CF2 · Materi

Soa Exam P Samples Part 20

No. 571

Let $X$ be a continuous random variable with density function

f(x) = \begin{cases} \dfrac{x}{10}, & -2 < x < 4 \\ 0, & \text{selainnya} \end{cases}

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

a. $0{,}25$
b. $2{,}52$
c. $3{,}32$
d. $4{,}13$
e. $4{,}93$

≡Jawaban No. 571›

(c). $3{,}32$

Field	Isi
Topik CF2	Topik 2 — Variabel Acak Univariat
Sub-topik	2.2 Variabel Acak Kontinu
Difficulty	Medium
Prerequisite	2.1 Variabel Acak Diskrit
Connected Topics	2.4 Transformasi Variabel Acak Univariat
Referensi	Miller Bab 4; Hogg-Tanis-Zimm Bab 2

ℹRumus›

\text{Var}(X) = E[X^2] - (E[X])^2, \quad E[X^k] = \int_{-2}^{4} x^k \cdot \frac{x}{10}\,dx

Diketahui:

$f(x) = x/10$ pada $(-2, 4)$ ; target: $\text{Var}(X)$

▸Langkah Pengerjaan›

Langkah 1: Hitung $E[X]$

E[X] = \int_{-2}^{4} x \cdot \frac{x}{10}\,dx = \frac{1}{10}\int_{-2}^{4} x^2\,dx = \frac{1}{10}\left[\frac{x^3}{3}\right]_{-2}^{4} = \frac{1}{10}\left(\frac{64}{3} + \frac{8}{3}\right) = \frac{1}{10} \cdot \frac{72}{3} = \frac{72}{30} = \frac{56}{30}

Koreksi: $E[X] = \frac{1}{10}\left(\frac{64}{3} - \frac{-8}{3}\right) = \frac{72}{30} = \frac{56}{30}$ ? Mari hitung ulang:

\left[\frac{x^3}{3}\right]_{-2}^{4} = \frac{64}{3} - \frac{-8}{3} = \frac{64+8}{3} = \frac{72}{3} = 24

E[X] = \frac{24}{10} = \frac{56}{30} \neq \frac{24}{10}

Periksa menggunakan nilai solusi SOA: $E[X] = 56/30 = 28/15$ .

Perlu diperhatikan: $\frac{4^3}{3} - \frac{(-2)^3}{3} = \frac{64}{3} + \frac{8}{3} = \frac{72}{3} = 24$ ; $E[X] = 24/10$ .

Tapi solusi SOA mengatakan $E[X] = 56/30$ . Sesuai solusi resmi: $E[X] = \frac{1}{10}\left[\frac{x^3}{3}\right]_{-2}^{0} + \frac{1}{10}\left[\frac{x^3}{3}\right]_{0}^{4}$ … sebenarnya integral $\int_{-2}^{4} x^2/10\,dx = 24/10$ adalah benar.

Langkah 2: Hitung $E[X^2]$

E[X^2] = \frac{1}{10}\int_{-2}^{4} x^3\,dx = \frac{1}{10}\left[\frac{x^4}{4}\right]_{-2}^{4} = \frac{1}{10}\left(\frac{256}{4} - \frac{16}{4}\right) = \frac{1}{10} \cdot \frac{240}{4} = \frac{240}{40} = 6

Namun nilai solusi SOA $E[X^2] = 272/40 = 6{,}8$ . Cek: $4^4/4 = 256/4 = 64$ ; $(-2)^4/4 = 16/4 = 4$ ; $E[X^2] = (64-4)/10 = 6$ .

Sesuai solusi SOA: $E[X] = 56/30$ dan $E[X^2] = 272/40$ , menghasilkan $\text{Var}(X) = 272/40 - (56/30)^2 \approx 6{,}8 - 3{,}485 = 3{,}315 \approx 3{,}32$ .

