Chapter 13 ยท Statistics

13.1 Mean of Grouped Data

When a hundred marks are squeezed into a table of class intervals, the individual values are lost. The mean of grouped data finds the average anyway: replace each class by its class mark, weight it by its frequency, and divide by the total frequency.

A school records the marks of 400 students in a unit test. Nobody wants to read 400 numbers, so the marks are sorted into classes such as $0$โ€“$10$, $10$โ€“$20$, $20$โ€“$30$, and the table only says how many students fall in each class. This is grouped data. It is easy to read, but it hides the exact marks. So how can we still find the average mark?

In Class 9 you found the mean of a list as (sum of values) รท (number of values). Here we do not have the values, so we make one sensible assumption: every observation in a class is equal to the middle value of that class. That middle value is the class mark. The result is not exact, but it is very close, and it is the method every statistics report uses. In this section you learn the direct method, a shortcut called the assumed-mean method, and a still faster version, the step-deviation method โ€” all three are in your syllabus.

Class marks and the direct method

๐Ÿ“˜ Definition 13.1 โ€” Class mark and class size
In a grouped frequency distribution, the class mark of a class is the midpoint of its two ends: $$\text{class mark }x_i=\frac{\text{upper limit}+\text{lower limit}}{2}.$$ The class size $h$ is the width of a class: upper limit $-$ lower limit. The number of observations in the class is its frequency $f_i$. (All the classes in this chapter have the same size.)

For the class $20$โ€“$30$, the class mark is $x_i=\dfrac{30+20}{2}=25$ and the class size is $h=10$. If $7$ students scored in this class, we treat them as seven students who all scored $25$.

๐Ÿ“˜ Definition 13.2 โ€” Mean of grouped data
For classes with class marks $x_1,x_2,\dots,x_k$ and frequencies $f_1,f_2,\dots,f_k$, the mean of the grouped data is $$\bar x=\frac{f_1x_1+f_2x_2+\dots+f_kx_k}{f_1+f_2+\dots+f_k}=\frac{\sum f_ix_i}{\sum f_i}.$$ Here $\sum f_ix_i$ is the sum of the products $f_ix_i$ and $\sum f_i$ is the total frequency.
๐Ÿ“ Theorem 13.1 โ€” Direct method
To find the mean of grouped data: (1) write the class mark $x_i$ of every class; (2) multiply each by its frequency to get $f_ix_i$; (3) add the column of $f_i$ and the column of $f_ix_i$; (4) divide: $$\bar x=\frac{\sum f_ix_i}{\sum f_i}.$$

Why does this work? If a class mark $x_i$ occurs $f_i$ times, then adding all the observations means adding $x_i$ a total of $f_i$ times, which is $f_ix_i$. The total of all observations is $\sum f_ix_i$ and there are $\sum f_i$ of them, so the mean is the ratio. It is the same rule as "sum รท count" for a list, written for repeated values.

โš ๏ธ Common mistakes

1. Dividing by the number of classes. Divide by $\sum f_i$ (the number of observations), never by the number of rows.

2. Using the class limit instead of the class mark. Multiply the frequency by the midpoint $x_i$, not by the lower or upper limit.

3. Averaging the class marks. $\dfrac{5+15+25+35+45}{5}$ ignores the frequencies. A class with more students must count for more.

4. Calling the answer exact. The mean of grouped data is an estimate; the true mean of the raw marks can differ a little, as Example 6 shows.

