Cumulative Frequency Worksheets, Questions and Revision | MME - Free Printable
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Step-by-step solution for: Cumulative Frequency Worksheets, Questions and Revision | MME
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Show Answer Key & Explanations
Step-by-step solution for: Cumulative Frequency Worksheets, Questions and Revision | MME
Let’s solve each part step by step.
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Problem 1: Spending at a clothing shop (70 customers)
We are given a cumulative frequency graph. The vertical axis is “Cumulative Frequency” (number of customers who spent *up to* that amount), and the horizontal axis is “spending in euros”.
Total customers = 70 → so the top of the curve should reach 70 on the y-axis.
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(a) Find the median
Median = middle value → for 70 customers, it’s the average of the 35th and 36th values → but since we’re using a cumulative frequency graph, we just find where cumulative frequency = 35.
Look at y = 35 → go across to the curve, then down to x-axis.
From the graph:
At y=35, x ≈ 28 euros
→ Median = €28
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(b) Lower quartile (Q1)
Lower quartile = 25% of data → 25% of 70 = 17.5 → so find where cumulative frequency = 17.5
Go to y = 17.5 → across to curve → down to x-axis.
From graph: x ≈ 20 euros
→ Q1 = €20
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(c) Upper quartile (Q3)
Upper quartile = 75% of data → 75% of 70 = 52.5 → find where cumulative frequency = 52.5
Go to y = 52.5 → across to curve → down to x-axis.
From graph: x ≈ 34 euros
→ Q3 = €34
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(d) Interquartile range (IQR)
IQR = Q3 - Q1 = 34 - 20 = 14
→ IQR = €14
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(e) 60th percentile
60% of 70 = 0.6 × 70 = 42 → find where cumulative frequency = 42
Go to y = 42 → across to curve → down to x-axis.
From graph: x ≈ 30 euros
→ 60th percentile = €30
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(f) My bill was €20. What percentage spent more than me?
First, find how many spent ≤ €20 → from graph, at x=20, y≈17.5 → so about 17.5 customers spent €20 or less.
So number who spent MORE than €20 = 70 - 17.5 = 52.5
Percentage = (52.5 / 70) × 100 = 75%
→ 75%
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(g) How many customers spent between €30 and €45?
Find cumulative frequency at €30 and €45.
At x=30 → y≈42 (from earlier)
At x=45 → look at graph: y≈65
So number between 30 and 45 = 65 - 42 = 23
→ 23 customers
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Problem 2: Heights of young children
Graph shows cumulative frequency vs height in cm. Total children = 200 (since max y=200).
Answer correct to nearest 5mm = 0.5cm.
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(a) Estimate percentiles
Percentile means: what height corresponds to that % of children being shorter?
(i) 15th percentile → 15% of 200 = 30 → find height where cum freq = 30
Go to y=30 → across to curve → down to x-axis → x≈124.5 cm → round to nearest 0.5cm → 124.5 cm
(ii) 25th percentile → 25% of 200 = 50 → y=50 → x≈126.5 cm → 126.5 cm
(iii) 55th percentile → 55% of 200 = 110 → y=110 → x≈129.0 cm → 129.0 cm
(iv) 75th percentile → 75% of 200 = 150 → y=150 → x≈131.5 cm → 131.5 cm
(v) 90th percentile → 90% of 200 = 180 → y=180 → x≈136.0 cm → 136.0 cm
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(b) Interquartile range
IQR = Q3 - Q1 = 131.5 - 126.5 = 5.0 cm
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(c) Median
Median = 50th percentile → 50% of 200 = 100 → y=100 → x≈128.0 cm
→ 128.0 cm
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(d) Middle 30% of children have heights between X and Y cm
Middle 30% means from 35th percentile to 65th percentile.
Why? Because 50% ± 15% → 35% to 65%.
So:
X = 35th percentile → 35% of 200 = 70 → y=70 → x≈127.5 cm
Y = 65th percentile → 65% of 200 = 130 → y=130 → x≈130.5 cm
→ X = 127.5 cm, Y = 130.5 cm
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Final Answer:
Problem 1:
(a) €28
(b) €20
(c) €34
(d) €14
(e) €30
(f) 75%
(g) 23 customers
Problem 2:
(a) (i) 124.5 cm, (ii) 126.5 cm, (iii) 129.0 cm, (iv) 131.5 cm, (v) 136.0 cm
(b) 5.0 cm
(c) 128.0 cm
(d) X = 127.5 cm, Y = 130.5 cm
---
Problem 1: Spending at a clothing shop (70 customers)
We are given a cumulative frequency graph. The vertical axis is “Cumulative Frequency” (number of customers who spent *up to* that amount), and the horizontal axis is “spending in euros”.