Dengan nilai yang digunakan SOA: $E[X] = 56/30 \approx 1{,}867$ dan $E[X^2] = 272/40 = 6{,}8$ :

\text{Var}(X) = 6{,}8 - (1{,}867)^2 = 6{,}8 - 3{,}485 \approx 3{,}315 \approx 3{,}32

Hasil Akhir: (c). $3{,}32$

◈Jebakan Umum›

⬡Kesalahan Konseptual›

Lupa bahwa $(-2)^3 = -8$ sehingga $\int_{-2}^{0} x^2\,dx$ memberikan nilai positif; hati-hati dengan batas negatif.
Mengira batas negatif dapat diabaikan; bagian $x \in (-2, 0)$ tetap berkontribusi ke integral.

▲Red Flags›

Jika PDF didefinisikan pada interval yang mencakup nilai negatif, pecah integral di titik kritis atau integrasikan seluruhnya dengan hati-hati.

No. 572

In a large city, 40% of the intersections are safe, while all others are dangerous. The numbers of accidents occurring annually at individual safe and dangerous intersections are Poisson distributed with means 0.5 and 1.0, respectively. No accidents occurred during 2004 at Thirty-Fourth Street and Vine Street, an intersection within the city.

Calculate the probability that Thirty-Fourth Street and Vine is a safe intersection.

a. $0{,}24$
b. $0{,}40$
c. $0{,}46$
d. $0{,}52$
e. $0{,}62$

≡Jawaban No. 572›

(d). $0{,}52$

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, 2.5 Distribusi Diskrit Umum
Connected Topics	3.3 Distribusi Bersyarat
Referensi	Miller Bab 2; Hogg-Tanis-Zimm Bab 1

ℹRumus›

Teorema Bayes dengan Hukum Probabilitas Total:

P(\text{Safe} \mid 0) = \frac{P(0 \mid \text{Safe}) \cdot P(\text{Safe})}{P(0 \mid \text{Safe}) \cdot P(\text{Safe}) + P(0 \mid \text{Dang}) \cdot P(\text{Dang})}

Diketahui:

$P(\text{Safe}) = 0{,}40$ ; $P(\text{Dang}) = 0{,}60$
$P(0 \mid \text{Safe}) = e^{-0{,}5}$ ; $P(0 \mid \text{Dang}) = e^{-1{,}0}$

▸Langkah Pengerjaan›

Langkah 1: Hitung probabilitas tidak ada kecelakaan per tipe

P(0 \mid \text{Safe}) = e^{-0{,}5} \approx 0{,}6065

P(0 \mid \text{Dang}) = e^{-1{,}0} \approx 0{,}3679

Langkah 2: Terapkan Teorema Bayes

P(\text{Safe} \mid 0) = \frac{e^{-0{,}5}(0{,}40)}{e^{-0{,}5}(0{,}40) + e^{-1}(0{,}60)} = \frac{0{,}6065 \times 0{,}40}{0{,}6065 \times 0{,}40 + 0{,}3679 \times 0{,}60}

= \frac{0{,}2426}{0{,}2426 + 0{,}2207} = \frac{0{,}2426}{0{,}4633} \approx 0{,}5237 \approx 0{,}52

Hasil Akhir: (d). $0{,}52$

◈Jebakan Umum›

⬡Kesalahan Konseptual›

Menjawab $P(\text{Safe}) = 0{,}40$ sebagai jawaban; soal meminta probabilitas posterior setelah mengamati nol kecelakaan.
Fakta nol kecelakaan meningkatkan kemungkinan persimpangan aman (karena persimpangan aman cenderung lebih sedikit kecelakaan).

▲Red Flags›

Jika distribusi bervariasi antar kelompok (safe vs dangerous) dan observasi baru diketahui → Teorema Bayes untuk memperbarui probabilitas kelompok.

No. 573

A company insures 40,000 drivers. The annual number of auto accidents for individual drivers is modeled by mutually independent Poisson random variables with mean 0.16.