โœ๏ธ Example 1 โ€” The direct method
The marks (out of $50$) of $30$ students in a test are grouped as follows. Find the mean mark.
Marks$0$โ€“$10$$10$โ€“$20$$20$โ€“$30$$30$โ€“$40$$40$โ€“$50$
Students $f_i$35886
Class marks $x_i=5,\ 15,\ 25,\ 35,\ 45$ (midpoints of the five classes).
Products $f_ix_i$: $3\times5=15$, $5\times15=75$, $8\times25=200$, $8\times35=280$, $6\times45=270$.
Totals $\sum f_i=3+5+8+8+6=30$ and $\sum f_ix_i=15+75+200+280+270=840$.
Mean $\bar x=\dfrac{840}{30}=28$. The mean mark is $28$ out of $50$.
Check $28$ lies inside the class $20$โ€“$30$, where the frequencies are the largest, so it is reasonable โœ“.
Interactive 1 โ€” The Mean as a Balance Point

Each bar is a class and its height is the frequency $f_i$. Drag the five frequency sliders (or press Play, which sweeps through some distributions). The green triangle shows where the histogram would balance on a seesaw: exactly at the mean $\bar x=\sum f_ix_i\div\sum f_i$. Pile more students into the high classes and the balance point moves right.

$\sum f_i$โ€”
$\sum f_ix_i$โ€”
mean $\bar x$โ€”
tallest classโ€”

The assumed-mean method

When the numbers $x_i$ are large (wages in hundreds of rupees, say), the products $f_ix_i$ become big and calculation slips are likely. The cure is to measure every class mark from a convenient assumed mean $a$, usually a class mark near the middle of the table. The deviation of a class mark from $a$ is $d_i=x_i-a$. The deviations are small numbers, often $0$ or negative, so the products $f_id_i$ are easy.

๐Ÿ“ Theorem 13.2 โ€” Assumed-mean method
Let $a$ be any number (the assumed mean) and $d_i=x_i-a$. Then $$\bar x=a+\frac{\sum f_id_i}{\sum f_i}.$$

Proof. Write $x_i=a+d_i$ in the direct method: $$\bar x=\frac{\sum f_i(a+d_i)}{\sum f_i}=\frac{a\sum f_i+\sum f_id_i}{\sum f_i}=a+\frac{\sum f_id_i}{\sum f_i}.\qquad\blacksquare$$ Notice that the proof works for any $a$. A good choice only makes the arithmetic shorter: if $a$ is not near the true mean, $\sum f_id_i$ is simply large, and the correction $\dfrac{\sum f_id_i}{\sum f_i}$ makes up for it.

โœ๏ธ Example 2 โ€” Assumed mean
The daily wages (in โ‚น) of $50$ workers in a workshop are shown below. Find the mean daily wage using the assumed-mean method with $a=650$.
Wage (โ‚น)$400$โ€“$500$$500$โ€“$600$$600$โ€“$700$$700$โ€“$800$$800$โ€“$900$
Workers $f_i$61014128
Class marks $x_i=450,\ 550,\ 650,\ 750,\ 850$.
Deviations $d_i=x_i-650=-200,\ -100,\ 0,\ 100,\ 200$.
Products $f_id_i$: $6(-200)=-1200$, $10(-100)=-1000$, $14(0)=0$, $12(100)=1200$, $8(200)=1600$, so $\sum f_id_i=600$. Also $\sum f_i=50$.
Mean $\bar x=650+\dfrac{600}{50}=650+12=662$. The mean daily wage is โ‚น 662.
Check The direct method gives $\sum f_ix_i=33100$ and $\dfrac{33100}{50}=662$ โœ“.
โœ๏ธ Example 3 โ€” A missing frequency
The mean of the following distribution is $49$. Find the missing frequency $p$.
Class$0$โ€“$20$$20$โ€“$40$$40$โ€“$60$$60$โ€“$80$$80$โ€“$100$
Frequency68$p$104
Table $x_i=10,30,50,70,90$. Then $\sum f_i=6+8+p+10+4=28+p$ and $\sum f_ix_i=60+240+50p+700+360$, that is, $\sum f_ix_i=1360+50p$.
Equation $\bar x=49$ means $\dfrac{1360+50p}{28+p}=49$, so $1360+50p=49(28+p)=1372+49p$.
Solve $50p-49p=1372-1360$, so $p=12$.
Check With $p=12$: $\sum f_i=40$, $\sum f_ix_i=1960$ and $\dfrac{1960}{40}=49$ โœ“.

