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6th grade questions for box plots, histograms, dot plots, and frequency tables
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Here are graded practice questions (with answers) you can use for 6th grade students on box plots, histograms, dot plots, and frequency tables. Each section has problems that ask students to make plots and to read/interpret them.
Instructions for teachers: For drawing questions, students should draw the required plot on graph paper or digitally. Answers below include the computations and interpretations.
BOX PLOTS (making and interpreting)
1) Data: 8, 11, 13, 15, 17, 18, 20, 22, 24, 27
a) Find the five-number summary (min, Q1, median, Q3, max).
b) Draw the box plot.
Answer:
- Sorted already. n = 10
- Min = 8; Max = 27
- Median = average of 17 and 18 = 17.5
- Lower half = 8,11,13,15,17 → Q1 = 13
- Upper half = 18,20,22,24,27 → Q3 = 22
- Five-number summary: 8, 13, 17.5, 22, 27
2) Data: 4, 6, 7, 7, 8, 9, 11, 12
a) Find the five-number summary and draw the box plot.
b) Which half (lower or upper) is more spread out?
Answer:
- n = 8; Min = 4; Max = 12; Median = average of 7 and 8 = 7.5
- Lower half = 4,6,7,7 → Q1 = average of 6 and 7 = 6.5
- Upper half = 8,9,11,12 → Q3 = average of 9 and 11 = 10
- Five-number summary: 4, 6.5, 7.5, 10, 12
- Spread: Upper half IQR = 10 − 6.5 = 3.5, lower half span is smaller; upper half more spread out.
3) Data: 3, 5, 6, 6, 7, 8, 30
a) Compute Q1, median, Q3, IQR and determine if 30 is an outlier using 1.5×IQR rule.
Answer:
- Sorted. n = 7. Median = 6
- Lower half (below median) = 3,5,6 → Q1 = 5
- Upper half = 6,7,8 → Q3 = 7
- IQR = 7 − 5 = 2. 1.5×IQR = 3
- Upper fence = Q3 + 3 = 10. Lower fence = Q1 − 3 = 2
- 30 > 10 → 30 is an outlier.
4) A teacher gives these test scores: 62 (2 students), 68 (1), 70 (3), 75 (2), 88 (1). (Numbers show scores and how many students got that score.)
a) Build the full list of scores, then find the five-number summary and draw a box plot.
Answer:
- Full list (sorted): 62,62,68,70,70,70,75,75,88 (n=9)
- Median = 70 (middle value)
- Lower half = 62,62,68, median of lower half = average of 62 and 68? Wait: for n=9, lower half is first 4 values: 62,62,68,70 → Q1 = average of middle two = (62+68)/2 = 65
- Upper half = 70,75,75,88 → Q3 = average of middle two = (75+75)/2 = 75
- Min = 62, Max = 88
- Five-number summary: 62, 65, 70, 75, 88
5) You see a box plot with min = 10, Q1 = 14, median = 18, Q3 = 22, max = 30.
a) What percent of the data is between Q1 and Q3?
b) If there are 20 data points, how many are at or below the median (approximately)?
Answer:
- a) Between Q1 and Q3 is the middle 50% of the data.
- b) At or below median ≈ 50% of 20 = 10 students.
6) Two box plots compare math scores of Class A and Class B. Class A: median 75, IQR 20. Class B: median 70, IQR 10.
a) Which class shows more variability?
b) Which class has a higher middle score?
Answer:
- a) Class A (IQR 20) is more variable.
- b) Class A has higher median (75 vs. 70).
HISTOGRAMS (making and interpreting)
7) Data (shoe sizes): 6, 7, 6, 8, 7, 6, 9, 7, 8, 6, 7, 9, 8, 6
a) Make a histogram with bins for 6, 7, 8, 9 (each whole number).
b) Which shoe size is most common?
Answer:
- Counts: 6 → 5, 7 → 4, 8 → 3, 9 → 2
- Most common: size 6.
8) Data: bus arrival delays (minutes): 0, 1, 2, 2, 3, 3, 3, 5, 6, 8, 10, 12
a) Make a histogram with bins 0–2, 3–5, 6–8, 9–11, 12–14.
b) Which bin has the highest frequency?
Answer:
- Bin counts: 0–2: {0,1,2,2} → 4; 3–5: {3,3,3,5} → 4; 6–8: {6,8} → 2; 9–11: {10} → 1; 12–14: {12} → 1
- Highest frequency bins: 0–2 and 3–5 (tie).
