Year 7 Mathematics | Victorian Curriculum 2.0
Summary statistics
Topic 14 | Statistics & Probability | Answer key

Tier 1: basic skills

Fluency

Fluency

    1. 88
    2. 66
    3. 1616
    4. 55
    5. 66. Sorted: 2,4,6,8,102, 4, 6, 8, 10.
    6. 6.56.5. Sorted: 3,5,8,103, 5, 8, 10; mean of 55 and 88.
    7. 15.515.5. Mean of middle two 1414 and 1717.
    8. 33 (appears three times)
    9. 44 and 66 (bimodal)
    10. 77. Method: 9−29 - 2.
    11. 3434. Method: 72−3872 - 38.
    12. 1515
    13. Mean 77; median 66; mode 55; range 66.

Tier 2: mixed practice

Reasoning

Mixed practice

    1. Sorted: 12,14,15,15,15,16,17,18,18,2012, 14, 15, 15, 15, 16, 17, 18, 18, 20; median =15+162=15.5= \dfrac{15 + 16}{2} = 15.5.
    2. 1616. Method: sum =160= 160; 160÷10160 \div 10.
    3. 1515 (appears three times).
    4. 88. Method: 20−1220 - 12.
    5. Range changes most (from 88 to 8888). The mode is unaffected. The median shifts only slightly; the mean goes up by about 7.67.6.
    6. 7070. Method: total required =5×70=350= 5 \times 70 = 350; subtract 65+72+68+75=28065 + 72 + 68 + 75 = 280.
    7. 2121. Method: total =6×18=108= 6 \times 18 = 108; sum of five given =87= 87; 108−87=21108 - 87 = 21.
    8. 1616. Method: original total =15×10=150= 15 \times 10 = 150; correction +10+10 gives 160160; new mean 160÷10160 \div 10.
    9. Any data set with all values equal to 1010, e.g. {10,10,10,10}\{10, 10, 10, 10\}.
    10. Many possible. Example: 1,2,3,4,401, 2, 3, 4, 40. Mean =10= 10, median =3= 3.

Tier 3: explain and spot the mistake

Reasoning

Explain and spot the mistake

    1. The mode is the most frequently occurring value, not the largest one. The student has confused mode with maximum (the upper end of the range). Correct: the mode is whichever value appears most often; a data set can have no mode, one mode, or multiple modes.
    2. The mean uses every value, so a single extreme number can pull it noticeably up or down. The median depends only on position in the sorted list, so one outlier only shifts the middle by one rank at most - hence the median remains close to the bulk of the data when there are extreme values.
    3. Yes to both. Categorical data (e.g. eye colours) can have a mode (most common colour) but no mean - you cannot average “red”, “blue”, “green”. A data set with all distinct numerical values (e.g. 1,2,3,41, 2, 3, 4) has a mean (2.52.5) but no mode, since no value repeats.
    4. Not always. For an even-count numerical set, the median is the average of the two middle values, which may not itself be in the data. Example: median of {2,4}\{2, 4\} is 33, which is not in the set.

Tier 4: real-world problems

Problem solving

Real-world problems

    1. Mean ≈37.1\approx 37.1; median 3838. Method: sum =260= 260; mean =260/7= 260 / 7; sort and take the 44th value (3838).
    2. Mean age =10= 10. Method: sum =4+7+10+13+16=50= 4 + 7 + 10 + 13 + 16 = 50; 50÷550 \div 5.
    3. Mean ≈24.7\approx 24.7 degC (exactly 173/7173/7); range 99 degC. Method: 173÷7≈24.71173 \div 7 \approx 24.71; 30−2130 - 21.
    4. New mean =62= 62. Method: previous total =7×60=420= 7 \times 60 = 420; new total =420+76=496= 420 + 76 = 496; 496÷8=62496 \div 8 = 62.
    5. Mean of middle 88 is 48.7548.75. Method: total of all 1010 is 500500; remove 20+90=11020 + 90 = 110; remaining total 390390; 390÷8390 \div 8.
Year 7 Mathematics study companion | Answer key