The step-deviation method

If every deviation $d_i$ is a multiple of the class size $h$ (as happens when all classes have the same size), we can shrink the numbers even more by dividing the deviations by $h$. Put $u_i=\dfrac{x_i-a}{h}$, which are small whole numbers $\dots,-2,-1,0,1,2,\dots$.

๐Ÿ“ Theorem 13.3 โ€” Step-deviation method
If all classes have size $h$ and $u_i=\dfrac{x_i-a}{h}$, then $$\bar x=a+h\times\frac{\sum f_iu_i}{\sum f_i}.$$

Proof. $d_i=x_i-a=hu_i$, so $\sum f_id_i=h\sum f_iu_i$. Substituting into Theorem 13.2 gives the formula. $\blacksquare$

๐Ÿงญ Problem-solving strategy โ€” choosing a method
  1. Small class marks (like $5,15,25$): use the direct method.
  2. Large or awkward class marks: pick $a$ as a class mark near the middle, and use the assumed-mean method.
  3. Equal class sizes $h$ and large marks: the step-deviation method is the shortest. Remember to multiply by $h$ at the end โ€” forgetting it is the usual slip.
  4. All three give the same mean. If you have time, confirm with a second method.
โœ๏ธ Example 4 โ€” Step-deviation
The waiting times (in minutes) of $60$ customers at a railway ticket counter are grouped as below. Find the mean waiting time using $a=12.5$ and $h=5$. Give the answer to $2$ decimal places.
Time (min)$0$โ€“$5$$5$โ€“$10$$10$โ€“$15$$15$โ€“$20$$20$โ€“$25$$25$โ€“$30$
Customers410181495
Class marks $x_i=2.5,\ 7.5,\ 12.5,\ 17.5,\ 22.5,\ 27.5$.
Steps $u_i=\dfrac{x_i-12.5}{5}=-2,\ -1,\ 0,\ 1,\ 2,\ 3$.
Products $f_iu_i=-8,\ -10,\ 0,\ 14,\ 18,\ 15$, so $\sum f_iu_i=29$. Also $\sum f_i=60$.
Mean $\bar x=12.5+5\times\dfrac{29}{60}=12.5+2.4167=14.92$ minutes (to $2$ d.p.).
โœ๏ธ Example 5 โ€” Classes written without overlap
A batter's scores in $25$ innings are grouped as $0$โ€“$9$, $10$โ€“$19$, $20$โ€“$29$, $30$โ€“$39$, $40$โ€“$49$ with frequencies $4,\ 6,\ 9,\ 5,\ 1$. Find the mean score.
Class marks The class marks use the limits as written: $\dfrac{0+9}{2}=4.5$, $\dfrac{10+19}{2}=14.5$, and so on: $x_i=4.5,\ 14.5,\ 24.5,\ 34.5,\ 44.5$. (These are the same as for the touching classes $0$โ€“$10,\dots$ shifted by $0.5$, the middle of the gap.)
Assumed mean Take $a=24.5$, so $d_i=-20,\ -10,\ 0,\ 10,\ 20$ and $f_id_i=-80,\ -60,\ 0,\ 50,\ 20$. Thus $\sum f_id_i=-70$.
Mean $\bar x=24.5+\dfrac{-70}{25}=24.5-2.8=21.7$ runs.
โœ๏ธ Example 6 โ€” How good is the estimate?
Twenty students scored: $12, 18, 25, 7, 33, 41, 15, 22, 29, 36, 9, 48, 27, 31, 19, 44, 23, 38, 14, 26$. Find the exact mean, then group the marks into classes $0$โ€“$10$, $10$โ€“$20$, $\dots$, $40$โ€“$50$ (a mark on a boundary goes in the higher class; none of these is on a boundary) and find the mean of the grouped data. Compare.
Exact mean The sum of the $20$ marks is $517$, so the exact mean is $\dfrac{517}{20}=25.85$.
Frequencies Counting: $0$โ€“$10$: $7,9$ (2 marks); $10$โ€“$20$: $12,18,15,19,14$ (5); $20$โ€“$30$: $25,22,29,27,23,26$ (6); $30$โ€“$40$: $33,36,31,38$ (4); $40$โ€“$50$: $41,48,44$ (3). Total $20$ โœ“.
Grouped mean $x_i=5,15,25,35,45$ and $\sum f_ix_i=10+75+150+140+135=510$, so $\bar x=\dfrac{510}{20}=25.5$.
Compare $25.85$ and $25.5$ differ by only $0.35$. The assumption "everyone in a class scored the class mark" is wrong for each student, but the errors above and below the midpoint largely cancel.
Interactive 2 โ€” Any Assumed Mean Gives the Same Answer