9) Given this grouped frequency table for hours studied last week, draw a histogram (bins are the ranges shown):
- 0–2 hours: 3 students
- 3–5 hours: 8 students
- 6–8 hours: 6 students
- 9–11 hours: 3 students
a) Which range has the most students?
b) How many students in total?
Answer:
- a) 3–5 hours has the most (8 students).
- b) Total = 3 + 8 + 6 + 3 = 20 students.
10) Interpretation: Two histograms show heights of plants. Histogram A is concentrated around 10–14 cm, histogram B is spread from 6–18 cm. Which set is more consistent (less variable)?
Answer:
- Histogram A is more consistent (less spread).
11) Create a histogram: Data (number of books read this year): 0,1,1,2,2,2,3,4,4,5,6,6,7,8
- Use bins 0–1, 2–3, 4–5, 6–7, 8–9.
- Which bin contains the median?
Answer:
- Sorted; n = 14; median is average of 7th and 8th values: 7th = 3, 8th = 4 → median = 3.5
- Bin counts: 0–1: {0,1,1}→3; 2–3: {2,2,2,3}→4; 4–5: {4,4,5}→3; 6–7: {6,6,7}→3; 8–9: {8}→1
- Median 3.5 falls in bin 2–3 or 4–5? Since 3.5 > 3, it belongs in 4–5 bin because that bin contains 4; but median value 3.5 sits between bins—if asked which bin contains the middle values, the two middle numbers are 3 and 4 which are in bins 2–3 and 4–5. So answer: the median is between the 2–3 and 4–5 bins (middle values are 3 and 4).
DOT PLOTS (making and interpreting)
12) Data: 3, 4, 4, 5, 5, 5, 6, 6, 7
a) Make a dot plot.
b) Find mode, median, and range.
Answer:
- Mode = 5 (appears 3 times)
- Median = 5 (middle value of 9 numbers)
- Range = 7 − 3 = 4
13) Data (number of pets students have): 0,0,1,1,1,2,2,3
a) Draw a dot plot.
b) What is the relative frequency of students with 1 pet?
Answer:
- Total students = 8
- Number with 1 pet = 3 → relative frequency = 3/8 = 0.375 (or 37.5%)
14) Create a dot plot from these data and answer questions: scores out of 10: 10,9,9,8,8,8,7,6,6,5
a) How many students scored 8 or higher?
b) What is the median?
Answer:
- Scores 8 or higher: 10,9,9,8,8,8 → 6 students
- Sorted n=10 → median is average of 5th and 6th values. Sorted: 5,6,6,7,8,8,8,9,9,10 → 5th=8, 6th=8 → median = 8
15) Given this dot plot (imagine dots at values 1:••, 2:•••, 3:•, 4:••), how many data points are there and what is the mode?
Answer:
- Counts: 1→2, 2→3, 3→1, 4→2; total = 8 points; mode = 2
FREQUENCY TABLES (building, relative & cumulative frequency)
16) Build a frequency table for the data: 2, 3, 3, 4, 4, 4, 5, 5, 6
a) Include counts and relative frequencies.
Answer:
- Values and counts: 2→1, 3→2, 4→3, 5→2, 6→1. Total = 9.
- Relative frequencies: 2→1/9 ≈ 0.111, 3→2/9 ≈ 0.222, 4→3/9 = 0.333, 5→2/9 ≈ 0.222, 6→1/9 ≈ 0.111
17) Given grouped frequencies for ages of animals at a shelter:
- 0–1 year: 5
- 2–3 years: 12
- 4–5 years: 9
- 6–7 years: 4
a) Create the cumulative frequency table.
b) How many animals are 3 years old or younger?
Answer:
- Cumulative: 0–1 → 5; 0–3 → 5+12 = 17; 0–5 → 17+9 = 26; 0–7 → 26+4 = 30
- Animals 3 years old or younger = 17
18) Build a frequency table and cumulative frequency for test grades (letters) from this list: A, B, A, C, B, A, B, B, C
Answer:
- Counts: A→3, B→4, C→2; total = 9
- Cumulative (by alphabetical order A then B then C): A cumulative = 3; A+B = 7; A+B+C = 9
19) A survey of how many texts students sent yesterday gives this frequency table:
- 0 texts: 6 students
- 1 text: 9 students
- 2 texts: 4 students
- 3 or more: 1 student
a) What is the relative frequency for 0 texts?
b) What percentage sent at least 1 text?
