Statistics exercise for Grade 8 - Free Printable

Final Answer:
(b) 5
(c)(i) 132
(c)(ii) 4.4
(d) 7/30

──────────────────────────────────────

Explanation:
Let’s go step by step.

First, count how many times each number of books (1 to 7) appears in the list.

The full list of 30 numbers is:
5, 4, 6, 3, 2, 1, 7, 4, 5, 3,
6, 5, 4, 3, 7, 6, 2, 5, 4, 5,
5, 7, 5, 4, 3, 2, 1, 6, 3, 4

Now count:

- 1 appears: positions 6 and 27 → 2 times
- 2 appears: positions 5, 17, 26 → 3 times
- 3 appears: positions 4, 10, 14, 25, 29 → wait, let’s list all 3s:
4th, 10th, 14th, 25th, 29th — that’s 5? Let’s double-check carefully.

Better: write all numbers in order and tally:

1: ✔ (2 times)
2: ✔✔✔ (3 times)
3: look at list:
→ 4th = 3
→ 10th = 3
→ 14th = 3
→ 25th = 3
→ 29th = 3
That’s 5 times.

4:
→ 2nd = 4
→ 8th = 4
→ 12th = 4
→ 19th = 4
→ 24th = 4
→ 30th = 4
That’s 6 times.

5:
→ 1st = 5
→ 9th = 5
→ 12? no, 12 is 4
Let’s list all 5s:
1, 9, 12? no — better go one by one:

Index: 1→5
2→4
3→6
4→3
5→2
6→1
7→7
8→4
9→5
10→3
11→6
12→
13→4
14→3
15→7
16→6
17→2
18→5
19→4
20→5
21→5
22→7
23→5
24→4
25→3
26→2
27→1
28→6
29→3
30→4

Now pick all 5s: indices 1, 9, 12, 18, 20, 21, 23 → that’s 7 times.

6: indices 3, 11, 16, 28 → 4 times? Wait:
3→6
11→6
16→6
28→6
Also check: any more? No — that’s 4.

7: indices 7, 15, 22 → 3 times.

Let’s verify total:
1:2
2:3
3:5
4:6
5:7
6:4
7:3
Sum = 2+3=5; +5=10; +6=16; +7=23; +4=27; +3=30 ✔ correct.

So frequency table:

x | f
1 | 2
2 | 3
3 | 5
4 | 6
5 | 7
6 | 4
7 | 3

Tally:
1: ||
2: |||
3: |||||
4: |||| || (6)
5: |||| ||| (7)
6: ||||
7: |||

f × x:
1×2 = 2
2×3 = 6
3×5 = 15
4×6 = 24
5×7 = 35
6×4 = 24
7×3 = 21

Now (b) modal number = the one with highest frequency → 5 appears 7 times → mode = 5

(c)(i) total books = sum of f×x = 2+6+15+24+35+24+21
2+6=8; +15=23; +24=47; +35=82; +24=106; +21=132

(c)(ii) mean = total books ÷ number of students = 132 ÷ 30 = 4.4

(d) Less than 4 books means 1, 2, or 3 books.
Frequency for 1:2, 2:3, 3:5 → total = 2+3+5 = 10
Probability = 10 / 30 = 1/3? Wait — 10/30 = 1/3, but earlier I wrote 7/30. That was a mistake.

Wait! Recheck:
Less than 4 → x = 1, 2, 3
f(1)=2, f(2)=3, f(3)=5 → total = 10
Total students = 30
So probability = 10/30 = 1/3 = 0.333...

But my Final Answer said 7/30 — that’s wrong.

Let me double-check frequencies again — maybe I miscounted 3s or others.

List again clearly (30 numbers):

Row 1: 5, 4, 6, 3, 2, 1, 7, 4, 5, 3
→ 5,4,6,3,2,1,7,4,5,3
Count so far:
1:1
2:1
3:2
4:2
5:2
6:1
7:1

Row 2: 6, 5, 4, 3, 7, 6, 2, 5, 4, 5
→ 6,5,4,3,7,6,2,5,4,5
Add:
1:1
2:2
3:3
4:4
5:5
6:3
7:2

Row 3: 5, 7, 5, 4, 3, 2, 1, 6, 3, 4
→ 5,7,5,4,3,2,1,6,3,4
Add:
1:2
2:3
3:5
4:6
5:8? Wait — previous 5 was 5, now +3 more (positions: 1st, 3rd, 8th in row 3) → 5+3 = 8? But earlier I had 7.

Let’s count 5s explicitly:

From row1: positions 1,9 → two 5s
Row2: positions 2,8,10 → three more → total 5
Row3: positions 1,3,8? Row3: [5,7,5,4,3,2,1,6,3,4] → 5 at index1 and index3 → two more → total 7.

Yes, 2+3+2 = 7 fives.

Now 3s:
Row1: positions 4,10 → 2
Row2: position 4 → 1 more → 3
Row3: positions 5 and 9 → 2 more → total 5. Correct.

So f(1)=2, f(2)=3, f(3)=5 → 10 students with <4 books.

Thus probability = 10/30 = 1/3.

But the problem says “LESS THAN 4”, so 1,2,3 — yes.

So my initial Final Answer was wrong on (d). Correct is 10/30 = 1/3.

Wait — maybe the question expects fraction simplified? 1/3.

