CF2 · Materi

Soa Exam P Samples Part 9

No. 241

An investor invests 100 dollars in a stock. Each month, the investment has probability 0.5 of increasing by 1.10 dollars and probability 0.5 of decreasing by 0.90 dollars. The changes in price in different months are mutually independent.

Calculate the probability that the investment has a value greater than 91 dollars at the end of month 100.

a. $0{,}63$
b. $0{,}75$
c. $0{,}82$
d. $0{,}94$
e. $0{,}97$

≡Jawaban No. 241›

(e). $0{,}97$

Field	Isi
Topik CF2	Topik 4 — Inferensi Statistik
Sub-topik	4.3 Teorema Limit Pusat
Difficulty	Medium
Prerequisite	2.1 Variabel Acak Diskrit, 4.2 Distribusi Sampel
Connected Topics	2.5 Distribusi Diskrit Umum, 4.4 Hukum Bilangan Besar
Referensi	Hogg-Tanis-Zimm Bab 5.5–5.6; Miller Bab 6

ℹRumus›

Teorema Limit Pusat: jika $X_1, X_2, \ldots, X_n$ i.i.d. dengan $E[X_k] = \mu$ dan $\text{Var}(X_k) = \sigma^2$ , maka

S = \sum_{k=1}^{n} X_k \approx N(n\mu,\, n\sigma^2) \quad \text{untuk } n \text{ besar}

sehingga

P(S > c) \approx P\!\left(Z > \frac{c - n\mu}{\sqrt{n\sigma^2}}\right), \quad Z \sim N(0,1)

Diketahui:

Investasi awal = 100; setiap bulan naik $+1{,}10$ (prob. $0{,}5$ ) atau turun $-0{,}90$ (prob. $0{,}5$ )
Perubahan antar bulan saling independen; $n = 100$ bulan
Target: $P(\text{nilai akhir} > 91) = P(100 + S_{100} > 91) = P(S_{100} > -9)$

▸Langkah Pengerjaan›

Langkah 1: Hitung $E[X_k]$ dan $\text{Var}(X_k)$

Setiap $X_k$ bernilai $+1{,}10$ atau $-0{,}90$ masing-masing dengan prob. $0{,}5$ .

E[X_k] = 0{,}5(1{,}10) + 0{,}5(-0{,}90) = 0{,}55 - 0{,}45 = 0{,}10

E[X_k^2] = 0{,}5(1{,}10)^2 + 0{,}5(-0{,}90)^2 = 0{,}5(1{,}21) + 0{,}5(0{,}81) = 1{,}01

\text{Var}(X_k) = E[X_k^2] - (E[X_k])^2 = 1{,}01 - (0{,}10)^2 = 1{,}01 - 0{,}01 = 1{,}00

Langkah 2: Distribusi $S_{100} = \sum_{k=1}^{100} X_k$

Dengan CLT:

E[S_{100}] = 100 \times 0{,}10 = 10

\text{Var}(S_{100}) = 100 \times 1{,}00 = 100 \implies \sigma_{S} = 10

Langkah 3: Hitung probabilitas

Nilai akhir investasi $= 100 + S_{100}$ . Syarat nilai $> 91$ :

P(100 + S_{100} > 91) = P(S_{100} > -9)

Standarisasi:

P\!\left(Z > \frac{-9 - 10}{10}\right) = P(Z > -1{,}90) = \Phi(1{,}90) \approx 0{,}9713

Hasil Akhir: (e). $0{,}97$

◈Jebakan Umum›

⬡Kesalahan Konseptual›

Lupa menyertakan nilai awal (100) saat mengonversi kondisi nilai akhir $> 91$ ke kondisi pada $S_{100}$ .
Mengira $\text{Var}(X_k) = E[X_k^2]$ ; selalu kurangkan $(E[X_k])^2$ .

⬡Kesalahan Interpretasi Soal›

“nilai investasi $> 91$ ” bukan " $S_{100} > 91$ " — perlu dikurangi nilai awal 100 terlebih dahulu.

▲Red Flags›

Jika $n$ besar dan distribusi per periode tidak disebutkan → gunakan CLT.
Jika soal menyebut perubahan independen identik → langsung identifikasi sebagai setting CLT.

No. 242

Let $X$ denote the loss amount sustained by an insurance company’s policyholder in an auto collision. Let $Z$ denote the portion of $X$ that the insurance company will have to pay. An actuary determines that $X$ and $Z$ are independent with respective density and probability functions

f_X(x) = \frac{1}{8}e^{-x/8}, \quad x > 0

P(Z = z) = \begin{cases} 0{,}45, & z = 1 \\ 0{,}55, & z = 0 \end{cases}

Calculate the variance of the insurance company’s claim payment $ZX$ .

a. $13{,}0$
b. $15{,}8$
c. $28{,}8$
d. $35{,}2$
e. $44{,}6$

≡Jawaban No. 242›

(e). $44{,}6$

Field	Isi
Topik CF2	Topik 3 — Variabel Acak Multivariat
Sub-topik	3.5 Independensi dan Korelasi
Difficulty	Medium
Prerequisite	2.6 Distribusi Kontinu Umum, 2.1 Variabel Acak Diskrit
Connected Topics	3.1 Distribusi Gabungan, 3.7 Distribusi Majemuk
Referensi	Hogg-Tanis-Zimm Bab 4.1; Miller Bab 3–4

ℹRumus›

Untuk $X$ dan $Z$ independen:

E[ZX] = E[Z] \cdot E[X]

E[(ZX)^2] = E[Z^2] \cdot E[X^2]

\text{Var}(ZX) = E[(ZX)^2] - (E[ZX])^2

Untuk $X \sim \text{Exp}(\beta = 8)$ (kontinu, $x > 0$ ; $\beta$ = parameter scale):

E[X] = 8, \quad E[X^2] = \text{Var}(X) + (E[X])^2 = 64 + 64 = 128

Diketahui:

$X \sim \text{Exp}(\beta = 8)$ : $E[X] = 8$ , $\text{Var}(X) = 64$ , $E[X^2] = 128$
$Z$ : diskrit, $P(Z=1) = 0{,}45$ , $P(Z=0) = 0{,}55$
$X$ dan $Z$ independen
Target: $\text{Var}(ZX)$

▸Langkah Pengerjaan›

Langkah 1: Hitung $E[Z]$ dan $E[Z^2]$

E[Z] = 1(0{,}45) + 0(0{,}55) = 0{,}45

E[Z^2] = 1^2(0{,}45) + 0^2(0{,}55) = 0{,}45

Langkah 2: Hitung $E[ZX]$ menggunakan independensi

E[ZX] = E[Z] \cdot E[X] = 0{,}45 \times 8 = 3{,}6

Langkah 3: Hitung $E[(ZX)^2]$

E[(ZX)^2] = E[Z^2] \cdot E[X^2] = 0{,}45 \times 128 = 57{,}6

Langkah 4: Hitung Variansi

\text{Var}(ZX) = E[(ZX)^2] - (E[ZX])^2 = 57{,}6 - (3{,}6)^2 = 57{,}6 - 12{,}96 = 44{,}64

Hasil Akhir: (e). $44{,}6$

◈Jebakan Umum›

⬡Kesalahan Konseptual›

Mengira $\text{Var}(ZX) = \text{Var}(Z)\cdot\text{Var}(X)$ ; rumus ini tidak berlaku umum — gunakan momen kedua.
Lupa bahwa $E[X^2] = \text{Var}(X) + (E[X])^2 = 64 + 64 = 128$ untuk distribusi Eksponensial.

⬡Kesalahan Interpretasi Soal›

Mengira $Z$ kontinu karena disebut “porsi”; perhatikan bahwa $Z$ diberikan sebagai distribusi diskrit dengan dua nilai.

▲Red Flags›

Jika ada produk dua variabel independen → gunakan $E[Z^2]\cdot E[X^2]$ untuk momen kedua produk, bukan $E[Z]^2 \cdot E[X]^2$ .

No. 243

A couple takes out a medical insurance policy that reimburses them for days of work missed due to illness. Let $X$ and $Y$ denote the number of days missed during a given month by the wife and husband, respectively. The policy pays a monthly benefit of 50 times the maximum of $X$ and $Y$ , subject to a benefit limit of 100. $X$ and $Y$ are independent, each with a discrete uniform distribution on the set $\{0, 1, 2, 3, 4\}$ .

Calculate the expected monthly benefit for missed days of work that is paid to the couple.

a. $70$
b. $90$
c. $92$
d. $95$
e. $140$

≡Jawaban No. 243›

(b). $90$

Field	Isi
Topik CF2	Topik 3 — Variabel Acak Multivariat
Sub-topik	3.1 Distribusi Gabungan
Difficulty	Medium
Prerequisite	2.1 Variabel Acak Diskrit, 3.5 Independensi dan Korelasi
Connected Topics	3.2 Distribusi Marginal, 3.3 Distribusi Bersyarat
Referensi	Hogg-Tanis-Zimm Bab 4.1; Miller Bab 3

ℹRumus›

Manfaat $B = \min(50 \cdot \max(X, Y),\, 100)$ .

Karena $X, Y \in \{0,1,2,3,4\}$ independen dengan peluang sama $1/5$ masing-masing, setiap pasangan $(x,y)$ memiliki peluang $\frac{1}{25}$ .

E[B] = \sum_{x=0}^{4}\sum_{y=0}^{4} B(x,y) \cdot \frac{1}{25}

Diketahui:

$X, Y \sim U\{0,1,2,3,4\}$ , independen; $P(X=x) = P(Y=y) = \frac{1}{5}$
Manfaat $= 50 \cdot \max(X,Y)$ , dibatasi maksimum 100
Target: $E[B]$

▸Langkah Pengerjaan›

Langkah 1: Identifikasi nilai manfaat yang mungkin

$\max(X,Y)$ dapat bernilai $0, 1, 2, 3, 4$ , menghasilkan $50 \cdot \max = 0, 50, 100, 150, 200$ . Namun manfaat dibatasi 100, sehingga nilai manfaat yang mungkin hanya: $0$ , $50$ , dan $100$ .

Langkah 2: Hitung probabilitas setiap manfaat

$B = 0$ : hanya terjadi jika $\max(X,Y) = 0$ , yaitu $(X,Y) = (0,0)$ . Probabilitas $= \frac{1}{25}$ .
$B = 50$ : terjadi jika $\max(X,Y) = 1$ , yaitu pasangan $(0,1), (1,0), (1,1)$ . Probabilitas $= \frac{3}{25}$ .
$B = 100$ : semua kasus lainnya, probabilitas $= \frac{21}{25}$ .

Langkah 3: Hitung nilai harapan

E[B] = 0 \cdot \frac{1}{25} + 50 \cdot \frac{3}{25} + 100 \cdot \frac{21}{25} = \frac{0 + 150 + 2100}{25} = \frac{2250}{25} = 90

Hasil Akhir: (b). $90$

◈Jebakan Umum›

⬡Kesalahan Konseptual›

Lupa menerapkan batas manfaat (cap) 100; menghitung $E[50 \cdot \max(X,Y)]$ tanpa pembatasan akan memberikan hasil berbeda.
Menghitung $E[\max(X,Y)]$ secara terpisah tanpa mengelompokkan nilai yang sudah dicap.

