S2 Mathematics Practice Book · Unit 10: Statistics with grouped data

Unit 10 Statistics with grouped data

Key unit competence: Utilize grouped data and related measures of central tendency in drawing simple conclusions.

Key ideas

Large data sets are grouped into classes (class intervals) such as 50–59. The lower and upper class limits are 50 and 59; the class boundaries are 49.5 and 59.5.
Class width $=$ upper boundary $-$ lower boundary (10 for 50–59). Midpoint (class mark) $= \frac{lower limit + upper limit}{2}$ . Range = highest value − lowest value.
A frequency distribution table lists classes with their frequencies; the cumulative frequency column gives running totals.
Mean of grouped data: $\overset{x}{ˉ} = \frac{\sum f x}{\sum f}$ where $x$ is the class midpoint (an estimate).
Modal class: the class with the highest frequency. Mode $= L + \frac{f _{1} - f _{0}}{2 f _{1} - f _{0} - f _{2}} \times h$ ( $L$ : lower boundary of the modal class; $f_{1}$ its frequency; $f_{0}$ , $f_{2}$ frequencies of the classes before and after; $h$ class width).
Median $= L + \frac{\frac{N}{2} - c f}{f _{m}} \times h$ ( $L$ : lower boundary of the median class; $N$ total frequency; $c f$ cumulative frequency before the median class; $f_{m}$ its frequency).
Graphs: histogram (bars touching, on class boundaries), frequency polygon (joining midpoints of bar tops), pie chart.

Formulas to know

\overset{x}{ˉ} = \frac{\sum f x}{\sum f}

Mode = L + \frac{f _{1} - f _{0}}{2 f _{1} - f _{0} - f _{2}} \times h

Median = L + \frac{\frac{N}{2} - c f}{f _{m}} \times h

Common misconceptions

"The mean of grouped data uses the class limits." No: use the class midpoints.
"The modal class is the class with the largest midpoint." No: it is the class with the largest frequency.
"Histogram bars have gaps between them." No: the bars touch because the data are continuous.

Worked examples

Example 1. Marks: 10–19 (f=5), 20–29 (f=12), 30–39 (f=8). Estimate the mean.

Solution

Midpoints 14.5, 24.5, 34.5. $\sum f x = 72.5 + 294 + 276 = 642.5$ , $\sum f = 25$ ; $\overset{x}{ˉ} = 25.7$ .

Example 2. For the same data, estimate the mode.

Solution

Modal class 20–29: $L = 19.5$ , $f_{1} = 12$ , $f_{0} = 5$ , $f_{2} = 8$ , $h = 10$ . Mode $= 19.5 + \frac{7}{11} \times 10 \approx 25.9$ .

Practice questions

Easier and harder questions are mixed, just like in a real exam.

Group these ages into classes 10–14, 15–19, 20–24: 12, 15, 18, 21, 11, 16, 19, 22, 14, 17, 20, 13. Make a frequency table with cumulative frequencies.
Show answer
10–14: 4 (cf 4); 15–19: 5 (cf 9); 20–24: 3 (cf 12).
Common mistake: Putting 14.5-type values in two classes or missing a value.
What is the difference between class limits and class boundaries? Give an example.
Show answer
Limits are the smallest and largest values written in the class (20–29); boundaries close the gaps between classes (19.5–29.5).
Common mistake: Using limits for the width, giving 9 instead of 10.
Using the table above, estimate the mean age.
Show answer
Midpoints 12, 17, 22: $\frac{48 + 85 + 66}{12} = \frac{199}{12} \approx 16.6$ years.
Common mistake: Using class limits instead of midpoints.
Draw a pie chart (give the angles) for classes A: 10, B: 25, C: 15, D: 10.
Show answer
Total 60: A $6 0^{\circ}$ , B $15 0^{\circ}$ , C $9 0^{\circ}$ , D $6 0^{\circ}$ .
Common mistake: Angles not adding up to $36 0^{\circ}$ .
Marks: 30–39 (3), 40–49 (8), 50–59 (12), 60–69 (5), 70–79 (2). Find the modal class and estimate the mode.
Show answer
Modal class 50–59: $49.5 + \frac{12 - 8}{24 - 8 - 5} \times 10 = 49.5 + \frac{4}{11} \times 10 \approx 53.1$ .
Common mistake: Using the class limit 50 for $L$ instead of the boundary 49.5.
Describe how to draw a histogram and a frequency polygon for the marks table.
Show answer
Mark class boundaries 29.5, 39.5, ..., 79.5 on the horizontal axis and frequency on the vertical axis; draw touching bars of heights 3, 8, 12, 5, 2. Join the midpoints of the bar tops (34.5, 3), (44.5, 8), ... to get the frequency polygon.
Common mistake: Leaving gaps between bars.
For the same marks, estimate the median.
Show answer
$N = 30$ , $\frac{N}{2} = 15$ ; cumulative 3, 11, 23: median class 50–59. $49.5 + \frac{15 - 11}{12} \times 10 \approx 52.8$ .
Common mistake: Using $c f$ of the median class itself (23) instead of the class before (11).
For the same marks, estimate the mean.
Show answer
Midpoints 34.5, 44.5, 54.5, 64.5, 74.5: $\sum f x = 103.5 + 356 + 654 + 322.5 + 149 = 1585$ ; $\overset{x}{ˉ} = \frac{1585}{30} \approx 52.8$ .
Common mistake: Dividing by the number of classes (5) instead of 30.