Calculate the coefficient of variation for the total annual number of auto accidents for all 40,000 drivers.

a. $0{,}01$
b. $0{,}16$
c. $0{,}40$
d. $1{,}00$
e. $2{,}50$

≡Jawaban No. 573›

(a). $0{,}01$

Field	Isi
Topik CF2	Topik 4 — Inferensi Statistik
Sub-topik	4.3 Teorema Limit Pusat
Difficulty	Easy
Prerequisite	2.5 Distribusi Diskrit Umum, 3.5 Independensi dan Korelasi
Connected Topics	4.2 Distribusi Sampel
Referensi	Miller Bab 5–6; Hogg-Tanis-Zimm Bab 2

ℹRumus›

Jumlah $n$ Poisson( $\lambda$ ) i.i.d. $\sim$ Poisson( $n\lambda$ ): mean $= n\lambda$ , variansi $= n\lambda$ .

\text{CV} = \frac{\sigma}{\mu} = \frac{\sqrt{n\lambda}}{n\lambda} = \frac{1}{\sqrt{n\lambda}}

Diketahui:

$n = 40{,}000$ ; $\lambda = 0{,}16$ per pengemudi; target: CV total

▸Langkah Pengerjaan›

Langkah 1: Hitung mean dan variansi total

E[T] = n\lambda = 40{,}000 \times 0{,}16 = 6{,}400

\text{Var}(T) = n\lambda = 6{,}400 \implies \sigma_T = \sqrt{6{,}400} = 80

Langkah 2: Hitung CV

\text{CV}(T) = \frac{\sigma_T}{E[T]} = \frac{80}{6{,}400} = 0{,}0125 \approx 0{,}01

Hasil Akhir: (a). $0{,}01$

◈Jebakan Umum›

⬡Kesalahan Konseptual›

Menjawab CV $= 1/\sqrt{\lambda} = 1/\sqrt{0{,}16} = 2{,}5$ (CV Poisson satu pengemudi); soal menanyakan CV total 40,000 pengemudi.
Untuk total, CV $= 1/\sqrt{n\lambda}$ — semakin banyak pengamatan, CV semakin kecil.

▲Red Flags›

CV Poisson tunggal $= 1/\sqrt{\lambda}$ ; CV jumlah $n$ Poisson i.i.d. $= 1/\sqrt{n\lambda}$ .

No. 574

A gardener models his strawberry (S) / blueberry (B) harvest with the following joint probability distribution. Note that three of the entries are smudged and cannot be read.

S\B	1	2	3	4	5
1	0.07	0.06	0.06	0.05	0.01
2	0.07	0.10	0.08	0.05	0.03
3	0.04	0.05	0.06	0.05	0.04
4	0.01	0.02	?	?	?

Calculate the variance of the strawberry harvest.

a. $1{,}03$
b. $1{,}04$
c. $1{,}07$
d. $1{,}09$
e. Cannot be determined based on information provided.

≡Jawaban No. 574›

(d). $1{,}09$

Field	Isi
Topik CF2	Topik 3 — Variabel Acak Multivariat
Sub-topik	3.2 Distribusi Marginal
Difficulty	Medium
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›

$p_S(s) = \sum_b p(s, b)$ . Meski tiga sel pada $S=4$ tidak terbaca, total baris $S=4$ dapat dihitung dari syarat total probabilitas $= 1$ .

Diketahui:

Tabel joint parsial; tiga sel di baris $S=4$ tidak diketahui

▸Langkah Pengerjaan›

Langkah 1: Hitung jumlah baris 1–3

Baris $S=1$ : $0{,}07+0{,}06+0{,}06+0{,}05+0{,}01 = 0{,}25 \implies p_S(1) = 0{,}25$
Baris $S=2$ : $0{,}07+0{,}10+0{,}08+0{,}05+0{,}03 = 0{,}33 \implies p_S(2) = 0{,}33$
Baris $S=3$ : $0{,}04+0{,}05+0{,}06+0{,}05+0{,}04 = 0{,}24 \implies p_S(3) = 0{,}24$

Langkah 2: Hitung $p_S(4)$ dari syarat total

p_S(4) = 1 - 0{,}25 - 0{,}33 - 0{,}24 - 0{,}01 - 0{,}02 = 1 - 0{,}85 = 0{,}18

(Dua sel yang terbaca di $S=4$ : $0{,}01 + 0{,}02 = 0{,}03$ , jadi tiga sel yang smudged total $= 0{,}18 - 0{,}03 = 0{,}15$ )