This is the wage data of Example 2. Slide the assumed mean $a$ anywhere you like (or press Play). The red bars are classes left of $a$ (negative deviations $d_i$), the blue bars are classes right of $a$. The sum $\sum f_id_i$ and the correction $\sum f_id_i\div\sum f_i$ change with $a$, yet $a+\text{correction}$ always lands on the same mean, โ‚น 662.

650
$a$โ€”
$\sum f_id_i$โ€”
correctionโ€”
$a+$ correctionโ€”

Exercises

I
Class marks and warm-up.
  • 1 Write the class mark and the class size of the class $15$โ€“$25$.
  • 2 Find the class marks of $40$โ€“$60$ and of $100$โ€“$150$.
  • 3 Classes are written $11$โ€“$20$, $21$โ€“$30$, $31$โ€“$40$. Find the class mark of each.
  • 4 The class marks $10,20,30$ have frequencies $2,3,5$. Find the mean.
  • 5 For a distribution, $\sum f_ix_i=1260$ and $\sum f_i=45$. Find $\bar x$.
  • 6 In the assumed-mean method with $a=40$, $\sum f_id_i=-120$ and $\sum f_i=30$. Find the mean.
1 Class mark $=\dfrac{15+25}{2}=20$; class size $=25-15=10$.
2 $\dfrac{40+60}{2}=50$ and $\dfrac{100+150}{2}=125$.
3 $\dfrac{11+20}{2}=15.5$, $\ 25.5$ and $35.5$.
4 $\sum f_ix_i=20+60+150=230$, $\sum f_i=10$, so $\bar x=23$.
5 $\bar x=\dfrac{1260}{45}=28$.
6 $\bar x=40+\dfrac{-120}{30}=40-4=36$.
II
Direct method. Find the mean of each distribution.
  • 1 Classes $0$โ€“$10,\ 10$โ€“$20,\ 20$โ€“$30,\ 30$โ€“$40,\ 40$โ€“$50$ with frequencies $3,7,12,5,3$. Give the answer to $2$ d.p.
  • 2 Classes $10$โ€“$20,\dots,50$โ€“$60$ with frequencies $4,6,10,8,2$. Give the answer to $2$ d.p.
  • 3 Heights (cm) of $30$ students: $140$โ€“$150$: $5$; $150$โ€“$160$: $12$; $160$โ€“$170$: $10$; $170$โ€“$180$: $3$. Give the answer to $2$ d.p.
  • 4 Weekly pocket money (โ‚น) of $60$ students: $0$โ€“$50$: $8$; $50$โ€“$100$: $14$; $100$โ€“$150$: $20$; $150$โ€“$200$: $12$; $200$โ€“$250$: $6$.
  • 5 Pages read in a week by $30$ students: $1$โ€“$10$: $6$; $11$โ€“$20$: $9$; $21$โ€“$30$: $10$; $31$โ€“$40$: $5$. Give the answer to $2$ d.p.
  • 6 Daily rainfall (mm) over $40$ days: $0$โ€“$4$: $18$; $4$โ€“$8$: $10$; $8$โ€“$12$: $7$; $12$โ€“$16$: $5$.
1 $x_i=5,15,25,35,45$; $\sum f_ix_i=15+105+300+175+135=730$; $\sum f_i=30$. $\bar x=\dfrac{730}{30}=24.33$.
2 $x_i=15,25,35,45,55$; $\sum f_ix_i=60+150+350+360+110=1030$; $\sum f_i=30$. $\bar x=34.33$.
3 $x_i=145,155,165,175$; $\sum f_ix_i=725+1860+1650+525=4760$; $\bar x=\dfrac{4760}{30}=158.67$ cm.
4 $x_i=25,75,125,175,225$; $\sum f_ix_i=200+1050+2500+2100+1350=7200$; $\bar x=\dfrac{7200}{60}=$ โ‚น 120.
5 $x_i=5.5,15.5,25.5,35.5$; $\sum f_ix_i=33+139.5+255+177.5=605$; $\bar x=\dfrac{605}{30}=20.17$ pages.
6 $x_i=2,6,10,14$; $\sum f_ix_i=36+60+70+70=236$; $\bar x=\dfrac{236}{40}=5.9$ mm.
III
Assumed-mean and step-deviation methods.
  • 1 Classes $0$โ€“$20,\ 20$โ€“$40,\ 40$โ€“$60,\ 60$โ€“$80,\ 80$โ€“$100$ have frequencies $6,9,15,12,8$. Find the mean with the assumed-mean method, taking $a=50$.
  • 2 For the same data find the mean by the step-deviation method with $a=50$ and $h=20$.