Answer:
- Total = 6+9+4+1 = 20
- a) Relative frequency for 0 texts = 6/20 = 0.30
- b) At least 1 text = 1 − 6/20 = 14/20 = 0.70 = 70%
MIXED INTERPRETATION PRACTICE
20) You have two data sets of 12 numbers each. Data set X histogram is centered around 50 with tight bins. Data set Y histogram is centered around 60 but spread widely from 40 to 80. Which set has a higher median? Which is more spread out?
Answer:
- Higher median: Y (centered around 60)
- More spread out: Y
21) A dot plot of lunchtime minutes shows most dots at 18–22 minutes and a few at 30 minutes. What might those few at 30 represent? How would that affect the mean vs median?
Answer:
- The few at 30 are outliers (students who took longer). Outliers pull the mean higher more than the median. The median is less affected.
Quick answer key summary (for teacher reference):
- Q1 box plot five-number summary: 8,13,17.5,22,27
- Q2 five-number summary: 4,6.5,7.5,10,12; upper half more spread
- Q3 outlier: 30 is outlier
- Q4 five-number summary: 62,65,70,75,88
- Q5 middle 50%; 10 at/below median (for 20)
- Q6 class A more variable and higher median
- Q7 histogram counts: 6→5,7→4,8→3,9→2; mode 6
- Q8 bin counts: 0–2→4,3–5→4,6–8→2,9–11→1,12–14→1; highest 0–2 and 3–5
- Q9 most students 3–5; total 20
- Q10 Histogram A more consistent
- Q11 bin counts: 0–1→3,2–3→4,4–5→3,6–7→3,8–9→1; median between 2–3 and 4–5 bins
- Q12 mode=5 median=5 range=4
- Q13 relative frequency for 1 pet = 3/8 = 0.375
- Q14 six students scored 8 or higher; median = 8
- Q15 total 8; mode=2
- Q16 counts 2→1,3→2,4→3,5→2,6→1; relative frequencies as listed
- Q17 cumulative: 5,17,26,30; 17 animals ≤3
- Q18 counts A→3,B→4,C→2; cumulative 3,7,9
- Q19 relative freq 0 texts = 0.30; at least 1 text = 70%
- Q20 Y higher median and more spread out
- Q21 outliers at 30; mean affected more than median
If you want, I can:
- Generate printable worksheets with empty graphs for students to fill in,
- Make multiple choice versions,
- Produce answer sheets formatted for grading,
- Or adapt questions to a specific textbook or state standard. Which would you like next?
Instructions for teachers: For drawing questions, students should draw the required plot on graph paper or digitally. Answers below include the computations and interpretations.
BOX PLOTS (making and interpreting)
1) Data: 8, 11, 13, 15, 17, 18, 20, 22, 24, 27
a) Find the five-number summary (min, Q1, median, Q3, max).
b) Draw the box plot.
Answer:
- Sorted already. n = 10
- Min = 8; Max = 27
- Median = average of 17 and 18 = 17.5
- Lower half = 8,11,13,15,17 → Q1 = 13
- Upper half = 18,20,22,24,27 → Q3 = 22
- Five-number summary: 8, 13, 17.5, 22, 27
2) Data: 4, 6, 7, 7, 8, 9, 11, 12
a) Find the five-number summary and draw the box plot.
b) Which half (lower or upper) is more spread out?
Answer:
- n = 8; Min = 4; Max = 12; Median = average of 7 and 8 = 7.5
- Lower half = 4,6,7,7 → Q1 = average of 6 and 7 = 6.5
- Upper half = 8,9,11,12 → Q3 = average of 9 and 11 = 10
- Five-number summary: 4, 6.5, 7.5, 10, 12
- Spread: Upper half IQR = 10 − 6.5 = 3.5, lower half span is smaller; upper half more spread out.
3) Data: 3, 5, 6, 6, 7, 8, 30
a) Compute Q1, median, Q3, IQR and determine if 30 is an outlier using 1.5×IQR rule.
Answer:
- Sorted. n = 7. Median = 6
- Lower half (below median) = 3,5,6 → Q1 = 5
- Upper half = 6,7,8 → Q3 = 7
- IQR = 7 − 5 = 2. 1.5×IQR = 3
- Upper fence = Q3 + 3 = 10. Lower fence = Q1 − 3 = 2
- 30 > 10 → 30 is an outlier.
4) A teacher gives these test scores: 62 (2 students), 68 (1), 70 (3), 75 (2), 88 (1). (Numbers show scores and how many students got that score.)
a) Build the full list of scores, then find the five-number summary and draw a box plot.