⬡Kesalahan Interpretasi Soal›

“Subject to a benefit limit of 100” berarti manfaat aktual $= \min(50\cdot\max(X,Y), 100)$ , bukan $\max + 100$ .

▲Red Flags›

Jika benefit memakai “subject to maximum/limit” → selalu terapkan $\min(\cdot, \text{limit})$ sebelum menghitung ekspektasi.

No. 244

The table below shows the joint probability function of a sailor’s number of boating accidents and number of hospitalizations from these accidents this year.

	Hospitalizations
Accidents	0	1	2	3
0	0.700
1	0.150	0.050
2	0.060	0.020	0.010
3	0.005	0.002	0.002	0.001

Calculate the sailor’s expected number of hospitalizations from boating accidents this year.

a. $0{,}085$
b. $0{,}099$
c. $0{,}410$
d. $1{,}000$
e. $1{,}500$

≡Jawaban No. 244›

(b). $0{,}099$

Field	Isi
Topik CF2	Topik 3 — Variabel Acak Multivariat
Sub-topik	3.2 Distribusi Marginal
Difficulty	Easy
Prerequisite	3.1 Distribusi Gabungan, 2.1 Variabel Acak Diskrit
Connected Topics	3.4 Nilai Harapan dan Variansi Bersyarat
Referensi	Hogg-Tanis-Zimm Bab 4.1; Miller Bab 3

ℹRumus›

Distribusi marginal variabel $H$ (hospitalisasi):

p_H(h) = \sum_{a} p(a, h)

Nilai harapan:

E[H] = \sum_{h=0}^{3} h \cdot p_H(h)

Diketahui:

Tabel distribusi gabungan $(A, H)$ diberikan
Target: $E[H]$ (marginal atas $H$ )

▸Langkah Pengerjaan›

Langkah 1: Hitung distribusi marginal $H$ (jumlahkan kolom)

p_H(0) = 0{,}700 + 0{,}150 + 0{,}060 + 0{,}005 = 0{,}915

p_H(1) = 0{,}050 + 0{,}020 + 0{,}002 = 0{,}072

p_H(2) = 0{,}010 + 0{,}002 = 0{,}012

p_H(3) = 0{,}001

Verifikasi: $0{,}915 + 0{,}072 + 0{,}012 + 0{,}001 = 1{,}000$ ✓

Langkah 2: Hitung nilai harapan

E[H] = 0(0{,}915) + 1(0{,}072) + 2(0{,}012) + 3(0{,}001)

= 0 + 0{,}072 + 0{,}024 + 0{,}003 = 0{,}099

Hasil Akhir: (b). $0{,}099$

◈Jebakan Umum›

⬡Kesalahan Konseptual›

Menjumlahkan baris (marginal $A$ ) bukannya kolom (marginal $H$ ) — pastikan kita mencari nilai harapan hospitalisasi, bukan kecelakaan.
Menggunakan distribusi gabungan langsung tanpa marginalisasi.

▲Red Flags›

Jika ditanya nilai harapan satu variabel dari tabel joint → jumlahkan sepanjang dimensi variabel yang lain untuk mendapat distribusi marginal.

No. 245

On Main Street, a driver’s speed just before an accident is uniformly distributed on $[5, 20]$ . Given the speed, the resulting loss from the accident is exponentially distributed with mean equal to three times the speed.

Calculate the variance of a loss due to an accident on Main Street.

a. $525$
b. $1463$
c. $1575$
d. $1632$
e. $1744$

≡Jawaban No. 245›

(e). $1744$

Field	Isi
Topik CF2	Topik 3 — Variabel Acak Multivariat
Sub-topik	3.4 Nilai Harapan dan Variansi Bersyarat
Difficulty	Hard
Prerequisite	2.6 Distribusi Kontinu Umum, 3.3 Distribusi Bersyarat
Connected Topics	3.7 Distribusi Majemuk, 2.2 Variabel Acak Kontinu
Referensi	Hogg-Tanis-Zimm Bab 4.4; Miller Bab 5

ℹRumus›

Hukum Variansi Total:

\text{Var}(X) = E[\text{Var}(X \mid S)] + \text{Var}(E[X \mid S])

Untuk $X \mid S = s \sim \text{Exp}(\beta = 3s)$ : $E[X \mid S] = 3S$ dan $\text{Var}(X \mid S) = (3S)^2 = 9S^2$ .

Untuk $S \sim U(5, 20)$ : $E[S] = 12{,}5$ , $\text{Var}(S) = \frac{(20-5)^2}{12} = \frac{225}{12} = 18{,}75$ , $E[S^2] = \text{Var}(S) + (E[S])^2 = 18{,}75 + 156{,}25 = 175$ .

Diketahui:

$S \sim U(5, 20)$ ; $E[S] = 12{,}5$ ; $E[S^2] = 175$
$X \mid S = s \sim \text{Exp}(\beta = 3s)$ : $E[X \mid S] = 3S$ , $\text{Var}(X \mid S) = 9S^2$
Target: $\text{Var}(X)$

▸Langkah Pengerjaan›

Langkah 1: Hitung $E[\text{Var}(X \mid S)]$

E[\text{Var}(X \mid S)] = E[9S^2] = 9 \cdot E[S^2] = 9 \times 175 = 1575

Langkah 2: Hitung $\text{Var}(E[X \mid S])$

\text{Var}(E[X \mid S]) = \text{Var}(3S) = 9 \cdot \text{Var}(S) = 9 \times 18{,}75 = 168{,}75

Langkah 3: Gabungkan dengan Hukum Variansi Total

\text{Var}(X) = 1575 + 168{,}75 = 1743{,}75 \approx 1744

Hasil Akhir: (e). $1744$

◈Jebakan Umum›

⬡Kesalahan Konseptual›

Lupa bahwa $\text{Var}(X \mid S) = (E[X \mid S])^2 = 9S^2$ untuk distribusi Eksponensial (variansi = kuadrat mean).
Mengira $E[S^2] = (E[S])^2$ ; selalu gunakan $E[S^2] = \text{Var}(S) + (E[S])^2$ .

▲Red Flags›

Jika distribusi bersyarat bergantung pada variabel acak lain → gunakan Hukum Variansi Total (Eve’s Law), bukan menghitung langsung.

No. 246

Let $X$ be the annual number of hurricanes hitting Florida, and let $Y$ be the annual number of hurricanes hitting Texas. $X$ and $Y$ are independent Poisson variables with respective means 1.70 and 2.30.

Calculate $\text{Var}(X - Y \mid X + Y = 3)$ .

a. $1{,}71$
b. $1{,}77$
c. $2{,}93$
d. $3{,}14$
e. $4{,}00$

≡Jawaban No. 246›

(c). $2{,}93$

Field	Isi
Topik CF2	Topik 3 — Variabel Acak Multivariat
Sub-topik	3.3 Distribusi Bersyarat, 3.4 Nilai Harapan dan Variansi Bersyarat
Difficulty	Hard
Prerequisite	2.5 Distribusi Diskrit Umum, 3.5 Independensi dan Korelasi
Connected Topics	3.1 Distribusi Gabungan, 3.2 Distribusi Marginal
Referensi	Hogg-Tanis-Zimm Bab 4.4; Miller Bab 5

ℹRumus›

Jika $X \sim \text{Poisson}(\lambda_X)$ dan $Y \sim \text{Poisson}(\lambda_Y)$ independen, maka $X \mid (X + Y = n) \sim B\!\left(n,\, \frac{\lambda_X}{\lambda_X + \lambda_Y}\right)$ .

Kemudian $X - Y \mid (X+Y=3) = 2X - 3 \mid (X+Y=3)$ , sehingga:

\text{Var}(X - Y \mid X+Y=3) = 4\,\text{Var}(X \mid X+Y=3)

Diketahui:

$X \sim \text{Poisson}(1{,}70)$ , $Y \sim \text{Poisson}(2{,}30)$ , independen
$\lambda_X + \lambda_Y = 4{,}00$ ; $p = \frac{1{,}70}{4{,}00} = 0{,}425$
Kondisi: $X + Y = 3$

▸Langkah Pengerjaan›

Langkah 1: Distribusi bersyarat $X \mid (X+Y=3)$

Karena $X$ dan $Y$ Poisson independen, distribusi bersyarat $X \mid (X+Y=3) \sim B(3, p)$ dengan

p = \frac{\lambda_X}{\lambda_X + \lambda_Y} = \frac{1{,}70}{4{,}00} = 0{,}425

Langkah 2: Variansi bersyarat $X$

\text{Var}(X \mid X+Y=3) = n \cdot p(1-p) = 3 \times 0{,}425 \times 0{,}575 = 0{,}7331

Langkah 3: Hubungkan $X - Y$ dengan $X$

Karena $Y = 3 - X$ ketika $X + Y = 3$ :

X - Y = X - (3 - X) = 2X - 3

Oleh karena itu:

\text{Var}(X - Y \mid X+Y=3) = \text{Var}(2X - 3 \mid X+Y=3) = 4 \cdot \text{Var}(X \mid X+Y=3)

= 4 \times 0{,}7331 = 2{,}9325 \approx 2{,}93

Hasil Akhir: (c). $2{,}93$

◈Jebakan Umum›

⬡Kesalahan Konseptual›

Menghitung $\text{Var}(X-Y)$ tanpa kondisi, yaitu menggunakan $\text{Var}(X) + \text{Var}(Y)$ secara langsung.
Tidak memanfaatkan sifat bahwa jumlah dua Poisson independen yang dikondisikan mengikuti distribusi Binomial.

▲Red Flags›

Jika $X, Y \sim \text{Poisson}$ independen dan diminta variansi/ekspektasi bersyarat pada $X+Y=n$ → selalu gunakan sifat Binomial bersyarat.

No. 247

Random variables $X$ and $Y$ have joint distribution

	$X = 0$	$X = 1$	$X = 2$
$Y = 0$	$1/15$	$a$	$2/15$
$Y = 1$	$a$	$b$	$a$
$Y = 2$	$2/15$	$a$	$1/15$

Let $a$ be the value that minimizes the variance of $X$ .

Calculate the variance of $Y$ .

a. $2/5$
b. $8/15$
c. $16/25$
d. $2/3$
e. $7/10$

≡Jawaban No. 247›

(a). $2/5$

Field	Isi
Topik CF2	Topik 3 — Variabel Acak Multivariat
Sub-topik	3.2 Distribusi Marginal
Difficulty	Hard
Prerequisite	3.1 Distribusi Gabungan, 2.1 Variabel Acak Diskrit
Connected Topics	3.5 Independensi dan Korelasi
Referensi	Hogg-Tanis-Zimm Bab 4.1; Miller Bab 3

ℹRumus›

Distribusi marginal: $p_X(x) = \sum_y p(x,y)$ .

Variansi: $\text{Var}(X) = E[X^2] - (E[X])^2$ .

Total probabilitas: $\sum_{x,y} p(x,y) = 1$ .