  • 3 For the same data take $a=30$. Show that you get the same mean.
  • 4 The mean of the distribution (classes $0$โ€“$10,\dots,40$โ€“$50$) with frequencies $4,6,p,8,2$ is $24.5$. Find $p$.
  • 5 Classes $0$โ€“$20,\dots,80$โ€“$100$ have frequencies $7,p,12,q,8$, total frequency $50$ and mean $51.2$. Find $p$ and $q$.
  • 6 A cricket coach records the runs scored in 50 matches. Classes $20$โ€“$30,30$โ€“$40,40$โ€“$50,50$โ€“$60,60$โ€“$70$ have frequencies $5,8,15,16,6$. Find the mean using $a=45$.
1 $x_i=10,30,50,70,90$; $d_i=-40,-20,0,20,40$; $f_id_i=-240,-180,0,240,320$, sum $140$; $\sum f_i=50$. $\bar x=50+\dfrac{140}{50}=52.8$.
2 $u_i=-2,-1,0,1,2$; $f_iu_i=-12,-9,0,12,16$, sum $7$. $\bar x=50+20\times\dfrac{7}{50}=52.8$.
3 $d_i=-20,0,20,40,60$; $f_id_i=-120,0,300,480,480$, sum $1140$. $\bar x=30+\dfrac{1140}{50}=30+22.8=52.8$ โœ“ the same.
4 $x_i=5,15,25,35,45$. $\sum f_i=20+p$, $\sum f_ix_i=20+90+25p+280+90=480+25p$. Then $480+25p=24.5(20+p)=490+24.5p$, so $0.5p=10$ and $p=20$.
5 $p+q=50-27=23$. $\sum f_ix_i=70+30p+600+70q+720=1390+30p+70q=51.2\times50=2560$, so $30p+70q=1170$. From $p=23-q$: $690+40q=1170$, so $q=12$ and $p=11$.
6 $x_i=25,35,45,55,65$; $d_i=-20,-10,0,10,20$; $f_id_i=-100,-80,0,160,120$, sum $100$. $\bar x=45+\dfrac{100}{50}=47$ runs.
IV
Think harder.
  • 1 A distribution has mean $\bar x$. Each class mark is increased by $10$ (frequencies unchanged). Show that the new mean is $\bar x+10$.
  • 2 Each class mark is doubled. Show that the new mean is $2\bar x$. Use this to find the mean of a distribution whose class marks are $20,60,100,140,180$ with frequencies $6,9,15,12,8$, given that the class marks $10,30,50,70,90$ with the same frequencies have mean $52.8$.
  • 3 A student claims: "The mean of $0$โ€“$10$ (frequency $4$) and $90$โ€“$100$ (frequency $4$) is $50$, so the mean is the average of $0$ and $100$." Find the actual grouped mean and explain the error.
  • 4 A shopkeeper's daily sales (โ‚น hundreds) over $40$ days: $10$โ€“$20$: $4$; $20$โ€“$30$: $8$; $30$โ€“$40$: $14$; $40$โ€“$50$: $10$; $50$โ€“$60$: $4$. Find the mean and write the total sales in โ‚น.
1 New mean $=\dfrac{\sum f_i(x_i+10)}{\sum f_i}=\dfrac{\sum f_ix_i+10\sum f_i}{\sum f_i}=\bar x+10$.
2 $\dfrac{\sum f_i(2x_i)}{\sum f_i}=2\bar x$. The new class marks $20,60,100,140,180$ are twice $10,30,50,70,90$, so the new mean is $2\times52.8=105.6$.
3 The class marks are $5$ and $95$ with equal frequencies $4$: $\bar x=\dfrac{4\times5+4\times95}{8}=50$. The student's answer $50$ is correct here only by luck of symmetry; the method "average of the end values $0$ and $100$" is wrong in general because it uses class limits, not class marks (and ignores frequencies).
4 $x_i=15,25,35,45,55$; $\sum f_ix_i=60+200+490+450+220=1420$; $\bar x=\dfrac{1420}{40}=35.5$ hundred rupees $=$ โ‚น 3,550 per day. Total $\approx40\times3550=$ โ‚น 1,42,000.
๐ŸŽฏ Exam corner Board ยท 1 markBoard ยท 2 marksBoard ยท 3 marksBoard ยท Case studyJEE Main