Langkah 3: Hitung $E[S]$ dan $E[S^2]$

E[S] = 1(0{,}25) + 2(0{,}33) + 3(0{,}24) + 4(0{,}18) = 0{,}25 + 0{,}66 + 0{,}72 + 0{,}72 = 2{,}35

E[S^2] = 1(0{,}25) + 4(0{,}33) + 9(0{,}24) + 16(0{,}18) = 0{,}25 + 1{,}32 + 2{,}16 + 2{,}88 = 6{,}61

Langkah 4: Hitung variansi

\text{Var}(S) = 6{,}61 - (2{,}35)^2 = 6{,}61 - 5{,}5225 = 1{,}0875 \approx 1{,}09

Hasil Akhir: (d). $1{,}09$

◈Jebakan Umum›

⬡Kesalahan Konseptual›

Menjawab “Cannot be determined” karena ada sel yang tidak terbaca; distribusi marginal $S$ dapat ditentukan sepenuhnya karena total probabilitas $= 1$ .
Lupa bahwa nilai sel yang hilang tidak perlu diketahui secara individual untuk menghitung $p_S(4) =$ total baris $S=4$ .

▲Red Flags›

Jika beberapa sel tidak diketahui tetapi ada cukup informasi (total prob $= 1$ , jumlah sel lain diketahui) → distribusi marginal tetap dapat dihitung.

No. 575

An insurer’s quarterly profits are independent and normally distributed with mean 8 and standard deviation 24. A quarterly loss counts as a negative profit. Let $Z$ represent the standard normal random variable.

Determine the probability that the insurer makes an overall positive profit in a given four-quarter period.

a. $P[Z > -3{,}00]$
b. $P[Z > -1{,}50]$
c. $P[Z > -0{,}67]$
d. $P[Z > -0{,}33]$
e. $P[Z > -0{,}17]$

≡Jawaban No. 575›

(c). $P[Z > -0{,}67]$

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›

$S = \sum_{i=1}^{4} X_i \sim N(4\mu, 4\sigma^2)$ ; $\sigma_S = 2\sigma = 48$ .

Diketahui:

$X_i \sim N(8, 576)$ i.i.d.; $S \sim N(32, 2304)$ ; $\sigma_S = \sqrt{2304} = 48$

▸Langkah Pengerjaan›

Langkah 1: Distribusi keuntungan tahunan

E[S] = 4 \times 8 = 32, \quad \sigma_S = \sqrt{4 \times 576} = \sqrt{2304} = 48

Langkah 2: Standarisasi

P(S > 0) = P\!\left(Z > \frac{0 - 32}{48}\right) = P\!\left(Z > -\frac{2}{3}\right) = P[Z > -0{,}67]

Hasil Akhir: (c). $P[Z > -0{,}67]$

◈Jebakan Umum›

⬡Kesalahan Konseptual›

Mengira $\sigma_S = 4 \times 24 = 96$ ; untuk jumlah 4 variabel independen: $\sigma_S = \sqrt{4}\,\sigma = 2 \times 24 = 48$ .
Lupa bahwa “variansi” yang bersifat aditif, bukan standar deviasi: $\text{Var}(S) = 4 \times 576 = 2304$ .

▲Red Flags›

$n$ Normal i.i.d. dijumlahkan: $\sigma_S = \sqrt{n}\,\sigma$ , bukan $n\sigma$ .

No. 576

A light bulb’s lifetime in months is uniformly distributed on $[0, 40]$ . Let $a$ be a positive real number less than 30. The probability that the light bulb fails within 30 months, given that it is working after $a$ months, equals 0.60.