Answer:
- Full list (sorted): 62,62,68,70,70,70,75,75,88 (n=9)
- Median = 70 (middle value)
- Lower half = 62,62,68, median of lower half = average of 62 and 68? Wait: for n=9, lower half is first 4 values: 62,62,68,70 → Q1 = average of middle two = (62+68)/2 = 65
- Upper half = 70,75,75,88 → Q3 = average of middle two = (75+75)/2 = 75
- Min = 62, Max = 88
- Five-number summary: 62, 65, 70, 75, 88
5) You see a box plot with min = 10, Q1 = 14, median = 18, Q3 = 22, max = 30.
a) What percent of the data is between Q1 and Q3?
b) If there are 20 data points, how many are at or below the median (approximately)?
Answer:
- a) Between Q1 and Q3 is the middle 50% of the data.
- b) At or below median ≈ 50% of 20 = 10 students.
6) Two box plots compare math scores of Class A and Class B. Class A: median 75, IQR 20. Class B: median 70, IQR 10.
a) Which class shows more variability?
b) Which class has a higher middle score?
Answer:
- a) Class A (IQR 20) is more variable.
- b) Class A has higher median (75 vs. 70).
HISTOGRAMS (making and interpreting)
7) Data (shoe sizes): 6, 7, 6, 8, 7, 6, 9, 7, 8, 6, 7, 9, 8, 6
a) Make a histogram with bins for 6, 7, 8, 9 (each whole number).
b) Which shoe size is most common?
Answer:
- Counts: 6 → 5, 7 → 4, 8 → 3, 9 → 2
- Most common: size 6.
8) Data: bus arrival delays (minutes): 0, 1, 2, 2, 3, 3, 3, 5, 6, 8, 10, 12
a) Make a histogram with bins 0–2, 3–5, 6–8, 9–11, 12–14.
b) Which bin has the highest frequency?
Answer:
- Bin counts: 0–2: {0,1,2,2} → 4; 3–5: {3,3,3,5} → 4; 6–8: {6,8} → 2; 9–11: {10} → 1; 12–14: {12} → 1
- Highest frequency bins: 0–2 and 3–5 (tie).
9) Given this grouped frequency table for hours studied last week, draw a histogram (bins are the ranges shown):
- 0–2 hours: 3 students
- 3–5 hours: 8 students
- 6–8 hours: 6 students
- 9–11 hours: 3 students
a) Which range has the most students?
b) How many students in total?
Answer:
- a) 3–5 hours has the most (8 students).
- b) Total = 3 + 8 + 6 + 3 = 20 students.
10) Interpretation: Two histograms show heights of plants. Histogram A is concentrated around 10–14 cm, histogram B is spread from 6–18 cm. Which set is more consistent (less variable)?
Answer:
- Histogram A is more consistent (less spread).
11) Create a histogram: Data (number of books read this year): 0,1,1,2,2,2,3,4,4,5,6,6,7,8
- Use bins 0–1, 2–3, 4–5, 6–7, 8–9.
- Which bin contains the median?
Answer:
- Sorted; n = 14; median is average of 7th and 8th values: 7th = 3, 8th = 4 → median = 3.5
- Bin counts: 0–1: {0,1,1}→3; 2–3: {2,2,2,3}→4; 4–5: {4,4,5}→3; 6–7: {6,6,7}→3; 8–9: {8}→1
- Median 3.5 falls in bin 2–3 or 4–5? Since 3.5 > 3, it belongs in 4–5 bin because that bin contains 4; but median value 3.5 sits between bins—if asked which bin contains the middle values, the two middle numbers are 3 and 4 which are in bins 2–3 and 4–5. So answer: the median is between the 2–3 and 4–5 bins (middle values are 3 and 4).
DOT PLOTS (making and interpreting)
12) Data: 3, 4, 4, 5, 5, 5, 6, 6, 7
a) Make a dot plot.
b) Find mode, median, and range.
Answer:
- Mode = 5 (appears 3 times)
- Median = 5 (middle value of 9 numbers)
- Range = 7 − 3 = 4
13) Data (number of pets students have): 0,0,1,1,1,2,2,3
a) Draw a dot plot.
b) What is the relative frequency of students with 1 pet?
Answer:
- Total students = 8
- Number with 1 pet = 3 → relative frequency = 3/8 = 0.375 (or 37.5%)
14) Create a dot plot from these data and answer questions: scores out of 10: 10,9,9,8,8,8,7,6,6,5
a) How many students scored 8 or higher?
b) What is the median?