Diketahui:

Tabel joint dengan parameter $a$ dan $b$ ; total prob. = 1
Cari $a$ yang meminimumkan $\text{Var}(X)$ , lalu hitung $\text{Var}(Y)$

▸Langkah Pengerjaan›

Langkah 1: Distribusi marginal $X$

p_X(0) = \frac{1}{15} + a + \frac{2}{15} = \frac{3}{15} + a = \frac{1}{5} + a

p_X(1) = a + b + a = 2a + b

p_X(2) = \frac{2}{15} + a + \frac{1}{15} = \frac{3}{15} + a = \frac{1}{5} + a

Langkah 2: Ekspektasi dan Variansi $X$

Berdasarkan simetri tabel ( $p_X(0) = p_X(2)$ ), maka $E[X] = 1$ .

\text{Var}(X) = E[(X-1)^2] = (0-1)^2\!\left(\frac{1}{5}+a\right) + (1-1)^2(2a+b) + (2-1)^2\!\left(\frac{1}{5}+a\right)

= 2\left(\frac{1}{5} + a\right) = \frac{2}{5} + 2a

Langkah 3: Minimumkan $\text{Var}(X)$

$\text{Var}(X) = \frac{2}{5} + 2a$ adalah fungsi linear-naik terhadap $a$ . Karena $a \geq 0$ (probabilitas tidak negatif), minimum tercapai di $a = 0$ .

Langkah 4: Hitung $\text{Var}(Y)$ dengan $a = 0$

Dengan $a = 0$ , distribusi marginal $Y$ (yang simetris dengan $X$ dari struktur tabel) identik dengan marginal $X$ :

p_Y(0) = \frac{1}{5} + 0 = \frac{1}{5}, \quad p_Y(1) = 0 + b, \quad p_Y(2) = \frac{1}{5}

Dari total probabilitas: $\frac{1}{15} + 0 + \frac{2}{15} + 0 + b + 0 + \frac{2}{15} + 0 + \frac{1}{15} = 1$ , sehingga $b = 1 - \frac{6}{15} = \frac{3}{5}$ .

Maka $\text{Var}(Y) = \text{Var}(X)\big|_{a=0} = \frac{2}{5} + 2(0) = \frac{2}{5}$ .

Hasil Akhir: (a). $\dfrac{2}{5}$

◈Jebakan Umum›

⬡Kesalahan Konseptual›

Tidak menyadari bahwa tabel memiliki simetri, sehingga distribusi marginal $X$ dan $Y$ identik (ketika $a=0$ ).
Mengira minimisasi variansi memerlukan kalkulus; fungsi linear dalam $a$ cukup diperiksa di batas $a = 0$ .

▲Red Flags›

Jika tabel joint simetris terhadap diagonal → cek apakah marginal $X$ dan $Y$ identik sebelum menghitung terpisah.

No. 248

Let $X$ be a random variable that takes on the values $-1$ , $0$ , and $1$ with equal probabilities. Let $Y = X^2$ .

Which of the following is true?

a. $\text{Cov}(X, Y) > 0$ ; the random variables $X$ and $Y$ are dependent.
b. $\text{Cov}(X, Y) > 0$ ; the random variables $X$ and $Y$ are independent.
c. $\text{Cov}(X, Y) = 0$ ; the random variables $X$ and $Y$ are dependent.
d. $\text{Cov}(X, Y) = 0$ ; the random variables $X$ and $Y$ are independent.
e. $\text{Cov}(X, Y) < 0$ ; the random variables $X$ and $Y$ are dependent.

≡Jawaban No. 248›

(c). $\text{Cov}(X, Y) = 0$ ; $X$ dan $Y$ dependen.

Field	Isi
Topik CF2	Topik 3 — Variabel Acak Multivariat
Sub-topik	3.5 Independensi dan Korelasi
Difficulty	Medium
Prerequisite	2.1 Variabel Acak Diskrit, 3.1 Distribusi Gabungan
Connected Topics	3.6 Matriks Variansi-Kovariansi
Referensi	Hogg-Tanis-Zimm Bab 4.1; Miller Bab 4

ℹRumus›

\text{Cov}(X, Y) = E[XY] - E[X]\cdot E[Y]

Independensi mensyaratkan: $P(X = x, Y = y) = P(X = x)\cdot P(Y = y)$ untuk semua $x, y$ .

Penting: $\text{Cov}(X,Y) = 0$ tidak mengimplikasikan independensi secara umum.

Diketahui:

$P(X = -1) = P(X = 0) = P(X = 1) = \frac{1}{3}$
$Y = X^2$ , sehingga $Y \in \{0, 1\}$

▸Langkah Pengerjaan›

Langkah 1: Hitung $E[X]$ , $E[Y]$ , dan $E[XY]$

E[X] = \frac{1}{3}(-1) + \frac{1}{3}(0) + \frac{1}{3}(1) = 0

E[Y] = E[X^2] = \frac{1}{3}(1) + \frac{1}{3}(0) + \frac{1}{3}(1) = \frac{2}{3}

E[XY] = E[X \cdot X^2] = E[X^3] = \frac{1}{3}(-1)^3 + \frac{1}{3}(0)^3 + \frac{1}{3}(1)^3 = \frac{-1+0+1}{3} = 0

Langkah 2: Hitung Kovarians

\text{Cov}(X,Y) = E[XY] - E[X]\cdot E[Y] = 0 - 0 \cdot \frac{2}{3} = 0

Langkah 3: Periksa Independensi

Jika $X = 0$ maka $Y = 0$ pasti, sehingga $P(X=0, Y=1) = 0$ .

Namun $P(X=0)\cdot P(Y=1) = \frac{1}{3} \cdot \frac{2}{3} = \frac{2}{9} \neq 0$ .

Jadi $X$ dan $Y$ dependen meskipun kovariansnya nol.

Hasil Akhir: (c). $\text{Cov}(X,Y) = 0$ ; $X$ dan $Y$ dependen.

◈Jebakan Umum›

⬡Kesalahan Konseptual›

Menyimpulkan independensi dari $\text{Cov} = 0$ ; ini hanya berlaku untuk distribusi Normal bivariat, tidak umum.
Kovarians mengukur hubungan linear; $Y = X^2$ adalah hubungan nonlinear sehingga kovarians bisa nol meski ada ketergantungan.

▲Red Flags›

Jika $Y$ merupakan fungsi deterministik dari $X$ → mereka pasti dependen, terlepas dari nilai kovariansnya.
Jika soal menanyakan independensi → selalu verifikasi definisi probabilitas joint, jangan hanya periksa kovarians.

No. 249

Losses follow an exponential distribution with mean 1. Two independent losses are observed.

Calculate the expected value of the smaller loss.

a. $0{,}25$
b. $0{,}50$
c. $0{,}75$
d. $1{,}00$
e. $1{,}50$

≡Jawaban No. 249›

(b). $0{,}50$

Field	Isi
Topik CF2	Topik 3 — Variabel Acak Multivariat
Sub-topik	3.8 Transformasi Variabel Acak Gabungan
Difficulty	Medium
Prerequisite	2.6 Distribusi Kontinu Umum, 3.5 Independensi dan Korelasi
Connected Topics	2.4 Transformasi Variabel Acak Univariat
Referensi	Hogg-Tanis-Zimm Bab 4.4; Miller Bab 6

ℹRumus›

Untuk $X, Y$ i.i.d. $\text{Exp}(\beta = 1)$ , CDF minimum $Z = \min(X, Y)$ :

P(Z > z) = P(X > z)\cdot P(Y > z) = e^{-z} \cdot e^{-z} = e^{-2z}

Sehingga $Z \sim \text{Exp}\!\left(\beta = \frac{1}{2}\right)$ , dengan $E[Z] = \frac{1}{2}$ .

Diketahui:

$X, Y \sim \text{Exp}(\beta = 1)$ i.i.d.
$Z = \min(X, Y)$ ; target: $E[Z]$

▸Langkah Pengerjaan›

Langkah 1: CDF survival dari $Z$

P(Z > z) = P(X > z, Y > z) = P(X > z)\cdot P(Y > z) = e^{-z} \cdot e^{-z} = e^{-2z}, \quad z > 0

Langkah 2: Identifikasi distribusi $Z$

Survival function $e^{-2z}$ sesuai dengan $Z \sim \text{Exp}(\beta = \tfrac{1}{2})$ (rate = 2).

Langkah 3: Hitung nilai harapan

E[Z] = \frac{1}{2}

Hasil Akhir: (b). $0{,}50$

◈Jebakan Umum›

⬡Kesalahan Konseptual›

Menggunakan $E[\min] = \frac{E[X]}{2} = \frac{1}{2}$ secara intuitif tanpa justifikasi; derivasi via survival function wajib ditunjukkan.
Mengira minimum dua Eksponensial masih Eksponensial dengan mean sama; mean-nya adalah $\frac{1}{n}$ dari mean asal untuk $n$ i.i.d. Eksponensial.

▲Red Flags›

Jika soal minta statistik order ( $\min$ , $\max$ ) dari $n$ variabel i.i.d. Eksponensial → gunakan survival function, hasilnya selalu Eksponensial dengan rate $n\lambda$ .

No. 250

A delivery service owns two cars that consume 15 and 30 miles per gallon. Fuel costs 3 per gallon. On any given business day, each car travels a number of miles that is independent of the other and is normally distributed with mean 25 miles and standard deviation 3 miles.

Calculate the probability that on any given business day, the total fuel cost to the delivery service will be less than 7.

a. $0{,}13$
b. $0{,}23$
c. $0{,}29$
d. $0{,}38$
e. $0{,}47$

≡Jawaban No. 250›

(b). $0{,}23$

Field	Isi
Topik CF2	Topik 3 — Variabel Acak Multivariat
Sub-topik	3.5 Independensi dan Korelasi
Difficulty	Medium
Prerequisite	2.6 Distribusi Kontinu Umum, 3.1 Distribusi Gabungan
Connected Topics	4.3 Teorema Limit Pusat
Referensi	Miller Bab 6; Hogg-Tanis-Zimm Bab 5.5

ℹRumus›

Kombinasi linear variabel Normal independen:

C = aX + bY \implies C \sim N(aE[X] + bE[Y],\; a^2\text{Var}(X) + b^2\text{Var}(Y))

Diketahui:

Mobil 1: 15 mpg; Mobil 2: 30 mpg; harga BBM = 3/galon
$X, Y \sim N(25, 9)$ i.i.d. (jarak tempuh, dalam mil)
Biaya total: $C = 3 \cdot \frac{X}{15} + 3 \cdot \frac{Y}{30} = 0{,}2X + 0{,}1Y$

▸Langkah Pengerjaan›

Langkah 1: Ekspektasi dan variansi $C$

E[C] = 0{,}2(25) + 0{,}1(25) = 5 + 2{,}5 = 7{,}5

\text{Var}(C) = (0{,}2)^2(9) + (0{,}1)^2(9) = 0{,}36 + 0{,}09 = 0{,}45

\sigma_C = \sqrt{0{,}45} \approx 0{,}6708

Langkah 2: Standarisasi dan hitung probabilitas

P(C < 7) = P\!\left(Z < \frac{7 - 7{,}5}{0{,}6708}\right) = P(Z < -0{,}7454) \approx \Phi(-0{,}7454) \approx 0{,}23

Hasil Akhir: (b). $0{,}23$

◈Jebakan Umum›

⬡Kesalahan Konseptual›

Menggunakan efisiensi yang sama untuk kedua mobil; mobil 1 (15 mpg) mengonsumsi BBM lebih banyak per mil dibanding mobil 2 (30 mpg).
Lupa membagi jarak dengan efisiensi untuk mendapat konsumsi gallon: biaya = (mil/mpg) × harga.

