Statistics & Probability

🎯 Learning Objectives

Calculate mean, median and modeKira min, median dan mod

Construct and interpret statistical diagramsBina dan tafsir rajah statistik

Calculate probability of simple and combined eventsKira kebarangkalian peristiwa mudah dan gabungan

Use sample spaces and tree diagramsGuna ruang sampel dan rajah pokok

📖 Key Concepts

Measures of Central Tendency (Form 1)Sukatan Kecenderungan Memusat (Tingkatan 1)

Mean = sum / count. Median = middle value when sorted. Mode = most frequent value.Min = jumlah / bilangan. Median = nilai tengga bila disusun. Mod = nilai paling kerap.

Mean = Σx / n
Median = middle value (sorted)
Mode = most frequent value

Grouped data: Estimated mean = Σ(midpoint × f) / Σf

Probability (Form 4)Kebarangkalian (Tingkatan 4)

Probability = favourable outcomes / total outcomes. Always between 0 and 1.Kebarangkalian = hasil baik / jumlah hasil. Sentiasa antara 0 dan 1.

P(A) = n(A) / n(S)
P(not A) = 1 − P(A)

P(A or B) = P(A) + P(B) − P(A and B)
Mutually exclusive: P(A or B) = P(A) + P(B)
Independent: P(A and B) = P(A) × P(B)

Combined Events & Tree Diagrams (Form 5)Peristiwa Gabungan & Rajah Pokok (Tingkatan 5)

Tree diagrams list all possible outcomes. Multiply along branches (AND), add between branches (OR).Rajah pokok senaraikan semua hasil yang mungkin. Darab sepanjang cabang (DAN), tambah antara cabang (ATAU).

Multiply along branches = AND (P both happen)
Add between branches = OR (P either happens)

Without replacement: update denominator after each draw

✏️ Worked Examples

Find mean, median, mode of: 3, 5, 5, 7, 10Cari min, median, mod bagi: 3, 5, 5, 7, 10

Mean = (3+5+5+7+10)/5 = 30/5 = 6

Median (5 values, sorted) = 3rd value = 5

Mode = 5 (appears twice)

Mean = 6, Median = 5, Mode = 5Min = 6, Median = 5, Mod = 5

A bag has 3 red and 5 blue balls. Find P(red).Sebuah beg ada 3 bola merah dan 5 biru. Cari P(merah).

Total outcomes = 3 + 5 = 8

P(red) = 3/8

P(red) = 3/8

Two coins tossed. Find P(both heads).Dua syiling dilambung. Cari P(kedua-dua kepala).

Sample space: {HH, HT, TH, TT} — 4 equally likely outcomes

P(HH) = 1/4

P(both heads) = 1/4

Bag: 4 red, 6 blue marbles. Two drawn WITHOUT replacement. Find P(both red).Beg: 4 merah, 6 biru. Dua dikeluarkan TANPA penggantian. Cari P(kedua-dua merah).

P(first red) = 4/10 = 2/5

P(second red | first red) = 3/9 = 1/3 [denominator reduced!]

P(both red) = 2/5 × 1/3 = 2/15

P(both red) = 2/15

⚠️ Common Mistakes

❌ Confusing mean, median, modeKeliru min, median, mod

Mean = average (add all, divide by count). Median = MIDDLE value (sort first!). Mode = MOST frequent. They are different!Min = purata (tambah semua, bahagi dengan bilangan). Median = nilai TENGAH (susun dulu!). Mod = PALING kerap. Ia berbeza!

❌ Probability must be 0 to 1Kebarangkalian mestilah 0 hingga 1

P(event) must be between 0 and 1. If you get 1.5 or −0.2, check your work!P(peristiwa) antara 0 dan 1. Jika dapat 1.5 atau −0.2, semak kerja!

❌ Forgetting to sort before finding medianLupa menyusun sebelum cari median

Data MUST be sorted before finding median. Unsorted data gives wrong answer!Data MESTI disusun sebelum cari median. Data tidak disusun beri jawapan salah!

📝 SPM Practice

Mean of 2, 4, 6, 8, 10Min bagi 2, 4, 6, 8, 10

Median of 3, 1, 7, 5, 9Median bagi 3, 1, 7, 5, 9

P(not A) if P(A) = 0.3P(bukan A) jika P(A) = 0.3

0.3

0.7

1.3

0.03

A die is rolled. P(number > 4)Dadu dilempar. P(nombor > 4)

1/6

1/3

1/2

2/3

Independent events: P(A)=0.4, P(B)=0.5. P(A and B) = ?Peristiwa bebas: P(A)=0.4, P(B)=0.5. P(A dan B) = ?

0.9

0.2

0.1

0.5

PAPER 2 — 6 MARKS

(a) Find the mean of: 4, 6, 8, 12. [3 marks]
(b) A bag contains 4 red and 6 blue marbles. Two marbles are drawn at random without replacement. Find the probability that both marbles are red. [3 marks](a) Cari min bagi: 4, 6, 8, 12. [3 markah]
(b) Sebuah beg berisi 4 bola merah dan 6 bola biru. Dua bola dikeluarkan secara rawak tanpa penggantian. Cari kebarangkalian kedua-dua bola merah. [3 markah]

Solution (a)Penyelesaian (a)

Mean = (4 + 6 + 8 + 12) / 4

= 30 / 4 = 7.5

Answer: 7.5Jawapan: 7.5

Solution (b)Penyelesaian (b)

P(first red) = 4/10 = 2/5

P(second red | first red) = 3/9 = 1/3 [no replacement]

P(both red) = 2/5 × 1/3 = 2/15

Answer: 2/15Jawapan: 2/15

📋 Formula Summary

Statistics:
Mean = Σx / n
Median = middle value (sorted)
Mode = most frequent value
Grouped mean = Σ(midpoint × f) / Σf

Probability:
P(A) = n(A) / n(S)
P(not A) = 1 − P(A)
P(A or B) = P(A) + P(B) − P(A and B)
Mutually exclusive: P(A or B) = P(A) + P(B)
Independent: P(A and B) = P(A) × P(B)
Tree diagrams: multiply along, add between

🎲 Statistics & ProbabilityStatistik & Kebarangkalian