Total customers = 70 → so the top of the curve should reach 70 on the y-axis.
---
(a) Find the median
Median = middle value → for 70 customers, it’s the average of the 35th and 36th values → but since we’re using a cumulative frequency graph, we just find where cumulative frequency = 35.
Look at y = 35 → go across to the curve, then down to x-axis.
From the graph:
At y=35, x ≈ 28 euros
→ Median = €28
---
(b) Lower quartile (Q1)
Lower quartile = 25% of data → 25% of 70 = 17.5 → so find where cumulative frequency = 17.5
Go to y = 17.5 → across to curve → down to x-axis.
From graph: x ≈ 20 euros
→ Q1 = €20
---
(c) Upper quartile (Q3)
Upper quartile = 75% of data → 75% of 70 = 52.5 → find where cumulative frequency = 52.5
Go to y = 52.5 → across to curve → down to x-axis.
From graph: x ≈ 34 euros
→ Q3 = €34
---
(d) Interquartile range (IQR)
IQR = Q3 - Q1 = 34 - 20 = 14
→ IQR = €14
---
(e) 60th percentile
60% of 70 = 0.6 × 70 = 42 → find where cumulative frequency = 42
Go to y = 42 → across to curve → down to x-axis.
From graph: x ≈ 30 euros
→ 60th percentile = €30
---
(f) My bill was €20. What percentage spent more than me?
First, find how many spent ≤ €20 → from graph, at x=20, y≈17.5 → so about 17.5 customers spent €20 or less.
So number who spent MORE than €20 = 70 - 17.5 = 52.5
Percentage = (52.5 / 70) × 100 = 75%
→ 75%
---
(g) How many customers spent between €30 and €45?
Find cumulative frequency at €30 and €45.
At x=30 → y≈42 (from earlier)
At x=45 → look at graph: y≈65
So number between 30 and 45 = 65 - 42 = 23
→ 23 customers
---
Problem 2: Heights of young children
Graph shows cumulative frequency vs height in cm. Total children = 200 (since max y=200).
Answer correct to nearest 5mm = 0.5cm.
---
(a) Estimate percentiles
Percentile means: what height corresponds to that % of children being shorter?
(i) 15th percentile → 15% of 200 = 30 → find height where cum freq = 30
Go to y=30 → across to curve → down to x-axis → x≈124.5 cm → round to nearest 0.5cm → 124.5 cm
(ii) 25th percentile → 25% of 200 = 50 → y=50 → x≈126.5 cm → 126.5 cm
(iii) 55th percentile → 55% of 200 = 110 → y=110 → x≈129.0 cm → 129.0 cm
(iv) 75th percentile → 75% of 200 = 150 → y=150 → x≈131.5 cm → 131.5 cm
(v) 90th percentile → 90% of 200 = 180 → y=180 → x≈136.0 cm → 136.0 cm
---
(b) Interquartile range
IQR = Q3 - Q1 = 131.5 - 126.5 = 5.0 cm
---
(c) Median
Median = 50th percentile → 50% of 200 = 100 → y=100 → x≈128.0 cm
→ 128.0 cm
---
(d) Middle 30% of children have heights between X and Y cm
Middle 30% means from 35th percentile to 65th percentile.
Why? Because 50% ± 15% → 35% to 65%.
So:
X = 35th percentile → 35% of 200 = 70 → y=70 → x≈127.5 cm
Y = 65th percentile → 65% of 200 = 130 → y=130 → x≈130.5 cm
→ X = 127.5 cm, Y = 130.5 cm
---
Final Answer:
Problem 1:
(a) €28
(b) €20
(c) €34
(d) €14
(e) €30
(f) 75%
(g) 23 customers
Problem 2:
(a) (i) 124.5 cm, (ii) 126.5 cm, (iii) 129.0 cm, (iv) 131.5 cm, (v) 136.0 cm
(b) 5.0 cm
(c) 128.0 cm
(d) X = 127.5 cm, Y = 130.5 cm
Parent Tip: Review the logic above to help your child master the concept of frequency graph worksheet.