▲Red Flags›

Periksa satuan: “miles per gallon” → konsumsi gallon = jarak ÷ efisiensi, bukan jarak × efisiensi.

No. 251

Two independent estimates are to be made on a building damaged by fire. Each estimate is normally distributed with mean $10b$ and variance $b^2$ .

Calculate the probability that the first estimate is at least 20 percent higher than the second.

a. $0{,}023$
b. $0{,}100$
c. $0{,}115$
d. $0{,}221$
e. $0{,}444$

≡Jawaban No. 251›

(b). $0{,}100$

Field	Isi
Topik CF2	Topik 3 — Variabel Acak Multivariat
Sub-topik	3.5 Independensi dan Korelasi
Difficulty	Medium
Prerequisite	2.6 Distribusi Kontinu Umum, 2.4 Transformasi Variabel Acak Univariat
Connected Topics	3.8 Transformasi Variabel Acak Gabungan
Referensi	Miller Bab 6; Hogg-Tanis-Zimm Bab 5.5

ℹRumus›

Selisih dua variabel Normal independen:

W = X - cY \sim N(\mu_X - c\mu_Y,\; \sigma_X^2 + c^2\sigma_Y^2)

Diketahui:

$X, Y \sim N(10b, b^2)$ independen (estimasi pertama dan kedua)
Target: $P(X \geq 1{,}2Y) = P(X - 1{,}2Y \geq 0)$

▸Langkah Pengerjaan›

Langkah 1: Definisikan $W = X - 1{,}2Y$

E[W] = 10b - 1{,}2(10b) = 10b - 12b = -2b

\text{Var}(W) = b^2 + (1{,}2)^2 b^2 = b^2(1 + 1{,}44) = 2{,}44\,b^2

\sigma_W = b\sqrt{2{,}44}

Langkah 2: Standarisasi

P(W \geq 0) = P\!\left(Z \geq \frac{0 - (-2b)}{b\sqrt{2{,}44}}\right) = P\!\left(Z \geq \frac{2b}{b\sqrt{2{,}44}}\right) = P\!\left(Z \geq \frac{2}{\sqrt{2{,}44}}\right)

= P\!\left(Z \geq \frac{2}{1{,}5620}\right) = P(Z \geq 1{,}280)

Langkah 3: Baca tabel Normal

P(Z \geq 1{,}280) = 1 - \Phi(1{,}280) \approx 1 - 0{,}900 = 0{,}100

Hasil Akhir: (b). $0{,}100$

◈Jebakan Umum›

⬡Kesalahan Konseptual›

“20% lebih tinggi” berarti $X \geq 1{,}2Y$ , bukan $X \geq Y + 0{,}2$ .
Parameter $b$ menghilang dalam standarisasi — ini wajar karena soal tidak memberikan nilai numerik $b$ .

▲Red Flags›

Jika parameter soal tidak numerik tapi jawaban numerik tetap ada → parameter pasti saling hilang (cancel) dalam proses standarisasi.

No. 252

The independent random variables $X$ and $Y$ have the same mean. The coefficients of variation of $X$ and $Y$ are 3 and 4 respectively.

Calculate the coefficient of variation of $\frac{1}{2}(X + Y)$ .

a. $5/4$
b. $7/4$
c. $5/2$
d. $7/2$
e. $7$

≡Jawaban No. 252›

(c). $5/2$

Field	Isi
Topik CF2	Topik 3 — Variabel Acak Multivariat
Sub-topik	3.5 Independensi dan Korelasi
Difficulty	Medium
Prerequisite	2.2 Variabel Acak Kontinu, 3.1 Distribusi Gabungan
Connected Topics	4.2 Distribusi Sampel
Referensi	Miller Bab 3–4; Hogg-Tanis-Zimm Bab 3

ℹRumus›

Koefisien variasi: $\text{CV}(X) = \dfrac{\sigma_X}{\mu_X}$ .

Untuk $Z = \frac{1}{2}(X+Y)$ dengan $X, Y$ independen:

E[Z] = \frac{E[X]+E[Y]}{2}, \quad \text{Var}(Z) = \frac{\text{Var}(X)+\text{Var}(Y)}{4}

Diketahui:

$E[X] = E[Y] = \mu$ ; $\text{CV}(X) = 3 \Rightarrow \sigma_X = 3\mu$ ; $\text{CV}(Y) = 4 \Rightarrow \sigma_Y = 4\mu$

▸Langkah Pengerjaan›

Langkah 1: Ekspektasi dan deviasi standar $Z$

E[Z] = \frac{\mu + \mu}{2} = \mu

\text{Var}(Z) = \frac{(3\mu)^2 + (4\mu)^2}{4} = \frac{9\mu^2 + 16\mu^2}{4} = \frac{25\mu^2}{4}

\sigma_Z = \frac{5\mu}{2}

Langkah 2: Hitung koefisien variasi $Z$

\text{CV}(Z) = \frac{\sigma_Z}{E[Z]} = \frac{5\mu/2}{\mu} = \frac{5}{2}

Hasil Akhir: (c). $\dfrac{5}{2}$

◈Jebakan Umum›

⬡Kesalahan Konseptual›

Menjumlahkan CV secara langsung: $\text{CV}(Z) \neq \frac{\text{CV}(X) + \text{CV}(Y)}{2}$ .
Mengira standar deviasi jumlah = jumlah standar deviasi; yang benar: variansi yang bersifat aditif untuk variabel independen.

▲Red Flags›

CV adalah rasio antara standar deviasi dan mean; hitung keduanya secara terpisah, lalu bagi.

No. 253

Points scored by a game participant can be modeled by $Z = 3X + 2Y - 5$ . $X$ and $Y$ are independent random variables with $\text{Var}(X) = 3$ and $\text{Var}(Y) = 4$ .

Calculate $\text{Var}(Z)$ .

a. $12$
b. $17$
c. $38$
d. $43$
e. $68$

≡Jawaban No. 253›

(d). $43$

Field	Isi
Topik CF2	Topik 3 — Variabel Acak Multivariat
Sub-topik	3.5 Independensi dan Korelasi
Difficulty	Easy
Prerequisite	2.1 Variabel Acak Diskrit, 2.2 Variabel Acak Kontinu
Connected Topics	3.6 Matriks Variansi-Kovariansi
Referensi	Miller Bab 4; Hogg-Tanis-Zimm Bab 3

ℹRumus›

Untuk $X, Y$ independen dan konstanta $a, b, c$ :

\text{Var}(aX + bY + c) = a^2\,\text{Var}(X) + b^2\,\text{Var}(Y)

(Konstanta tidak mempengaruhi variansi.)

Diketahui:

$Z = 3X + 2Y - 5$ ; $\text{Var}(X) = 3$ ; $\text{Var}(Y) = 4$ ; $X, Y$ independen

▸Langkah Pengerjaan›

Langkah 1: Terapkan rumus variansi kombinasi linear

\text{Var}(Z) = 3^2 \cdot \text{Var}(X) + 2^2 \cdot \text{Var}(Y) + \text{Var}(-5)

= 9(3) + 4(4) + 0 = 27 + 16 = 43

Hasil Akhir: (d). $43$

◈Jebakan Umum›

⬡Kesalahan Konseptual›

Menambahkan konstanta $-5$ dalam perhitungan variansi: $\text{Var}(c) = 0$ untuk semua konstanta $c$ .
Menggunakan koefisien tanpa dikuadratkan: $a\,\text{Var}(X)$ bukan $a^2\,\text{Var}(X)$ .

▲Red Flags›

Jika $X$ dan $Y$ tidak independen → harus tambahkan suku $2ab\,\text{Cov}(X,Y)$ .

No. 254

An actuary is studying hurricane models. A year is classified as a high, medium, or low hurricane year with probabilities 0.1, 0.3, and 0.6, respectively. The numbers of hurricanes in high, medium, and low years follow Poisson distributions with means 20, 15, and 10, respectively.

Calculate the variance of the number of hurricanes in a randomly selected year.

a. $11{,}25$
b. $12{,}50$
c. $12{,}94$
d. $13{,}42$
e. $23{,}75$

≡Jawaban No. 254›

(e). $23{,}75$

Field	Isi
Topik CF2	Topik 3 — Variabel Acak Multivariat
Sub-topik	3.7 Distribusi Majemuk
Difficulty	Hard
Prerequisite	2.5 Distribusi Diskrit Umum, 3.4 Nilai Harapan dan Variansi Bersyarat
Connected Topics	1.6 Teorema Bayes dan Hukum Probabilitas Total
Referensi	Hogg-Tanis-Zimm Bab 4.4; Miller Bab 5

ℹRumus›

Hukum Variansi Total (Eve’s Law): misalkan $N$ = jumlah badai dan $T$ = tipe tahun:

\text{Var}(N) = E[\text{Var}(N \mid T)] + \text{Var}(E[N \mid T])

Untuk $N \mid T \sim \text{Poisson}(\lambda_T)$ : $E[N \mid T] = \lambda_T$ dan $\text{Var}(N \mid T) = \lambda_T$ .

Momen kedua Poisson: $E[N^2 \mid T] = \lambda_T^2 + \lambda_T$ .

Diketahui:

$P(T=\text{tinggi}) = 0{,}1$ , $\lambda_{\text{tinggi}} = 20$
$P(T=\text{sedang}) = 0{,}3$ , $\lambda_{\text{sedang}} = 15$
$P(T=\text{rendah}) = 0{,}6$ , $\lambda_{\text{rendah}} = 10$

▸Langkah Pengerjaan›

Langkah 1: Hitung $E[N]$ (rata-rata tertimbang)

E[N] = 0{,}1(20) + 0{,}3(15) + 0{,}6(10) = 2 + 4{,}5 + 6 = 12{,}5

Langkah 2: Hitung $E[N^2]$ (rata-rata momen kedua)

Untuk Poisson: $E[N^2 \mid T] = \lambda_T^2 + \lambda_T$

E[N^2] = 0{,}1(20^2 + 20) + 0{,}3(15^2 + 15) + 0{,}6(10^2 + 10)

= 0{,}1(420) + 0{,}3(240) + 0{,}6(110) = 42 + 72 + 66 = 180

Langkah 3: Hitung variansi

\text{Var}(N) = E[N^2] - (E[N])^2 = 180 - (12{,}5)^2 = 180 - 156{,}25 = 23{,}75

Hasil Akhir: (e). $23{,}75$

◈Jebakan Umum›

⬡Kesalahan Konseptual›

Hanya menghitung $\text{Var}(N \mid T)$ rata-rata tertimbang tanpa menambahkan komponen $\text{Var}(E[N\mid T])$ .
Menggunakan mean Poisson sebagai momen kedua; momen kedua Poisson adalah $\lambda^2 + \lambda$ , bukan $\lambda^2$ .

