Statistics and Data Displays: practice solutions, Grade 6

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Practice solutions Grade 6 : Statistics and Data Displays — Zyro the alien explorer of Planète Maths

Written solutions to the chapter problems. Check each step, then correct yourself.

Sommaire

1 Which questions are statistical? ★★★

A question is statistical when we expect different answers.

Not statistical: there is one person and one answer.
Statistical: each student reads a different amount of time.
Not statistical: there is exactly one answer.
Statistical: the players have different heights.

Questions b and d are statistical.

2 Reading a dot plot of shoe sizes ★★★

The tallest stack is at size 7 with 4 dots, so the mode is size 7.
Size 8 has 2 dots and size 9 has 1 dot: \(2+1 = 3\) students.

Check: \(2+3+4+2+1 = 12\) students, which matches the 12 dots (size 5: 2, size 6: 3).

3 Building a frequency table ★★★

Count each value.

Rainy days	1	2	3	4	Total
Frequency	2	3	4	1	10

The total is \(2+3+4+1 = 10\), the number of weeks, so no value was missed.

4 Mean of five numbers ★★★

Sum: \(8+6+9+7+10 = 40\). Divide by 5: \(40 \div 5 = 8\).

The mean is 8 laps.

5 Median of an odd list ★★★

In order: 41, 48, 53, 59, 66. There are 5 values, so the median is the 3rd value, 53.

The median temperature is 53 °F.

6 Find the mode ★★★

The value 4 appears 3 times, 7 appears 2 times, and the others once. The mode is 4.

It means the shop most often sold 4 sandwiches in a day.

7 Range of temperatures ★★★

Maximum: 71. Minimum: 49. Range \(= 71 - 49 = 22\).

The range is 22 °F.

8 Median of an even list ★★★

In order: 3, 5, 7, 8, 10, 12. Six values, so average the 3rd and 4th: \(\dfrac{7+8}{2} = 7.5\).

The median is 7.5 pages.

9 Fair share of savings ★★★

Total: \(12+20+8+16 = 56\) dollars. Fair share: \(56 \div 4 = 14\) dollars each.

The siblings with 12 and 8 dollars receive money; the ones with 20 and 16 give. Check the balance: gives \(6+2 = 8\); receives \(2+6 = 8\).

10 Find the missing score ★★★

Total needed: \(5 \times 84 = 420\). First four: \(80+92+78+85 = 335\). Fifth: \(420 - 335 = 85\).

The fifth score was 85.

11 Reading a histogram of plant heights ★★★

Bins 0–5 and 5–10: \(2+5 = 7\) plants.
The bin 10–15, with 9 plants.
Bins 10–15 and 15–20: \(9+6 = 15\) plants out of 24, so \(\dfrac{15}{24} = \dfrac{5}{8}\).
No. The histogram only shows that the tallest plant is in the bin 20–25 cm, not its exact height.

Check: \(2+5+9+6+2 = 24\).

12 Five-number summary ★★★

There are 9 values already in order. Median: the 5th value, 8. Lower half (without the median): 2, 4, 5, 7, so \(Q_1 = \dfrac{4+5}{2} = 4.5\). Upper half: 9, 12, 13, 15, so \(Q_3 = \dfrac{12+13}{2} = 12.5\).

Summary: 2, 4.5, 8, 12.5, 15. IQR \(= 12.5 - 4.5 = 8\).

13 The effect of an outlier ★★★

Before: sum \(= 63\), mean \(= 63 \div 9 = 7\); the median is the 5th value, 7.

After: sum \(= 93\), mean \(= 93 \div 10 = 9.3\). In order the 5th and 6th values are 7 and 7, so the median is 7.

The mean rose from 7 to 9.3, while the median did not change. The mean is more affected by the outlier.

14 Mean absolute deviation ★★★

Mean: \(35 \div 5 = 7\). Distances: 4, 2, 0, 2, 4. Sum: 12. \(\text{MAD} = 12 \div 5 = 2.4\).

On average the values are 2.4 units from the mean.

15 Choosing bins for run times ★★★

10–15: 11, 12, 14, so 3 values. 15–20: 15, 17, 18, 19, so 4 values. 20–25: 21, 22, 24, so 3 values.

Bin (seconds)	10–15	15–20	20–25	Total
Frequency	3	4	3	10

The total is 10, as expected. The bars would have heights 3, 4 and 3.

16 Same mean, different spread ★★★

Team A: mean \(70 \div 5 = 14\). Distances: 4, 2, 0, 2, 4, sum 12, \(\text{MAD} = 2.4\).

Team B: mean \(70 \div 5 = 14\). Distances: 8, 4, 0, 4, 8, sum 24, \(\text{MAD} = 4.8\).

Both teams average 14 points, but Team A has the smaller MAD, so Team A is more consistent.

17 Comparing two box plots ★★★

Class A: 65. Class B: 70.
Class A: range \(95-40 = 55\), IQR \(80-55 = 25\). Class B: range \(90-50 = 40\), IQR \(75-60 = 15\).
Class B, because its IQR (15) is smaller than the IQR of Class A (25): the middle half of its scores is packed in a narrower interval.

18 A missing value from the mean ★★★

The sum must be \(5 \times 9.6 = 48\). Known values: \(4+7+12+15 = 38\). So \(x = 48 - 38 = 10\).

The list 4, 7, 10, 12, 15 is in increasing order, and the median is 10.

19 Which center describes house prices? ★★★

Sum: \(210+225+230+240+250+255+900 = 2310\). Mean: \(2310 \div 7 = 330\), that is 330 thousand dollars.

Median: the 4th value, 240 thousand dollars.

The median is better: six of the seven houses cost less than 260 thousand, and the single 900 thousand house pulls the mean up. The data are skewed right.

20 Adding a new score ★★★

Before: mean \(50 \div 5 = 10\), median 10.

After: 6, 8, 10, 12, 14, 22. Sum \(= 72\), mean \(= 72 \div 6 = 12\). Median \(= \dfrac{10+12}{2} = 11\).

The mean increased by 2 and the median increased by 1. The large new score affects the mean more.

21 The balance point ★★★

Mean: \((2+2+3+9) \div 4 = 16 \div 4 = 4\).

Below 4: \(2+2+1 = 5\) units (the dots 2, 2, 3 are 2, 2 and 1 away). Above 4: the dot 9 is \(9-4 = 5\) away.

Both totals equal 5, so the number line balances at 4.

22 Describe the distribution ★★★

The bars rise to a peak at 40–60 and then fall, with similar tails on both sides: the shape is roughly symmetric, with the peak in the middle.

The median is the mean of the 10th and 11th values. Counting: \(2+4 = 6\) values are below 40, and \(6+8 = 14\) are below 60, so the 10th and 11th values are both in the bin 40–60.

With no outliers and a roughly symmetric shape, the mean and the MAD are a good choice (the median and IQR would also work).