Let me recompute total f×x to confirm:

x : f : f×x
1 : 2 → 2
2 : 3 → 6
3 : 5 → 15
4 : 6 → 24
5 : 7 → 35
6 : 4 → 24
7 : 3 → 21
Sum: 2+6=8; 8+15=23; 23+24=47; 47+35=82; 82+24=106; 106+21=127? Wait — 106+21 = 127, not 132!

Mistake! 82 + 24 = 106, yes. 106 + 21 = 127.

But earlier I said 132. That’s the error.

Let me add again carefully:

2 (for 1)
+6 = 8
+15 = 23
+24 = 47
+35 = 82
+24 = 106
+21 = 127

But total students = 30, and sum of frequencies = 2+3+5+6+7+4+3 = 30 ✔️
So total books should be 127?

Let me manually sum all 30 numbers:

Row1: 5+4+6+3+2+1+7+4+5+3 =
5+4=9; +6=15; +3=18; +2=20; +1=21; +7=28; +4=32; +5=37; +3=40

Row2: 6+5+4+3+7+6+2+5+4+5 =
6+5=11; +4=15; +3=18; +7=25; +6=31; +2=33; +5=38; +4=42; +5=47

Row3: 5+7+5+4+3+2+1+6+3+4 =
5+7=12; +5=17; +4=21; +3=24; +2=26; +1=27; +6=33; +3=36; +4=40

Now total: row1=40, row2=47, row3=40 → 40+47=87; 87+40=127

Yes! Total books = 127.

So mean = 127 / 30 = 4.233... = 127/30 ≈ 4.23, but usually they want exact fraction or decimal to 1 dp? Problem says “calculate”, and marks: 2 marks — likely accept 127/30 or 4.23 or 4.2? But let’s see.

Wait — maybe I misread the original data. The image might have different numbers. Since I can’t see it, I must rely on the text provided.

The user pasted:

"5 4 6 3 2 1 7 4 5 3
6 5 4 3 7 6 2 5 4 5
5 7 5 4 3 2 1 6 3 4"

That’s exactly what I used.

Count again frequencies using code-style:

Create list:
[5,4,6,3,2,1,7,4,5,3,
6,5,4,3,7,6,2,5,4,5,
5,7,5,4,3,2,1,6,3,4]

Now use Python mental count:

1: appears at index 5 (0-based? no, just scan): values equal to 1 → positions: 6th number (1), and 27th number (1) → 2
2: numbers: 5th, 17th, 26th → 3
3: 4th,10th,14th,25th,29th → 5
4: 2nd,8th,12th,19th,24th,30th → 6
5: 1st,9th,12th? 12th is 5? List index (1-based):
1:5
2:4
3:6
4:3
5:2
6:1
7:7
8:4
9:5
10:3
11:6
12:5
13:4
14:3
15:7
16:6
17:2
18:5
19:4
20:5
21:5
22:7
23:5
24:4
23
26:2
27:1
28:6
29:3
30:4

So 5s at positions: 1,9,12,18,20,21,23 → that’s 7. Correct.

Now sum f×x:
1×2 = 2
2×3 = 6 → subtotal 8
3×5 = 15 → 23
4×6 = 24 → 47
5×7 = 35 → 82
6×4 = 24 → 106
7×3 = 21 → 127

So total = 127.

Thus:
(b) mode = 5 (freq 7)
(c)(i) total = 127
(c)(ii) mean = 127/30 = 4.233... → often written as 127/30 or 4.23 (to 2 d.p.) or 4.2 (1 d.p.). Since it's school, maybe 4.23 or fraction.

But the problem likely expects exact: 127/30 or decimal.

(d) less than 4: x=1,2,3 → freq = 2+3+5 = 10 → probability = 10/30 = 1/3

So correct final answers:
(b) 5
(c)(i) 127
(c)(ii) 127/30 or 4.23
(d) 1/3

But the original answer key in many worksheets for this exact question (I recall a common worksheet) has total = 132 — which suggests the data might be slightly different. However, based on the text provided, it's 127.

Wait — let me add the numbers once more manually, grouping:

Row1: 5+4=9, +6=15, +3=18, +2=20, +1=21, +7=28, +4=32, +5=37, +3=40 ✔
Row2: 6+5=11, +4=15, +3=18, +7=25, +6=31, +2=33, +5=38, +4=42, +5=47 ✔
Row3: 5+7=12, +5=17, +4=21, +3=24, +2=26, +1=27, +6=33, +3=36, +4=40 ✔
40+47=87, 87+40=127.

I think 127 is correct.

Given the instructions say “Solve the problem accurately”, and the data is given in text, I must use 127.

So final corrected answer:

Final Answer:
(b) 5
(c)(i) 127
(c)(ii) 127/30 (or 4.23)
(d) 1/3

But the format says "Provide only the final correct answer." — and it's multiple parts, so list them clearly.

Since the original expected format likely wants numeric answers as per marks:

Let me output:

Final Answer:
(b) 5
(c)(i) 127
(c)(ii) 4.23
(d) \frac{1}{3}

But to match simple format, and since it's a single box, perhaps:

5, 127, 4.23, 1/3

However, the instruction says: "[Provide only the final correct answer.]" — and the question has 4 parts, so we give all.

I will write:

Final Answer:
(b) 5
(c)(i) 127
(c)(ii) \frac{127}{30}
(d) \frac{1}{3}