The Lumenary · Grades 6–8

Statistics: centre, spread and distributions

Name: ______________________

Date: ______________________

You will learn to

Distinguish statistical questions, which anticipate variability, from questions with a single answer
Compute the mean, median, mode, range and interquartile range, and choose a suitable summary
Draw and compare dot plots, histograms and box plots of two data sets, looking at both centre and spread
Prove that the deviations from the mean sum to zero, that the median resists outliers while the mean does not, and that the mean minimises the total squared distance

1.
Which of these are statistical questions? (Choose all that apply.)
- A.How many hours do students in my school sleep on a school night?
- B.How many hours did I sleep last night?
- C.How much rain fell in Dhaka on each day of July last year?
- D.How tall is the tallest building in Nairobi?
2.
Six friends picked these numbers of mangoes: $4, 9, 6, 3, 8, 6$ . What is the mean number of mangoes per friend?
3.
Find the median of $12, 7, 3, 20, 9, 15$ .
- A. $10.5$
- B. $3$ and $20$
- C. $11$
- D. $9$
4.
Using this lesson's rule for quartiles, find the interquartile range of $3, 5, 7, 8, 10, 12, 13, 15, 19$ .
5.
The data $3, 4, 5, 5, 6, 7, 8$ are minutes to walk to school. The $8$ is a typing error and should have been $80$ . Which summaries change when it is corrected? (Choose all that apply.)
- A.The mean
- B.The median
- C.The range
- D.The mode
6.
Two maize varieties are grown on 20 plots each. Variety P: median yield $3.1$ tonnes per hectare, IQR $0.4$ . Variety Q: median $3.2$ , IQR $1.8$ . Which statement is the best comparison?
- A.Typical yields are similar, but Q's yields are much more variable; P is more dependable
- B.Q is better, because its median is higher
- C.P is better, because its IQR is smaller
- D.They cannot be compared, because they are different varieties
7.
In a test, a class of $10$ students has a mean score of $60$ and a class of $30$ students has a mean score of $80$ . What is the mean score of all $40$ students?
8.
In the proof of Theorem 6 we expand $\sum (x_i - c)^2 = \sum (x_i - \bar x)^2 + 2(\bar x - c)\sum (x_i - \bar x) + n(\bar x - c)^2$ . Why does the middle term vanish?
- A.Because $\sum (x_i - \bar x) = 0$ : the mean is the balance point (Theorem 1)
- B.Because $\bar x - c = 0$
- C.Because squares are never negative
- D.Because the median minimises the total distance
9.
In Lean, median_resistant says: 3 ≤ l.length → medianSorted (l.set (l.length - 1) v) = medianSorted l. Why is the hypothesis 3 ≤ l.length needed?
- A.For a list of 2 values, the median uses the last position, so changing it changes the median
- B.Lean cannot compute medians of short lists
- C.Because a median needs an odd number of values
- D.It is not needed; it just makes the proof shorter
10.
Write five_number_summary(data) returning a tuple (minimum, Q1, median, Q3, maximum) using this lesson's rule: sort the data; the median is the middle value (or the mean of the two middle values); $Q_1$ and $Q_3$ are the medians of the lower and upper halves, leaving out the median when the number of values is odd. Use a helper median(values).
Computational — Practice

Answer key — Statistics: centre, spread and distributions

1.
The questions about sleep across a school and rainfall over many days anticipate variability, so they are statistical. The other two have a single answer.
2.
Total $= 4 + 9 + 6 + 3 + 8 + 6 = 36$ , so the mean is $36 \div 6 = 6$ mangoes. (Here the mode and the median are also $6$ .)
3.
Sorted: $3, 7, 9, 12, 15, 20$ . The 3rd and 4th values are $9$ and $12$ , so the median is $\tfrac{9 + 12}{2} = 10.5$ .
4.
Median $= 10$ . Lower half $3, 5, 7, 8$ gives $Q_1 = \tfrac{5 + 7}{2} = 6$ ; upper half $12, 13, 15, 19$ gives $Q_3 = \tfrac{13 + 15}{2} = 14$ . So $\mathrm{IQR} = 14 - 6 = 8$ .
5.
The mean (from $\tfrac{38}{7} \approx 5.4$ to $\tfrac{110}{7} \approx 15.7$ ) and the range (from $5$ to $77$ ) change. The median ( $5$ ) and the mode ( $5$ ) do not. This is the example outlier_example in the Lean file.
6.
The medians are almost equal (3.1 and 3.2 t/ha), but Q's IQR is more than four times P's. So typical yields are similar, and P is much more consistent. A full comparison always reports both centre and spread.
7.
Totals: $10 \times 60 = 600$ and $30 \times 80 = 2400$ . Together $3000$ points for $40$ students: mean $= \tfrac{3000}{40} = 75$ . The mean of the means, $70$ , would be right only if the classes were the same size.
8.
$\sum (x_i - \bar x) = \sum x_i - n\bar x = 0$ (Theorem 1), so $2(\bar x - c) \sum (x_i - \bar x) = 0$ for every $c$ . In Lean, this is where mean_mul_length enters the proof of sum_sq_dev.
9.
For a list of length $2$ the median is the mean of positions $0$ and $1$ , and position $1$ is the last one. Changing it changes the median: $[1, 2] \mapsto [1, 10]$ moves the median from $1.5$ to $5.5$ . For length $n \ge 3$ the middle positions $n/2$ (and $n/2 - 1$ ) are before $n - 1$ , which is what the proof uses (omega checks it).

10.

def median(values):
    n = len(values)
    if n % 2 == 1:
        return values[n // 2]
    return (values[n // 2 - 1] + values[n // 2]) / 2

def five_number_summary(data):
    s = sorted(data)
    n = len(s)
    lower = s[:n // 2]
    upper = s[(n + 1) // 2:]
    return (s[0], median(lower), median(s), median(upper), s[-1])

The slices implement the convention exactly: for $n = 9$ , s[:4] and s[5:] leave out the median s[4]; for $n = 6$ , s[:3] and s[3:] split the data in two. (Python compares 6.0 == 6 as equal, so the tests accept either.)