▲Red Flags›

Jika distribusi bergantung pada kondisi acak (mixture model) → selalu gunakan Eve’s Law atau hitung via momen kedua marginal.

No. 255

A dental insurance company pays 100% of the cost of fillings and 70% of the cost of root canals. Fillings and root canals cost 50 and 500 each, respectively.

The tables below show the probability distributions of the annual number of fillings and annual number of root canals for each of the company’s policyholders.

# of Fillings	0	1	2	3
Probability	0.60	0.20	0.15	0.05

# of Root Canals	0	1
Probability	0.80	0.20

Calculate the expected annual payment per policyholder for fillings and root canals.

a. $90{,}00$
b. $102{,}50$
c. $132{,}50$
d. $250{,}00$
e. $400{,}00$

≡Jawaban No. 255›

(b). $102{,}50$

Field	Isi
Topik CF2	Topik 2 — Variabel Acak Univariat
Sub-topik	2.1 Variabel Acak Diskrit
Difficulty	Easy
Prerequisite	1.2 Aksioma dan Perhitungan Probabilitas
Connected Topics	3.5 Independensi dan Korelasi
Referensi	Miller Bab 3; Hogg-Tanis-Zimm Bab 2

ℹRumus›

Linearitas ekspektasi: $E[aF + bR] = a\,E[F] + b\,E[R]$

Total klaim: $C = 50F + 0{,}7 \times 500 \cdot R = 50F + 350R$

Diketahui:

$F$ = jumlah tambal gigi; $R$ = jumlah root canal; keduanya independen
Pembayaran per tambal gigi: $50$ ; per root canal: $0{,}7 \times 500 = 350$

▸Langkah Pengerjaan›

Langkah 1: Hitung $E[F]$

E[F] = 0(0{,}60) + 1(0{,}20) + 2(0{,}15) + 3(0{,}05) = 0 + 0{,}20 + 0{,}30 + 0{,}15 = 0{,}65

Langkah 2: Hitung $E[R]$

E[R] = 0(0{,}80) + 1(0{,}20) = 0{,}20

Langkah 3: Hitung $E[C]$

E[C] = 50 \times 0{,}65 + 350 \times 0{,}20 = 32{,}50 + 70{,}00 = 102{,}50

Hasil Akhir: (b). $102{,}50$

◈Jebakan Umum›

⬡Kesalahan Konseptual›

Menggunakan biaya penuh 500 untuk root canal tanpa mengalikan dengan coverage 70%; perusahaan hanya membayar $350$ .
Mengira $E[C] = E[50F] \cdot E[350R]$ ; ekspektasi produk $\neq$ produk ekspektasi kecuali independen, dan ini penjumlahan bukan perkalian.

▲Red Flags›

Perhatikan persentase coverage: “pays X% of cost Y” → pembayaran = X% × Y, bukan Y.

No. 256

A loss under a liability policy is modeled by an exponential distribution. The insurance company will cover the amount of that loss in excess of a deductible of 2000. The probability that the reimbursement is less than 6000, given that the loss exceeds the deductible, is 0.50.

Calculate the probability that the reimbursement is greater than 3000 but less than 9000, given that the loss exceeds the deductible.

a. $0{,}28$
b. $0{,}35$
c. $0{,}50$
d. $0{,}65$
e. $0{,}72$

≡Jawaban No. 256›

(b). $0{,}35$

Field	Isi
Topik CF2	Topik 2 — Variabel Acak Univariat
Sub-topik	2.6 Distribusi Kontinu Umum
Difficulty	Hard
Prerequisite	1.4 Probabilitas Bersyarat, 2.2 Variabel Acak Kontinu
Connected Topics	3.3 Distribusi Bersyarat
Referensi	Hogg-Tanis-Zimm Bab 3.2; Miller Bab 5

ℹRumus›

Sifat memoryless distribusi Eksponensial: distribusi reimbursement (kerugian di atas deductible) adalah Eksponensial dengan parameter yang sama.

Untuk $R = \text{reimbursement} \sim \text{Exp}(\lambda)$ :

P(R < r \mid R > 0) = 1 - e^{-\lambda r}

Diketahui:

Loss $L \sim \text{Exp}(\lambda)$ ; deductible $d = 2000$
Reimbursement $R = L - 2000 \mid L > 2000$ ; karena sifat memoryless, $R \sim \text{Exp}(\lambda)$
$P(R < 6000) = 0{,}50 \Rightarrow 1 - e^{-6000\lambda} = 0{,}50 \Rightarrow \lambda = \frac{\ln 2}{6000}$
Target: $P(3000 < R < 9000)$

▸Langkah Pengerjaan›

Langkah 1: Tentukan parameter $\lambda$

1 - e^{-6000\lambda} = 0{,}50 \implies e^{-6000\lambda} = 0{,}50 \implies \lambda = \frac{\ln 2}{6000}

Langkah 2: Hitung $P(R < 9000)$ dan $P(R < 3000)$

P(R < 9000) = 1 - e^{-9000\lambda} = 1 - e^{-\frac{9000 \ln 2}{6000}} = 1 - e^{-1{,}5\ln 2} = 1 - 2^{-1{,}5} = 1 - \frac{1}{2\sqrt{2}}

= 1 - \frac{1}{2{,}8284} \approx 1 - 0{,}3536 = 0{,}6464

P(R < 3000) = 1 - e^{-3000\lambda} = 1 - e^{-0{,}5\ln 2} = 1 - 2^{-0{,}5} = 1 - \frac{1}{\sqrt{2}} \approx 1 - 0{,}7071 = 0{,}2929

Langkah 3: Hitung probabilitas target

P(3000 < R < 9000) = P(R < 9000) - P(R < 3000) = 0{,}6464 - 0{,}2929 = 0{,}3535 \approx 0{,}35

Hasil Akhir: (b). $0{,}35$

◈Jebakan Umum›

⬡Kesalahan Konseptual›

Tidak menerapkan sifat memoryless; reimbursement kondisional pada $L > 2000$ memiliki distribusi Eksponensial dengan parameter yang sama dengan $L$ asli.
Menghitung $\lambda$ dari $P(L - 2000 < 6000 \mid L > 2000)$ dengan cara yang berbeda dari sifat memoryless.

▲Red Flags›

Jika ada deductible dan loss Eksponensial → reimbursement mengikuti Eksponensial yang sama (memoryless property).

No. 257

Let $X$ be the percentage score on a college-entrance exam for students who did not participate in an exam-preparation seminar. $X$ is modeled by a uniform distribution on $[a, 100]$ .

Let $Y$ be the percentage score on a college-entrance exam for students who did participate in an exam-preparation seminar. $Y$ is modeled by a uniform distribution on $[1{,}25a, 100]$ .

It is given that $E[X^2] = \frac{19600}{3}$ .

Calculate the 80th percentile of $Y$ .

a. $80$
b. $85$
c. $90$
d. $92$
e. $95$

≡Jawaban No. 257›

(e). $95$

Field	Isi
Topik CF2	Topik 2 — Variabel Acak Univariat
Sub-topik	2.2 Variabel Acak Kontinu, 2.6 Distribusi Kontinu Umum
Difficulty	Medium
Prerequisite	2.1 Variabel Acak Diskrit
Connected Topics	4.5 Estimasi Parameter
Referensi	Miller Bab 5; Hogg-Tanis-Zimm Bab 3

ℹRumus›

Untuk $X \sim U(\alpha, \beta)$ :

E[X] = \frac{\alpha+\beta}{2}, \quad \text{Var}(X) = \frac{(\beta-\alpha)^2}{12}, \quad E[X^2] = \text{Var}(X) + (E[X])^2

Persentil ke- $p$ : $F^{-1}(p) = \alpha + p(\beta - \alpha)$

Diketahui:

$X \sim U(a, 100)$ ; $E[X^2] = \frac{19600}{3}$
$Y \sim U(1{,}25a, 100)$
Target: persentil ke-80 dari $Y$

▸Langkah Pengerjaan›

Langkah 1: Gunakan $E[X^2]$ untuk menemukan $a$

E[X^2] = \text{Var}(X) + (E[X])^2 = \frac{(100-a)^2}{12} + \left(\frac{a+100}{2}\right)^2

Sederhanakan:

= \frac{(100-a)^2}{12} + \frac{(a+100)^2}{4} = \frac{(100-a)^2 + 3(a+100)^2}{12}

Ekspansikan pembilang:

(100-a)^2 + 3(a+100)^2 = (10000 - 200a + a^2) + 3(a^2 + 200a + 10000)

= 10000 - 200a + a^2 + 3a^2 + 600a + 30000 = 4a^2 + 400a + 40000

Jadi:

\frac{4a^2 + 400a + 40000}{12} = \frac{19600}{3}

4a^2 + 400a + 40000 = 4 \times 19600 = 78400

4a^2 + 400a - 38400 = 0 \implies a^2 + 100a - 9600 = 0

a = \frac{-100 \pm \sqrt{10000 + 38400}}{2} = \frac{-100 \pm \sqrt{48400}}{2} = \frac{-100 \pm 220}{2}

Ambil solusi positif: $a = \frac{120}{2} = 60$ .

Langkah 2: Tentukan distribusi $Y$

1{,}25a = 1{,}25 \times 60 = 75 \implies Y \sim U(75, 100)

Langkah 3: Hitung persentil ke-80

F_Y^{-1}(0{,}80) = 75 + 0{,}80(100 - 75) = 75 + 0{,}80 \times 25 = 75 + 20 = 95

Hasil Akhir: (e). $95$

◈Jebakan Umum›

⬡Kesalahan Konseptual›

Menggunakan $E[X] = \frac{19600}{3}$ sebagai momen pertama; soal menyatakan ini adalah $E[X^2]$ .
Mengambil solusi negatif $a = -160$ dari persamaan kuadrat; pastikan $a > 0$ karena ini adalah batas bawah persentase.

▲Red Flags›

Persentil ke-80 dari $U(\alpha, \beta)$ : interpolasi linear dari $\alpha$ ke $\beta$ sebesar 80% dari rentang.

No. 258

In a study of driver safety, drivers were categorized according to three risk factors. Exactly 1000 drivers exhibited each individual risk factor. Also, for each of the risk factors, there were exactly 400 drivers exhibiting that risk factor and neither of the other two risk factors. Finally, there were exactly 300 drivers who exhibited all three risk factors and 500 who exhibited none of the three risk factors.