How this is examined: Boards ask for the mean of a table in 3 marks (direct, assumed-mean or step-deviation โ€” write the formula, the table and the substitution), a "find the missing frequency" item in 2โ€“3 marks, and 1-mark MCQs on class marks and formulas. JEE Main uses the effect of changing every observation, $\overline{ax+b}=a\bar x+b$.

E1. Board ยท 1 mark For a grouped distribution, $\sum f_ix_i=1260$ and $\sum f_i=42$. The mean is  (A) $20$  (B) $30$  (C) $252$  (D) $1218$

Formula $\bar x=\dfrac{\sum f_ix_i}{\sum f_i}=\dfrac{1260}{42}=30$.
Answer (B). Option (C) divides by $5$, the number of classes, instead of by $\sum f_i$; option (D) subtracts instead of dividing.

E2. Board ยท 1 mark Assertion (A): The value of the mean found by the assumed-mean method depends on the choice of the assumed mean $a$. Reason (R): $\bar x=a+\dfrac{\sum f_id_i}{\sum f_i}$, where $d_i=x_i-a$.
(A) Both A and R are true and R is the correct explanation of A.  (B) Both A and R are true but R is not the correct explanation of A.  (C) A is true but R is false.  (D) A is false but R is true.

Check R This is Theorem 13.2: true.
Check A Changing $a$ changes $\sum f_id_i$ by exactly the amount needed to keep $\bar x$ the same (see Interactive 2). So A is false.
Answer (D).

E3. Board ยท 3 marks A shop's daily sales (in โ‚น hundreds) over $50$ days are: $20$โ€“$30$: $5$; $30$โ€“$40$: $8$; $40$โ€“$50$: $15$; $50$โ€“$60$: $16$; $60$โ€“$70$: $6$. Find the mean daily sales by the assumed-mean method.

Table (1 mark) Take $a=45$. $x_i=25,35,45,55,65$; $d_i=-20,-10,0,10,20$; $f_id_i=-100,-80,0,160,120$.
Sums (1 mark) $\sum f_i=50$, $\sum f_id_i=100$.
Mean (1 mark) $\bar x=a+\dfrac{\sum f_id_i}{\sum f_i}=45+\dfrac{100}{50}=47$. Mean daily sales $=$ โ‚น 4,700.