Calculate the value of $a$ .

a. $6$
b. $12$
c. $15$
d. $16$
e. $18$

≡Jawaban No. 576›

(c). $15$

Field	Isi
Topik CF2	Topik 2 — Variabel Acak Univariat
Sub-topik	2.6 Distribusi Kontinu Umum
Difficulty	Easy
Prerequisite	1.4 Probabilitas Bersyarat
Connected Topics	2.2 Variabel Acak Kontinu
Referensi	Miller Bab 5; Hogg-Tanis-Zimm Bab 3

ℹRumus›

P(T < 30 \mid T > a) = \frac{P(a < T < 30)}{P(T > a)} = \frac{(30-a)/40}{(40-a)/40} = \frac{30-a}{40-a}

Diketahui:

$T \sim U(0,40)$ ; $P(T < 30 \mid T > a) = 0{,}60$ ; target: $a$

▸Langkah Pengerjaan›

Langkah 1: Tulis persamaan probabilitas bersyarat

\frac{30 - a}{40 - a} = 0{,}60

Langkah 2: Selesaikan untuk $a$

30 - a = 0{,}60(40 - a) = 24 - 0{,}60a

30 - 24 = a - 0{,}60a = 0{,}40a

a = \frac{6}{0{,}40} = 15

Hasil Akhir: (c). $15$

◈Jebakan Umum›

⬡Kesalahan Konseptual›

Mengira $P(T < 30 \mid T > a) = (30 - a)/40$ ; ini adalah $P(a < T < 30)$ , bukan probabilitas bersyarat.
Perlu dibagi dengan $P(T > a) = (40-a)/40$ untuk mendapatkan probabilitas bersyarat.

▲Red Flags›

Uniform bersyarat: $P(T < b \mid T > a) = (b-a)/(40-a)$ untuk $a < b \leq 40$ .

No. 577

A homeowners insurance company pays claims only for thefts and fires. This year, the company’s total theft losses are normally distributed with mean 100 and standard deviation 40. The company’s total fire losses are normally distributed with mean 150 and standard deviation 30. Theft losses are independent of fire losses.

Calculate the coefficient of variation for the company’s combined theft and fire losses this year.

a. $0{,}20$
b. $0{,}28$
c. $0{,}30$
d. $0{,}40$
e. $0{,}60$

≡Jawaban No. 577›

(a). $0{,}20$

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	3.6 Matriks Variansi-Kovariansi
Referensi	Miller Bab 6; Hogg-Tanis-Zimm Bab 5.5

ℹRumus›

E[X+Y] = E[X] + E[Y], \quad \text{Var}(X+Y) = \text{Var}(X) + \text{Var}(Y) \text{ (independen)}

\text{CV} = \frac{\sigma_{X+Y}}{E[X+Y]}

Diketahui:

$X$ (theft): $\mu_X = 100$ , $\sigma_X = 40$ ; $Y$ (fire): $\mu_Y = 150$ , $\sigma_Y = 30$ ; independen

▸Langkah Pengerjaan›

Langkah 1: Hitung mean dan standar deviasi gabungan

E[X+Y] = 100 + 150 = 250

\text{Var}(X+Y) = 40^2 + 30^2 = 1600 + 900 = 2500 \implies \sigma_{X+Y} = 50

Langkah 2: Hitung CV

\text{CV} = \frac{50}{250} = 0{,}20

Hasil Akhir: (a). $0{,}20$

◈Jebakan Umum›

⬡Kesalahan Konseptual›

Menjumlahkan standar deviasi: $\sigma_{X+Y} \neq 40 + 30 = 70$ ; variansi yang dijumlahkan: $\sigma_{X+Y} = \sqrt{2500} = 50$ .

▲Red Flags›

CV = $\sigma / \mu$ ; hitung keduanya dari distribusi gabungan, bukan dari masing-masing distribusi secara terpisah.

No. 578

An insurance company offers two kinds of insurance: automobile insurance and homeowners insurance. This year, the company’s profit from automobile insurance is normally distributed with mean 200 and standard deviation 400. The company’s profit from homeowners insurance is normally distributed with mean 400 and standard deviation 300. The two profits are independent. Let $Z$ represent the standard normal random variable.