Matching and drag-and-drop

Matching

Match each item in column A with the correct answer in column B. Write the letter next to each number.

Column A	Column B
1. Classes 0–9, 10–19, 20–29, 30–39 have frequencies 6, 10, 14, 10. Estimate the median (2 d.p.).	a. 44.5
2. What is the midpoint of the class 40–49?	b. 22.36
3. The highest value is 87 and the lowest 23. Find the range.	c. 64
4. A class of width 5 has frequency 20. Find its frequency density.	d. 4

Show answer

1 → b (22.36) · 2 → a (44.5) · 3 → c (64) · 4 → d (4)

Step 1. $\frac{N}{2} = 20$ ; cumulative frequencies 6, 16, 30, so the median class is 20–29 ( $L = 19.5$ , $F = 16$ , $f = 14$ ).
Step 2. Median $= 19.5 + \frac{20 - 16}{14} \times 10 \approx 22.36$ .
$\frac{40 + 49}{2} = 44.5$ .
$87 - 23 = 64$ .
Step 1. Frequency density $= \frac{frequency}{class width}$ .
Step 2. $\frac{20}{5} = 4$ .

Drag and drop
Choose the correct word or phrase from the box for each gap. One word in the box is not needed.
an estimate · 20 · $1 5^{\circ}$ · $9 0^{\circ}$
1. The mean of grouped data is .
2. In a pie chart of 60 values, a class with frequency 15 has angle .
3. For $N = 40$ , the median is about the value at position .
Show answer
1: an estimate · 2: $9 0^{\circ}$ · 3: 20
- We assume all values in a class are at the midpoint.
- $\frac{15}{60} \times 36 0^{\circ}$ .
- For grouped data use $\frac{N}{2} = 20$ .

Matching

Match each item in column A with the correct answer in column B. Write the letter next to each number.

Column A	Column B
1. Ages 10–14 and 15–19: what is the midpoint of 15–19?	a. 55 kg
2. 4 parcels weigh 50–54 kg and 6 parcels weigh 55–59 kg. Estimate the mean mass.	b. 17
3. In the mode formula, $f_{0}$ is:	c. the frequency of the class before the modal class
4. What are the class boundaries of 40–49?	d. 39.5 and 49.5

Show answer

1 → b (17) · 2 → a (55 kg) · 3 → c (the frequency of the class before the modal class) · 4 → d (39.5 and 49.5)

$\frac{15 + 19}{2} = 17$ .
Step 1. Midpoints 52 and 57: $\sum f x = 4 \times 52 + 6 \times 57 = 208 + 342 = 550$ .
Step 2. Mean $= \frac{550}{10} = 55$ kg.
$f_{0}$ is the frequency just before, $f_{2}$ just after.
Subtract 0.5 from the lower limit and add 0.5 to the upper limit.

Drag and drop
Choose the correct word or phrase from the box for each gap. One word in the box is not needed.
20 and 25 · 4, 11, 20, 25 · 19.5 and 24.5 · midpoint
1. Frequencies 4, 7, 9, 5 give cumulative frequencies .
2. A class 20–24 has boundaries .
3. To estimate the mean of grouped data, each class is represented by its .
Show answer
1: 4, 11, 20, 25 · 2: 19.5 and 24.5 · 3: midpoint
- Add each frequency to the previous total.
- Half a unit beyond each limit.
- The midpoint is the typical value of the class.

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