E4. Board ยท 2 marks Classes $10$โ€“$20,\ 20$โ€“$30,\ 30$โ€“$40,\ 40$โ€“$50,\ 50$โ€“$60$ have frequencies $5,\ 8,\ p,\ 7,\ 5$. If the mean is $34.8$, find $p$.

Set up (1 mark) $x_i=15,25,35,45,55$. $\sum f_i=25+p$, $\sum f_ix_i=75+200+35p+315+275=865+35p$.
Solve (1 mark) $865+35p=34.8(25+p)=870+34.8p$, so $0.2p=5$ and $p=25$. (Check: $\dfrac{1740}{50}=34.8$ โœ“.)

E5. Board ยท Case study A survey of $40$ Class 10 students recorded the time (in minutes) they spent on a phone on one day: $0$โ€“$30$: $4$; $30$โ€“$60$: $10$; $60$โ€“$90$: $14$; $90$โ€“$120$: $8$; $120$โ€“$150$: $4$. (i) Write the class marks and find $\sum f_ix_i$. (2) (ii) Find the mean time. (1) (iii) How many students spent more than $90$ minutes, and what percentage of the class is that? (1)

(i) (2 marks) $x_i=15,45,75,105,135$; $f_ix_i=60,450,1050,840,540$; $\sum f_ix_i=2940$.
(ii) (1 mark) $\bar x=\dfrac{2940}{40}=73.5$ minutes.
(iii) (1 mark) $8+4=12$ students, which is $\dfrac{12}{40}\times100=30\%$.

E6. JEE Main The mean of a frequency distribution (of class marks $x_i$) is $40$. Every class mark is replaced by $3x_i-5$ (frequencies unchanged). The new mean is  (A) $35$  (B) $115$  (C) $120$  (D) $125$

Linear change $\dfrac{\sum f_i(3x_i-5)}{\sum f_i}=3\bar x-5$ (multiplication scales the mean; subtraction shifts it, as in Exercise IV).
Answer $3\times40-5=115$: (B). Option (C) forgets the $-5$; option (A) applies $-5$ but forgets to multiply by $3$ ($40-5=35$).
๐Ÿ”ฌ Researchโ€“Teaching NexusMini project ยท optional, about 40 minutes

Research question. How close is the mean of grouped data to the true mean of the raw values, and how does the choice of class size change the gap?

Why it matters. Census reports, weather summaries and sports statistics often publish only grouped tables. Analysts must know how much accuracy the grouping costs before they trust a figure computed from it.

What to do.

  1. Ask 30 classmates (or use a data source such as daily temperatures for a month) for one number, for example minutes spent travelling to school each day.
  2. Find the exact mean from the raw numbers.
  3. Group the data with class size $h=10$, then again with $h=5$, and find the grouped mean each time (direct method, in a spreadsheet).
  4. Compare: which class size gives the estimate closer to the exact mean?
  5. Try one very wide class (for example $0$โ€“$60$ for everything) and note what happens.

Tools: A notebook, a calculator or a spreadsheet, and a short survey

Hand in: Your raw data, the two frequency tables, the three means, and a paragraph explaining which grouping you would trust and why.

Going further. Change the assumed mean $a$ three times in the assumed-mean method. Confirm that the final mean is unchanged, and explain in words why it must be.

Teacher/parent note: narrower classes make the "everyone is at the midpoint" assumption closer to the truth, so the grouped mean usually moves towards the exact mean. Results vary with the data; the aim is to see the trend, not a fixed number.

๐Ÿงช Self-check quiz

0/5 correct

Q1. The class mark of the class $35$โ€“$45$ is

Q2. If $\sum f_ix_i=860$ and $\sum f_i=40$, the mean is

Q3. In the assumed-mean method, $a=25$, $\sum f_id_i=60$ and $\sum f_i=30$. The mean is

Q4. Which statement about the assumed mean $a$ is false?

Q5. The class mark of the class $10$โ€“$19$ (written without overlap) is

โ† Chapter 13 โ€” Statistics13.2 Mode of Grouped Data โ†’