Calculate the number of drivers in the study.

a. $2000$
b. $2300$
c. $2450$
d. $2750$
e. $3500$

≡Jawaban No. 258›

(c). $2450$

Field	Isi
Topik CF2	Topik 1 — Dasar-Dasar Probabilitas
Sub-topik	1.2 Aksioma dan Perhitungan Probabilitas, 1.3 Metode Enumerasi
Difficulty	Medium
Prerequisite	1.1 Eksperimen Acak dan Ruang Sampel
Connected Topics	1.5 Kejadian Independen
Referensi	Miller Bab 2; Hogg-Tanis-Zimm Bab 1

ℹRumus›

Prinsip inklusi-eksklusi untuk tiga himpunan $A, B, C$ :

|A \cup B \cup C| = |A| + |B| + |C| - |A \cap B| - |A \cap C| - |B \cap C| + |A \cap B \cap C|

Diketahui:

$|A| = |B| = |C| = 1000$ ; $|A \cap B \cap C| = 300$
Eksklusif (hanya satu faktor): $|A \text{ saja}| = 400$ untuk masing-masing
$|\text{tidak ada}| = 500$

▸Langkah Pengerjaan›

Langkah 1: Tentukan komponen dalam diagram Venn

Setiap lingkaran memiliki 1000 elemen, dengan 400 eksklusif dan 300 berada di irisan tiga.

Sisakan: $1000 - 400 - 300 = 300$ untuk irisan dua-atau-lebih dalam setiap lingkaran.

Biarkan $X, Y, Z$ = jumlah elemen di irisan tepat dua himpunan (misal $X = |A \cap B \text{ saja}|$ , dll).

Dari lingkaran $A$ : $400 + 300 + X + Y = 1000 \implies X + Y = 300$

Dari lingkaran $B$ : $400 + 300 + X + Z = 1000 \implies X + Z = 300$

Dari lingkaran $C$ : $400 + 300 + Y + Z = 1000 \implies Y + Z = 300$

Langkah 2: Selesaikan sistem persamaan

Dari dua persamaan pertama: $Y = Z$ ; dari kedua dan ketiga: $X = Y$ . Jadi $X = Y = Z = 150$ .

Langkah 3: Hitung total peserta

\text{Total} = 3(400) + 3(150) + 300 + 500 = 1200 + 450 + 300 + 500 = 2450

Hasil Akhir: (c). $2450$

◈Jebakan Umum›

⬡Kesalahan Konseptual›

Menjumlahkan $3 \times 1000 + 300 + 500$ tanpa memperhatikan penghitungan ganda pada irisan dua dan irisan tiga.

▲Red Flags›

Jika soal menyebut “exactly 400 exhibiting only that factor” → ini adalah wilayah eksklusif dalam diagram Venn, bukan irisan.

No. 259

An insurance company examines its pool of auto insurance customers and gathers the following information:

(i) All customers insure at least one car.
(ii) 64% of the customers insure more than one car.
(iii) 20% of the customers insure a sports car.
(iv) Of those customers who insure more than one car, 15% insure a sports car.

Calculate the probability that a randomly selected customer insures exactly one car, and that the car is not a sports car.

a. $0{,}16$
b. $0{,}19$
c. $0{,}26$
d. $0{,}29$
e. $0{,}31$

≡Jawaban No. 259›

(c). $0{,}26$

Field	Isi
Topik CF2	Topik 1 — Dasar-Dasar Probabilitas
Sub-topik	1.4 Probabilitas Bersyarat, 1.2 Aksioma dan Perhitungan Probabilitas
Difficulty	Medium
Prerequisite	1.1 Eksperimen Acak dan Ruang Sampel
Connected Topics	1.5 Kejadian Independen
Referensi	Miller Bab 2; Hogg-Tanis-Zimm Bab 1

ℹRumus›

Hukum Penjumlahan: $P(A \cup B) = P(A) + P(B) - P(A \cap B)$

Hukum Perkalian: $P(A \cap B) = P(B \mid A) \cdot P(A)$

Diketahui:

$A$ = insure lebih dari satu mobil: $P(A) = 0{,}64$ , sehingga $P(A^c) = 0{,}36$
$B$ = insure sports car: $P(B) = 0{,}20$
$P(B \mid A) = 0{,}15$
Target: $P(A^c \cap B^c)$

▸Langkah Pengerjaan›

Langkah 1: Hitung $P(A \cap B)$

P(A \cap B) = P(B \mid A) \cdot P(A) = 0{,}15 \times 0{,}64 = 0{,}096

Langkah 2: Hitung $P(A \cup B)$

P(A \cup B) = P(A) + P(B) - P(A \cap B) = 0{,}64 + 0{,}20 - 0{,}096 = 0{,}744

Langkah 3: Hitung komplementnya

P(A^c \cap B^c) = 1 - P(A \cup B) = 1 - 0{,}744 = 0{,}256 \approx 0{,}26

Hasil Akhir: (c). $0{,}26$

◈Jebakan Umum›

⬡Kesalahan Konseptual›

Mengira $P(A^c \cap B^c) = P(A^c) \cdot P(B^c)$ ; ini hanya berlaku jika $A$ dan $B$ independen, yang tidak disebutkan.
Menggunakan De Morgan: $P(A^c \cap B^c) = P((A \cup B)^c) = 1 - P(A \cup B)$ .

▲Red Flags›

Target “tepat satu mobil AND bukan sports car” = $A^c \cap B^c$ = komplemen $(A \cup B)$ .

No. 260

An insurance company has found that 1% of all applicants for life insurance have diabetes.

Calculate the probability that five or fewer of 200 randomly selected applicants have diabetes.

a. $0{,}85$
b. $0{,}88$
c. $0{,}91$
d. $0{,}95$
e. $0{,}98$

≡Jawaban No. 260›

(e). $0{,}98$

Field	Isi
Topik CF2	Topik 2 — Variabel Acak Univariat
Sub-topik	2.5 Distribusi Diskrit Umum
Difficulty	Medium
Prerequisite	2.1 Variabel Acak Diskrit
Connected Topics	4.3 Teorema Limit Pusat
Referensi	Miller Bab 5; Hogg-Tanis-Zimm Bab 2

ℹRumus›

$X \sim B(n, p)$ dengan $n$ besar dan $p$ kecil dapat diaproksimasi dengan $\text{Poisson}(\lambda = np)$ :

P(X \leq k) \approx \sum_{x=0}^{k} \frac{e^{-\lambda}\lambda^x}{x!}

Diketahui:

$X \sim B(200, 0{,}01)$ ; $\lambda = np = 200 \times 0{,}01 = 2$
Target: $P(X \leq 5)$

▸Langkah Pengerjaan›

Langkah 1: Aproksimasi Poisson

Karena $n = 200$ besar dan $p = 0{,}01$ kecil, gunakan $X \approx \text{Poisson}(\lambda = 2)$ .

Langkah 2: Hitung $P(X \leq 5)$

P(X \leq 5) = e^{-2}\sum_{x=0}^{5} \frac{2^x}{x!} = e^{-2}\!\left(1 + 2 + 2 + \frac{4}{3} + \frac{2}{3} + \frac{4}{15}\right)

= e^{-2}\!\left(1 + 2 + 2 + \frac{4}{3} + \frac{2}{3} + \frac{4}{15}\right)

Hitung suku-suku:

\frac{2^0}{0!} = 1, \quad \frac{2^1}{1!} = 2, \quad \frac{2^2}{2!} = 2, \quad \frac{2^3}{3!} = \frac{8}{6} \approx 1{,}333, \quad \frac{2^4}{4!} = \frac{16}{24} \approx 0{,}667, \quad \frac{2^5}{5!} = \frac{32}{120} \approx 0{,}267

\sum = 1 + 2 + 2 + 1{,}333 + 0{,}667 + 0{,}267 = 7{,}267

P(X \leq 5) \approx e^{-2} \times 7{,}267 = 0{,}1353 \times 7{,}267 \approx 0{,}983

Hasil Akhir: (e). $0{,}98$

◈Jebakan Umum›

⬡Kesalahan Konseptual›

Menggunakan distribusi Binomial langsung dengan $\binom{200}{x}$ yang sangat kompleks; aproksimasi Poisson jauh lebih efisien untuk $n$ besar dan $p$ kecil.
Lupa bahwa kondisi aproksimasi Poisson yang baik: $n \geq 20$ , $p \leq 0{,}05$ , dan $np$ moderat.

▲Red Flags›

Jika $n$ besar, $p$ kecil, dan ditanya probabilitas ekor bawah → gunakan Poisson dengan $\lambda = np$ .

No. 261

The probability that an agent sells an insurance policy to a potential customer during a first appointment is 0.20. The events of selling an insurance policy to different potential customers during first appointments are mutually independent.

The agent has scheduled first appointments with five potential customers.

Calculate the probability that the agent sells an insurance policy during at least two of these appointments.

a. $0{,}04$
b. $0{,}20$
c. $0{,}26$
d. $0{,}40$
e. $0{,}74$

≡Jawaban No. 261›

(c). $0{,}26$

Field	Isi
Topik CF2	Topik 2 — Variabel Acak Univariat
Sub-topik	2.5 Distribusi Diskrit Umum
Difficulty	Easy
Prerequisite	2.1 Variabel Acak Diskrit, 1.5 Kejadian Independen
Connected Topics	1.3 Metode Enumerasi
Referensi	Miller Bab 5; Hogg-Tanis-Zimm Bab 2

ℹRumus›

$X \sim B(n, p)$ :

P(X = k) = \binom{n}{k} p^k (1-p)^{n-k}

P(X \geq 2) = 1 - P(X = 0) - P(X = 1)

Diketahui:

$X \sim B(5, 0{,}20)$
Target: $P(X \geq 2)$

▸Langkah Pengerjaan›

Langkah 1: Hitung $P(X = 0)$

P(X = 0) = \binom{5}{0}(0{,}20)^0(0{,}80)^5 = (0{,}80)^5 = 0{,}32768

Langkah 2: Hitung $P(X = 1)$

P(X = 1) = \binom{5}{1}(0{,}20)^1(0{,}80)^4 = 5 \times 0{,}20 \times 0{,}4096 = 0{,}4096

Langkah 3: Hitung $P(X \geq 2)$

P(X \geq 2) = 1 - 0{,}32768 - 0{,}4096 = 1 - 0{,}73728 = 0{,}26272 \approx 0{,}26

Hasil Akhir: (c). $0{,}26$

◈Jebakan Umum›

⬡Kesalahan Konseptual›

Menghitung $P(X \geq 2) = 1 - P(X \leq 1)$ adalah benar, tetapi sering lupa menyertakan $P(X=0)$ .

▲Red Flags›

“Setidaknya dua” ( $\geq 2$ ) → pendekatan komplemen lebih efisien dari menghitung $P(2) + P(3) + P(4) + P(5)$ .

No. 262

A manufacturer produces computers and releases them in shipments of 100. From a shipment of 100, the probability that exactly three computers are defective is twice the probability that exactly two computers are defective. The events that different computers are defective are mutually independent.