Answer:
- Scores 8 or higher: 10,9,9,8,8,8 → 6 students
- Sorted n=10 → median is average of 5th and 6th values. Sorted: 5,6,6,7,8,8,8,9,9,10 → 5th=8, 6th=8 → median = 8
15) Given this dot plot (imagine dots at values 1:••, 2:•••, 3:•, 4:••), how many data points are there and what is the mode?
Answer:
- Counts: 1→2, 2→3, 3→1, 4→2; total = 8 points; mode = 2
FREQUENCY TABLES (building, relative & cumulative frequency)
16) Build a frequency table for the data: 2, 3, 3, 4, 4, 4, 5, 5, 6
a) Include counts and relative frequencies.
Answer:
- Values and counts: 2→1, 3→2, 4→3, 5→2, 6→1. Total = 9.
- Relative frequencies: 2→1/9 ≈ 0.111, 3→2/9 ≈ 0.222, 4→3/9 = 0.333, 5→2/9 ≈ 0.222, 6→1/9 ≈ 0.111
17) Given grouped frequencies for ages of animals at a shelter:
- 0–1 year: 5
- 2–3 years: 12
- 4–5 years: 9
- 6–7 years: 4
a) Create the cumulative frequency table.
b) How many animals are 3 years old or younger?
Answer:
- Cumulative: 0–1 → 5; 0–3 → 5+12 = 17; 0–5 → 17+9 = 26; 0–7 → 26+4 = 30
- Animals 3 years old or younger = 17
18) Build a frequency table and cumulative frequency for test grades (letters) from this list: A, B, A, C, B, A, B, B, C
Answer:
- Counts: A→3, B→4, C→2; total = 9
- Cumulative (by alphabetical order A then B then C): A cumulative = 3; A+B = 7; A+B+C = 9
19) A survey of how many texts students sent yesterday gives this frequency table:
- 0 texts: 6 students
- 1 text: 9 students
- 2 texts: 4 students
- 3 or more: 1 student
a) What is the relative frequency for 0 texts?
b) What percentage sent at least 1 text?
Answer:
- Total = 6+9+4+1 = 20
- a) Relative frequency for 0 texts = 6/20 = 0.30
- b) At least 1 text = 1 − 6/20 = 14/20 = 0.70 = 70%
MIXED INTERPRETATION PRACTICE
20) You have two data sets of 12 numbers each. Data set X histogram is centered around 50 with tight bins. Data set Y histogram is centered around 60 but spread widely from 40 to 80. Which set has a higher median? Which is more spread out?
Answer:
- Higher median: Y (centered around 60)
- More spread out: Y
21) A dot plot of lunchtime minutes shows most dots at 18–22 minutes and a few at 30 minutes. What might those few at 30 represent? How would that affect the mean vs median?
Answer:
- The few at 30 are outliers (students who took longer). Outliers pull the mean higher more than the median. The median is less affected.
Quick answer key summary (for teacher reference):
- Q1 box plot five-number summary: 8,13,17.5,22,27
- Q2 five-number summary: 4,6.5,7.5,10,12; upper half more spread
- Q3 outlier: 30 is outlier
- Q4 five-number summary: 62,65,70,75,88
- Q5 middle 50%; 10 at/below median (for 20)
- Q6 class A more variable and higher median
- Q7 histogram counts: 6→5,7→4,8→3,9→2; mode 6
- Q8 bin counts: 0–2→4,3–5→4,6–8→2,9–11→1,12–14→1; highest 0–2 and 3–5
- Q9 most students 3–5; total 20
- Q10 Histogram A more consistent
- Q11 bin counts: 0–1→3,2–3→4,4–5→3,6–7→3,8–9→1; median between 2–3 and 4–5 bins
- Q12 mode=5 median=5 range=4
- Q13 relative frequency for 1 pet = 3/8 = 0.375
- Q14 six students scored 8 or higher; median = 8
- Q15 total 8; mode=2
- Q16 counts 2→1,3→2,4→3,5→2,6→1; relative frequencies as listed
- Q17 cumulative: 5,17,26,30; 17 animals ≤3
- Q18 counts A→3,B→4,C→2; cumulative 3,7,9
- Q19 relative freq 0 texts = 0.30; at least 1 text = 70%
- Q20 Y higher median and more spread out
- Q21 outliers at 30; mean affected more than median
If you want, I can:
- Generate printable worksheets with empty graphs for students to fill in,
- Make multiple choice versions,
- Produce answer sheets formatted for grading,
- Or adapt questions to a specific textbook or state standard. Which would you like next?
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