Every solution shows the ordered data, each calculation, and a clearly marked final answer. The solution figures mark the mean and median on the dot plots and the quartiles on the ordered data. Compare your work with them, and check that you ordered the data before finding the median or quartiles.
Problem 1: Find the mean
Sum: \(4 + 7 + 9 + 12 = 32\).
\(32 : 4 = 8\)
The mean is 8.
Problem 2: Median, odd count
Order: 9, 11, 12, 15, 20.
The middle value is the 3rd one.
The median is 12.
Problem 3: Median, even count
Order: 3, 5, 6, 8, 10, 12.
The two middle values are 6 and 8. \(\frac{6 + 8}{2} = 7\)
The median is 7.
Problem 4: Find the range
Greatest: 40. Least: 18.
\(40 – 18 = 22\)
The range is 22 centimeters.
Problem 5: Quartiles and IQR
Order: 3, 4, 6, 7, 9, 10, 12, 13, 15, 18.
Median: \(\frac{9 + 10}{2} = 9.5\).
Lower half 3, 4, 6, 7, 9 has median 6, so \(Q_1 = 6\). Upper half 10, 12, 13, 15, 18 has median 13, so \(Q_3 = 13\).
\(IQR = 13 – 6 = 7\)
Answers: \(Q_1 = 6\), median 9.5, \(Q_3 = 13\), IQR 7.
Problem 6: Find the MAD
Mean: \(\frac{4 + 6 + 6 + 8 + 11}{5} = \frac{35}{5} = 7\).
Distances from 7: 3, 1, 1, 1, 4. Sum: \(3 + 1 + 1 + 1 + 4 = 10\).
\(MAD = 10 : 5 = 2\)
The MAD is 2.
Problem 7: Missing value
The total of four numbers with mean 12 is \(4 \times 12 = 48\).
The known numbers add to \(10 + 15 + 9 = 34\).
\(48 – 34 = 14\)
The fourth number is 14.
Problem 8: Effect of an outlier
(a) Total: \(88 + 92 + 90 + 86 + 94 + 30 = 480\). \(480 : 6 = 80\). The mean is 80.
(b) Order: 30, 86, 88, 90, 92, 94. \(\frac{88 + 90}{2} = 89\). The median is 89.
(c) The median is better. The score 30 is an outlier that pulls the mean below five of the six scores.
Problem 9: Basketball consistency
(a) Maya: \(\frac{64}{4} = 16\). Diego: \(\frac{8 + 12 + 20 + 24}{4} = \frac{64}{4} = 16\). Both means are 16.
(b) Maya’s distances: 2, 0, 0, 2, so \(MAD = \frac{4}{4} = 1\). Diego’s distances: 8, 4, 4, 8, so \(MAD = \frac{24}{4} = 6\).
(c) Maya is more consistent because her MAD of 1 is much smaller than Diego’s MAD of 6.
Problem 10: Allowance survey
(a) Total: \(8 + 10 + 10 + 12 + 12 + 14 + 46 = 112\). \(112 : 7 = 16\). The mean is \(\$16\).
(b) The 4th ordered value is \(\$12\).
(c) The median is better, since \(\$46\) is an outlier and the mean is higher than six of the seven values.
(d) Without \(\$46\) the total is 66. \(66 : 6 = 11\). The mean is \(\$11\).
See every chapter on the Grade 6 math page.