Calculate the probability that a randomly selected computer is defective.

a. $0{,}040$
b. $0{,}042$
c. $0{,}058$
d. $0{,}060$
e. $0{,}072$

≡Jawaban No. 262›

(c). $0{,}058$

Field	Isi
Topik CF2	Topik 2 — Variabel Acak Univariat
Sub-topik	2.5 Distribusi Diskrit Umum
Difficulty	Medium
Prerequisite	2.1 Variabel Acak Diskrit, 1.5 Kejadian Independen
Connected Topics	1.3 Metode Enumerasi
Referensi	Miller Bab 5; Hogg-Tanis-Zimm Bab 2

ℹRumus›

$X \sim B(n, p)$ :

P(X = k) = \binom{n}{k} p^k (1-p)^{n-k}

Diketahui:

$X \sim B(100, p)$ ; $P(X = 3) = 2 \cdot P(X = 2)$
Target: nilai $p$

▸Langkah Pengerjaan›

Langkah 1: Tulis persamaan dari kondisi yang diberikan

\binom{100}{3} p^3 (1-p)^{97} = 2 \binom{100}{2} p^2 (1-p)^{98}

Langkah 2: Sederhanakan

Bagi kedua sisi dengan $p^2(1-p)^{97}$ (karena $p \neq 0$ dan $p \neq 1$ ):

\binom{100}{3} p = 2 \binom{100}{2} (1-p)

161700 \cdot p = 2 \times 4950 \times (1-p) = 9900(1-p)

161700p = 9900 - 9900p

171600p = 9900

p = \frac{9900}{171600} \approx 0{,}0577 \approx 0{,}058

Hasil Akhir: (c). $0{,}058$

◈Jebakan Umum›

⬡Kesalahan Konseptual›

Lupa menyederhanakan faktor $p^2(1-p)^{97}$ dari kedua sisi sebelum menyelesaikan; persamaan menjadi rumit tidak perlu.
Mengira “tiga kali defektif = dua kali lipat peluang dua defektif” adalah interpretasi yang berbeda dari $P(X=3) = 2P(X=2)$ .

▲Red Flags›

Jika kondisi berupa rasio probabilitas dua nilai distribusi → tulis rasio, sederhanakan, lalu selesaikan untuk parameter yang tidak diketahui.

No. 263

In any 12-month period, the probability that a home is damaged by fire is 20% and the probability of a theft loss at a home is 30%. The occurrences of fire damage and theft loss are independent events.

Calculate the probability that a randomly selected home will either be damaged by fire or will have a theft loss, but not both, during the next year.

a. $0{,}30$
b. $0{,}38$
c. $0{,}44$
d. $0{,}50$
e. $0{,}56$

≡Jawaban No. 263›

(b). $0{,}38$

Field	Isi
Topik CF2	Topik 1 — Dasar-Dasar Probabilitas
Sub-topik	1.5 Kejadian Independen, 1.2 Aksioma dan Perhitungan Probabilitas
Difficulty	Easy
Prerequisite	1.1 Eksperimen Acak dan Ruang Sampel
Connected Topics	1.4 Probabilitas Bersyarat
Referensi	Miller Bab 2; Hogg-Tanis-Zimm Bab 1

ℹRumus›

Probabilitas “salah satu tetapi tidak keduanya” (XOR):

P(A \triangle B) = P(A \cap B^c) + P(A^c \cap B) = P(A)(1-P(B)) + (1-P(A))P(B)

Diketahui:

$F$ = kebakaran: $P(F) = 0{,}20$ ; $T$ = pencurian: $P(T) = 0{,}30$ ; $F$ dan $T$ independen
Target: $P(F \triangle T)$

▸Langkah Pengerjaan›

Langkah 1: Hitung probabilitas “hanya kebakaran”

P(F \cap T^c) = P(F) \cdot P(T^c) = 0{,}20 \times 0{,}70 = 0{,}14

Langkah 2: Hitung probabilitas “hanya pencurian”

P(F^c \cap T) = P(F^c) \cdot P(T) = 0{,}80 \times 0{,}30 = 0{,}24

Langkah 3: Jumlahkan

P(F \triangle T) = 0{,}14 + 0{,}24 = 0{,}38

Hasil Akhir: (b). $0{,}38$

◈Jebakan Umum›

⬡Kesalahan Konseptual›

Menghitung $P(F \cup T) = P(F) + P(T) - P(F \cap T) = 0{,}44$ untuk menjawab “keduanya atau salah satu” — ini bukan “salah satu tapi tidak keduanya”.
“Either… or… but not both” = exclusive OR ( $\triangle$ ), bukan inclusive OR ( $\cup$ ).

▲Red Flags›

“But not both” → gunakan XOR, bukan $\cup$ . XOR = $P(A \cup B) - P(A \cap B)$ .

No. 264

In one company, 30% of males and 20% of females contribute to a supplemental retirement plan. Furthermore, 45% of the company’s employees are female.

Calculate the probability that a randomly selected employee is female, given that this employee contributes to a supplemental retirement plan.

a. $0{,}09$
b. $0{,}23$
c. $0{,}35$
d. $0{,}45$
e. $0{,}55$

≡Jawaban No. 264›

(c). $0{,}35$

Field	Isi
Topik CF2	Topik 1 — Dasar-Dasar Probabilitas
Sub-topik	1.6 Teorema Bayes dan Hukum Probabilitas Total
Difficulty	Easy
Prerequisite	1.4 Probabilitas Bersyarat
Connected Topics	1.5 Kejadian Independen
Referensi	Miller Bab 2; Hogg-Tanis-Zimm Bab 1

ℹRumus›

Teorema Bayes:

P(F \mid C) = \frac{P(C \mid F) \cdot P(F)}{P(C \mid F) \cdot P(F) + P(C \mid F^c) \cdot P(F^c)}

Diketahui:

$P(F) = 0{,}45$ , $P(F^c) = 0{,}55$
$P(C \mid F) = 0{,}20$ (wanita berkontribusi)
$P(C \mid F^c) = 0{,}30$ (pria berkontribusi)
Target: $P(F \mid C)$

▸Langkah Pengerjaan›

Langkah 1: Hitung $P(C)$ dengan Hukum Probabilitas Total

P(C) = P(C \mid F) \cdot P(F) + P(C \mid F^c) \cdot P(F^c) = 0{,}20(0{,}45) + 0{,}30(0{,}55) = 0{,}09 + 0{,}165 = 0{,}255

Langkah 2: Terapkan Teorema Bayes

P(F \mid C) = \frac{0{,}20 \times 0{,}45}{0{,}255} = \frac{0{,}09}{0{,}255} = 0{,}353 \approx 0{,}35

Hasil Akhir: (c). $0{,}35$

◈Jebakan Umum›

⬡Kesalahan Konseptual›

Menjawab $P(F) = 0{,}45$ karena itu data yang tersedia; tetapi soal bertanya probabilitas bersyarat $P(F \mid C)$ .
Menggunakan hanya pembilang $P(C \mid F) \cdot P(F)$ tanpa normalisasi dengan $P(C)$ .

▲Red Flags›

“Probabilitas bahwa… given that…” selalu mengisyaratkan penggunaan Teorema Bayes atau probabilitas bersyarat langsung.

No. 265

A health insurer sells policies to residents of territory X and territory Y. Past claims experience indicates the following:

(i) 20% of the total policyholders from territory X and territory Y combined filed no claims.
(ii) 15% of the policyholders from territory X filed no claims.
(iii) 40% of the policyholders from territory Y filed no claims.

Calculate the probability that a randomly selected policyholder was a resident of territory X, given that the policyholder filed no claims.

a. $0{,}09$
b. $0{,}27$
c. $0{,}50$
d. $0{,}60$
e. $0{,}80$

≡Jawaban No. 265›

(d). $0{,}60$

Field	Isi
Topik CF2	Topik 1 — Dasar-Dasar Probabilitas
Sub-topik	1.6 Teorema Bayes dan Hukum Probabilitas Total
Difficulty	Medium
Prerequisite	1.4 Probabilitas Bersyarat
Connected Topics	1.5 Kejadian Independen
Referensi	Miller Bab 2; Hogg-Tanis-Zimm Bab 1

ℹRumus›

Hukum Probabilitas Total:

P(C) = P(C \mid X) \cdot P(X) + P(C \mid Y) \cdot P(Y)

Teorema Bayes:

P(X \mid C) = \frac{P(C \mid X) \cdot P(X)}{P(C)}

Diketahui:

$C$ = tidak ada klaim; $P(C) = 0{,}20$
$P(C \mid X) = 0{,}15$ ; $P(C \mid Y) = 0{,}40$
$P(X) =$ tidak diketahui; $P(Y) = 1 - P(X)$

▸Langkah Pengerjaan›

Langkah 1: Tentukan $P(X)$ menggunakan Hukum Probabilitas Total

P(C) = 0{,}15 \cdot P(X) + 0{,}40 \cdot (1 - P(X))

0{,}20 = 0{,}15 \cdot P(X) + 0{,}40 - 0{,}40 \cdot P(X)

0{,}20 - 0{,}40 = -0{,}25 \cdot P(X)

P(X) = \frac{0{,}20}{0{,}25} = 0{,}80

Langkah 2: Terapkan Teorema Bayes

P(X \mid C) = \frac{P(C \mid X) \cdot P(X)}{P(C)} = \frac{0{,}15 \times 0{,}80}{0{,}20} = \frac{0{,}12}{0{,}20} = 0{,}60

Hasil Akhir: (d). $0{,}60$

◈Jebakan Umum›

⬡Kesalahan Konseptual›

Mengira $P(X)$ sudah diketahui atau sama dengan 0,5; soal ini memerlukan penentuan $P(X)$ melalui Hukum Probabilitas Total.
Lupa langkah awal: selalu cari $P(X)$ dulu sebelum menerapkan Bayes.

▲Red Flags›

Jika proporsi kelompok tidak diberikan secara langsung tetapi probabilitas marginal ( $P(C)$ ) diberikan → gunakan Hukum Probabilitas Total untuk mencari proporsi kelompok.

No. 266

Claim amounts are independent random variables with probability density function

f(x) = \begin{cases} \dfrac{10}{x^2}, & x > 10 \\ 0, & \text{selainnya} \end{cases}

Calculate the probability that the largest of three randomly selected claims is less than 25.

a. $\dfrac{8}{125}$
b. $\dfrac{12}{125}$
c. $\dfrac{27}{125}$
d. $\dfrac{2}{5}$
e. $\dfrac{3}{5}$

≡Jawaban No. 266›

(c). $\dfrac{27}{125}$

Field	Isi
Topik CF2	Topik 3 — Variabel Acak Multivariat
Sub-topik	3.8 Transformasi Variabel Acak Gabungan
Difficulty	Medium
Prerequisite	2.2 Variabel Acak Kontinu, 3.5 Independensi dan Korelasi
Connected Topics	2.4 Transformasi Variabel Acak Univariat
Referensi	Hogg-Tanis-Zimm Bab 4.4; Miller Bab 6

ℹRumus›

CDF untuk satu klaim: $F(x) = P(X \leq x) = \int_{10}^{x} \frac{10}{t^2}\,dt$

Untuk $n$ variabel i.i.d., CDF maksimum: $P(\max \leq x) = [F(x)]^n$

Diketahui:

$X_1, X_2, X_3$ i.i.d. dengan PDF di atas; $n = 3$
Target: $P(\max(X_1, X_2, X_3) < 25)$

▸Langkah Pengerjaan›

Langkah 1: Hitung $F(25)$

F(25) = \int_{10}^{25} \frac{10}{x^2}\,dx = \left[-\frac{10}{x}\right]_{10}^{25} = -\frac{10}{25} + \frac{10}{10} = -0{,}4 + 1 = 0{,}6 = \frac{3}{5}

Langkah 2: Gunakan CDF maksimum

P(\max(X_1, X_2, X_3) < 25) = [F(25)]^3 = \left(\frac{3}{5}\right)^3 = \frac{27}{125}

Hasil Akhir: (c). $\dfrac{27}{125}$

◈Jebakan Umum›

⬡Kesalahan Konseptual›

Menggunakan $P(\min < 25)$ alih-alih $P(\max < 25)$ ; untuk maksimum, semua variabel harus $< 25$ .
Menjawab $F(25) = 3/5$ tanpa mengkuadratkannya sebanyak $n = 3$ .

