class has the higher average score?

(2)

Two sets of exam scores are shown.
Set X: 55, 57, 58, 59, 60, 61, 62
Set Y: 40, 50, 60, 70, 80, 90, 100
State which set is more likely to be skewed and explain how the mean and median compare for that skew (one or two sentences).

(3)

Compare the spread of these two small data sets using range and give a one-sentence conclusion.
A: 3, 4, 5, 6, 7
B: 1, 5, 5, 6, 9

(3)

A cumulative frequency for test marks out of 20 is given as: marks 0-5: 3, 6-10: 9, 11-15: 15, 16-20: 20 (cumulative). Estimate the median mark.

(2)

A data set has mean 30 and range 12. If every value is increased by 5, what are the new mean and new range?

(3)

Two teachers compare their class test results out of 20.
Teacher P: scores 16, 15, 16, 17, 14, 16, 15, 16
Teacher Q: scores 10, 12, 14, 15, 16, 18, 19, 1
Find mean and range for both classes and state which class has the higher average and which is more consistent (smaller spread). Justify your answer with numbers.

(4)

A boxplot shows the five-number summary: min = 12, Q1 = 15, median = 18, Q3 = 22, max = 30.
State the interquartile range and give one sentence about how spread and skew are shown by this boxplot.

(3)

Which of these two datasets likely has the larger standard deviation? Give a short reason.
Set C: 49, 50, 51, 50, 49
Set D: 30, 40, 50, 60, 70

(3)

Grouped data: class widths 0-9, 10-19, 20-29 with frequencies 2, 5, 3. Use the midpoints to estimate the mean and then state which class (this or a class with mean 14 and same total frequency) has the higher average.

(4)

Two groups of runners have these times in minutes.
Group E: 7, 8, 7, 9, 8
Group F: 5, 9, 7, 8, 7
Calculate each group's IQR and say which group is more consistent using IQR as the measure.

(4)

Stretch question. A teacher compares two test score distributions out of 50.
Class G: 42, 43, 44, 45, 46, 47, 48, 49
Class H: 20, 30, 35, 40, 45, 46, 48, 50
Calculate mean and median for both classes, compute IQR for both, then recommend which class performed better overall and which was more consistent. Justify with the calculated statistics.

(5)

State the 90th percentile (the value above which 10% of data lie) for this sorted list of 10 values: 2, 4, 6, 8, 10, 12, 14, 16, 18, 20.

(2)

Explain in one sentence why the median is often a better measure than the mean for a distribution with outliers.

(2)

Mark scheme · KS3.M-S9 Comparing Distributions

Question 1

M1 sum of scores shown or stated (6+9+8+7+10 = 40) or method of adding
A1 mean = 8 cao
Answer: 8.0 (or 8)

Question 2

B1 median = 11 cao
Answer: 11

Question 3

B1 mode = 6 cao
Answer: 6

Question 4

M1 correct subtraction shown: max - min (52 - 45)
A1 range = 7 kg cao
Answer: 7 kg

Question 5

M1 correct use of (4*3 + 5*4 + 6*3) and total frequency 10 or equivalent
A1 mean = 5.0 minutes cao
Answer: 5.0 minutes (or 5)

Question 6

M1 comparison of means stated (72 > 68) or statement of Class A higher
A1 Class A has the higher average cao
Answer: Class A

Question 7

M1 identify Set Y as skewed (values spread with high outliers) or Set X as symmetric
M1 state effect on mean (pulled towards tail) or median less affected
A1 clear conclusion: for skewed Set Y mean is greater than median (because high values pull mean up) cao
Answer: Set Y is skewed; mean is greater than median because the high values pull the mean up

Question 8

M1 correct range for A shown (7 - 3 = 4) or for B (9 - 1 = 8)
M1 both ranges stated (A = 4, B = 8)
A1 conclusion that B has larger spread cao
Answer: Range A = 4, Range B = 8; B has the larger spread

Question 9

M1 recognise median position is 1/2 of 20 = 10th value and locate it in 11-15 group (cumulative reaches 9 then 15)
A1 estimate median = about 10 or 11 (choosing 10 as the lowest value in 11-15 group may be accepted) cao
Answer: Approximately 10 or 11

Question 10

M1 state new mean = old mean + 5
M1 state new range is unchanged (range = 12) or show reasoning
A1 new mean = 35 and new range = 12 cao
Answer: Mean = 35, Range = 12

Question 11

M1 correct mean for P or correct method shown (sum = 125, mean = 15.625)
M1 correct mean for Q or correct method shown (sum = 105, mean = 13.125)
M1 range for P = 17 - 14 = 3 and range for Q = 19 - 1 = 18 stated
A1 conclusion: P has higher average (15.625 > 13.125) and P is more consistent (smaller range) cao
Answer: P mean = 15.625, range = 3; Q mean = 13.125, range = 18. So P has the higher average and is more consistent.

Question 12

M1 correct IQR calculation Q3 - Q1 shown (22 - 15)
M1 IQR = 7 stated
A1 one-sentence comment linking box length and whisker lengths to spread or noting slight positive skew because upper whisker longer than lower oe
Answer: IQR = 7; the upper whisker is longer than the lower so the data show a slight positive skew and more spread above the median.

Question 13

M1 identify D as having larger spread (values further from mean)
M1 reference to values being more widely spread or giving approximate ranges (C range 2, D range 40)
A1 conclusion that D has the larger standard deviation with brief reason cao
Answer: Set D; values are spread much more widely so standard deviation will be larger

Question 14

M1 use midpoints correctly (4.5, 14.5, 24.5) and multiply by frequencies or equivalent
M1 correct total frequency = 10 stated
M1 correct estimated mean computed: (4.5*2 + 14.5*5 + 24.5*3)/10 = (9 + 72.5 + 73.5)/10 = 155/10
A1 estimated mean = 15.5 and conclusion that this class (15.5) has higher average than class with mean 14 cao
Answer: Estimated mean = 15.5; this class has the higher average than the class with mean 14

Question 15

M1 correct ordered lists shown or used for each group
M1 correct Q1 and Q3 for one group (E ordered 7,7,8,8,9 -> Q1=7, Q3=8.5)
M1 correct Q1 and Q3 for the other group (F ordered 5,7,7,8,9 -> Q1=6, Q3=8.5)
A1 IQR E = 1.5, IQR F = 2.5 so Group E is more consistent (smaller IQR) cao
Answer: IQR E = 1.5, IQR F = 2.5; Group E is more consistent (smaller IQR)

Question 16

M1 correct mean for G shown (sum 364, mean = 45.5) or method
M1 correct mean for H shown (sum 314, mean = 39.25) or method
M1 correct medians found (G median = 45.5, H median = 42.5)
M1 correct IQRs computed (G Q1=43.5, Q3=47.5 so IQR=4; H Q1=32.5, Q3=47 so IQR=14.5) or equivalent
A1 clear recommendation: G performed better on average (mean and median higher) and G is more consistent (smaller IQR) cao
Answer: G mean = 45.5, median = 45.5, IQR = 4; H mean = 39.25, median = 42.5, IQR = 14.5; recommend G: higher performance and more consistent.

Question 17

M1 correct method to find position: 0.9 x (n) = 0.9 x 10 = 9th value or appropriate percentile method
A1 90th percentile = 18 cao
Answer: 18

Question 18

M1 state that outliers pull the mean towards them
A1 state that median is less affected by extreme values so better represents centre for skewed data cao
Answer: Because outliers pull the mean towards them, the median is less affected by extreme values and so better represents the centre for skewed distributions.

Comparing Distributions - Worksheets, Questions and Revision