▲Red Flags›

$P(\max \leq x) = [F(x)]^n$ ; $P(\min > x) = [1-F(x)]^n$ untuk variabel i.i.d.

No. 267

The lifetime of a certain electronic device has an exponential distribution with mean 0.50.

Calculate the probability that the lifetime of the device is greater than 0.70, given that it is greater than 0.40.

a. $0{,}203$
b. $0{,}247$
c. $0{,}449$
d. $0{,}549$
e. $0{,}861$

≡Jawaban No. 267›

(d). $0{,}549$

Field	Isi
Topik CF2	Topik 2 — Variabel Acak Univariat
Sub-topik	2.6 Distribusi Kontinu Umum
Difficulty	Easy
Prerequisite	1.4 Probabilitas Bersyarat, 2.2 Variabel Acak Kontinu
Connected Topics	3.3 Distribusi Bersyarat
Referensi	Hogg-Tanis-Zimm Bab 3.2; Miller Bab 5

ℹRumus›

Sifat memoryless distribusi Eksponensial:

P(X > s + t \mid X > s) = P(X > t)

Untuk $X \sim \text{Exp}(\beta = 0{,}5)$ , rate $\lambda = 2$ :

P(X > t) = e^{-\lambda t} = e^{-2t}

Diketahui:

$X \sim \text{Exp}(\beta = 0{,}5)$ , rate $\lambda = 2$
Target: $P(X > 0{,}70 \mid X > 0{,}40)$

▸Langkah Pengerjaan›

Langkah 1: Terapkan sifat memoryless

P(X > 0{,}70 \mid X > 0{,}40) = P(X > 0{,}70 - 0{,}40) = P(X > 0{,}30)

Langkah 2: Hitung $P(X > 0{,}30)$

P(X > 0{,}30) = e^{-2 \times 0{,}30} = e^{-0{,}60} \approx 0{,}5488 \approx 0{,}549

Hasil Akhir: (d). $0{,}549$

◈Jebakan Umum›

⬡Kesalahan Konseptual›

Menghitung $\frac{P(X > 0{,}70)}{P(X > 0{,}40)}$ secara langsung (benar, tetapi lebih panjang) tanpa memanfaatkan memoryless property.
Menggunakan mean $\beta = 0{,}5$ sebagai rate; rate $\lambda = 1/\beta = 2$ .

▲Red Flags›

Jika soal mengkondisikan “lifetime > $s$ ” dan bertanya “probability > $s+t$ ” → selalu gunakan memoryless property untuk Eksponensial.

No. 268

A farmer purchases a five-year insurance policy that covers crop destruction due to hail. Over the five-year period, the farmer will receive a benefit of 20 for each year in which hail destroys his crop, subject to a maximum of three benefit payments. The probability that hail will destroy the farmer’s crop in any given year is 0.5, independent of any other year.

Calculate the expected benefit that the farmer will receive over the five-year period.

a. $30$
b. $34$
c. $40$
d. $46$
e. $50$

≡Jawaban No. 268›

(d). $46$

Field	Isi
Topik CF2	Topik 2 — Variabel Acak Univariat
Sub-topik	2.5 Distribusi Diskrit Umum, 2.1 Variabel Acak Diskrit
Difficulty	Hard
Prerequisite	1.5 Kejadian Independen, 1.3 Metode Enumerasi
Connected Topics	3.4 Nilai Harapan dan Variansi Bersyarat
Referensi	Hogg-Tanis-Zimm Bab 2; Miller Bab 5

ℹRumus›

$X \sim B(5, 0{,}5)$ = jumlah tahun dengan klaim. Manfaat aktual:

B = 20 \cdot \min(X, 3)

E[B] = \sum_{x=0}^{5} 20\min(x, 3) \cdot P(X = x)

Diketahui:

$X \sim B(5, 0{,}5)$ ; $P(X = k) = \binom{5}{k}(0{,}5)^5 = \frac{\binom{5}{k}}{32}$
Manfaat per pembayaran = 20; maksimum 3 pembayaran; total manfaat = $20\min(X, 3)$

▸Langkah Pengerjaan›

Langkah 1: Hitung probabilitas setiap nilai $X$

P(X=0) = \frac{1}{32}, \quad P(X=1) = \frac{5}{32}, \quad P(X=2) = \frac{10}{32}

P(X=3) = \frac{10}{32}, \quad P(X=4) = \frac{5}{32}, \quad P(X=5) = \frac{1}{32}

Langkah 2: Hitung $E[B] = E[20\min(X,3)]$

Manfaat aktual untuk setiap nilai:

$X$	$20\min(X,3)$	$P(X)$	Kontribusi
0	0	1/32	0
1	20	5/32	100/32
2	40	10/32	400/32
3	60	10/32	600/32
4	60	5/32	300/32
5	60	1/32	60/32

E[B] = \frac{0 + 100 + 400 + 600 + 300 + 60}{32} = \frac{1460}{32} = 45{,}625 \approx 46

Hasil Akhir: (d). $46$

◈Jebakan Umum›

⬡Kesalahan Konseptual›

Menghitung $E[20X] = 20 \times E[X] = 20 \times 2{,}5 = 50$ tanpa memperhatikan batas tiga pembayaran.
Lupa bahwa $X \geq 3$ semua memberikan manfaat yang sama (60), bukan $20X$ .

▲Red Flags›

“Subject to a maximum of $N$ payments” → gunakan $\min(X, N)$ dalam fungsi manfaat, bukan $X$ langsung.

No. 269

An insurance company has two divisions, auto and property. Total annual claims, $X$ , in the auto division follow a normal distribution with mean 10 and standard deviation 3. Total annual claims, $Y$ , in the property division follow a normal distribution with mean 12 and standard deviation 4.

Assume that $X$ and $Y$ are independent.

Calculate the probability that total overall claims, $X + Y$ , will not exceed 29.

a. $0{,}61$
b. $0{,}69$
c. $0{,}78$
d. $0{,}84$
e. $0{,}92$

≡Jawaban No. 269›

(e). $0{,}92$

Field	Isi
Topik CF2	Topik 3 — Variabel Acak Multivariat
Sub-topik	3.5 Independensi dan Korelasi
Difficulty	Easy
Prerequisite	2.6 Distribusi Kontinu Umum
Connected Topics	4.3 Teorema Limit Pusat
Referensi	Miller Bab 6; Hogg-Tanis-Zimm Bab 5.5

ℹRumus›

Jumlah dua Normal independen:

X + Y \sim N(\mu_X + \mu_Y,\; \sigma_X^2 + \sigma_Y^2)

Diketahui:

$X \sim N(10, 9)$ ; $Y \sim N(12, 16)$ ; independen
Target: $P(X + Y \leq 29)$

▸Langkah Pengerjaan›

Langkah 1: Distribusi $X + Y$

X + Y \sim N(10 + 12,\; 9 + 16) = N(22, 25)

\sigma_{X+Y} = \sqrt{25} = 5

Langkah 2: Standarisasi

P(X + Y \leq 29) = P\!\left(Z \leq \frac{29 - 22}{5}\right) = P(Z \leq 1{,}4) = \Phi(1{,}4) \approx 0{,}9192 \approx 0{,}92

Hasil Akhir: (e). $0{,}92$

◈Jebakan Umum›

⬡Kesalahan Konseptual›

Menjumlahkan deviasi standar: $\sigma_{X+Y} \neq \sigma_X + \sigma_Y = 7$ ; yang benar adalah variansi yang dijumlahkan.
Menggunakan $\sigma^2 = 3^2 + 4^2 = 9 + 16 = 25$ sudah benar; standar deviasi $= \sqrt{25} = 5$ , bukan 7.

▲Red Flags›

Untuk dua Normal independen: jumlahkan variansi, bukan standar deviasi.

No. 270

An industrial company provides health insurance to employees located at four different plants. Health insurance costs at each plant are independent of the costs at any other plant. Plant managers have calculated the following statistics:

Plant	Average Cost	Standard Deviation
W	2	1.0
X	2	1.0
Y	5	1.5
Z	7	2.0

Calculate the standard deviation of total company health insurance costs.

a. $1{,}4$
b. $2{,}1$
c. $2{,}9$
d. $5{,}5$
e. $6{,}3$

≡Jawaban No. 270›

(c). $2{,}9$

Field	Isi
Topik CF2	Topik 3 — Variabel Acak Multivariat
Sub-topik	3.5 Independensi dan Korelasi
Difficulty	Easy
Prerequisite	2.2 Variabel Acak Kontinu
Connected Topics	3.6 Matriks Variansi-Kovariansi
Referensi	Miller Bab 4; Hogg-Tanis-Zimm Bab 3

ℹRumus›

Untuk variabel-variabel independen, variansi total = jumlah variansi:

\text{Var}(W + X + Y + Z) = \text{Var}(W) + \text{Var}(X) + \text{Var}(Y) + \text{Var}(Z)

Diketahui:

$\sigma_W = 1{,}0$ , $\sigma_X = 1{,}0$ , $\sigma_Y = 1{,}5$ , $\sigma_Z = 2{,}0$
Semua biaya independen

▸Langkah Pengerjaan›

Langkah 1: Hitung variansi setiap plant

\text{Var}(W) = 1{,}0^2 = 1{,}00, \quad \text{Var}(X) = 1{,}0^2 = 1{,}00

\text{Var}(Y) = 1{,}5^2 = 2{,}25, \quad \text{Var}(Z) = 2{,}0^2 = 4{,}00

Langkah 2: Jumlahkan variansi

\text{Var}(\text{Total}) = 1{,}00 + 1{,}00 + 2{,}25 + 4{,}00 = 8{,}25

Langkah 3: Ambil akar kuadrat

\sigma_{\text{Total}} = \sqrt{8{,}25} \approx 2{,}872 \approx 2{,}9

Hasil Akhir: (c). $2{,}9$

◈Jebakan Umum›

⬡Kesalahan Konseptual›

Menjumlahkan standar deviasi langsung: $1{,}0 + 1{,}0 + 1{,}5 + 2{,}0 = 5{,}5$ ; ini adalah jawaban pengecoh (opsi d).
Mengingat kembali: standar deviasi tidak bersifat aditif; variansi yang aditif untuk variabel independen.

▲Red Flags›

Jika diminta standar deviasi total dari variabel-variabel independen → jumlahkan variansi, lalu ambil akar. Jangan jumlahkan standar deviasinya.