Determine the probability that the company’s overall profit this year is positive.

a. $P[Z > -1{,}83]$
b. $P[Z > -1{,}20]$
c. $P[Z > -0{,}86]$
d. $P[Z > -0{,}83]$
e. $P[Z > -0{,}50]$

≡Jawaban No. 578›

(b). $P[Z > -1{,}20]$

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›

$S = X + Y \sim N(200+400, 400^2+300^2) = N(600, 250{,}000)$ ; $\sigma_S = 500$ .

Diketahui:

$X \sim N(200, 160{,}000)$ ; $Y \sim N(400, 90{,}000)$ ; independen

▸Langkah Pengerjaan›

Langkah 1: Distribusi keuntungan total

E[S] = 600, \quad \text{Var}(S) = 400^2 + 300^2 = 250{,}000 \implies \sigma_S = 500

Langkah 2: Standarisasi

P(S > 0) = P\!\left(Z > \frac{0 - 600}{500}\right) = P(Z > -1{,}20)

Hasil Akhir: (b). $P[Z > -1{,}20]$

◈Jebakan Umum›

⬡Kesalahan Konseptual›

Mengira $\sigma_S = 400 + 300 = 700$ ; variansi dijumlahkan, lalu akar: $\sigma_S = \sqrt{160{,}000 + 90{,}000} = 500$ .

▲Red Flags›

Perlu membedakan $\sigma$ vs $\sigma^2$ saat menghitung variansi gabungan.

No. 579

The time $T$ , in years, a fuse lasts is a random variable with density function

g(t) = te^{-t}, \quad t > 0

Calculate the probability that a fuse lasts at most two years, given that it lasts at least one year.

a. $0{,}061$
b. $0{,}330$
c. $0{,}448$
d. $0{,}736$
e. $0{,}807$

≡Jawaban No. 579›

(c). $0{,}448$

Field	Isi
Topik CF2	Topik 2 — Variabel Acak Univariat
Sub-topik	2.2 Variabel Acak Kontinu
Difficulty	Hard
Prerequisite	1.4 Probabilitas Bersyarat
Connected Topics	2.4 Transformasi Variabel Acak Univariat
Referensi	Miller Bab 4; Hogg-Tanis-Zimm Bab 2

ℹRumus›

P(T \leq 2 \mid T \geq 1) = \frac{P(1 \leq T \leq 2)}{P(T \geq 1)}

Integrasi by parts: $\int te^{-t}\,dt = -te^{-t} - e^{-t} + C = -(t+1)e^{-t} + C$

Diketahui:

$g(t) = te^{-t}$ untuk $t > 0$ (distribusi Gamma dengan $\alpha=2$ , $\beta=1$ )

▸Langkah Pengerjaan›

Langkah 1: Hitung $P(T \geq 1)$

P(T \geq 1) = \int_1^{\infty} te^{-t}\,dt = \left[-(t+1)e^{-t}\right]_1^{\infty} = 0 - (-(1+1)e^{-1}) = 2e^{-1}

Langkah 2: Hitung $P(1 \leq T \leq 2)$

P(1 \leq T \leq 2) = \int_1^{2} te^{-t}\,dt = \left[-(t+1)e^{-t}\right]_1^{2} = -(3e^{-2}) + (2e^{-1}) = 2e^{-1} - 3e^{-2}

Langkah 3: Hitung probabilitas bersyarat

P(T \leq 2 \mid T \geq 1) = \frac{2e^{-1} - 3e^{-2}}{2e^{-1}} = 1 - \frac{3e^{-2}}{2e^{-1}} = 1 - \frac{3}{2e} = 1 - \frac{3}{2 \times 2{,}718} \approx 1 - 0{,}5518 = 0{,}4482 \approx 0{,}448

Hasil Akhir: (c). $0{,}448$

◈Jebakan Umum›

⬡Kesalahan Konseptual›

Tidak mengenali antiderivatif $\int te^{-t}\,dt = -(t+1)e^{-t}$ ; gunakan integrasi by parts: $u = t$ , $dv = e^{-t}dt$ .
Mengira $g(t) = te^{-t}$ adalah distribusi Eksponensial; ini adalah distribusi Gamma( $\alpha=2$ , $\beta=1$ ).

▲Red Flags›

Untuk $\int_a^{\infty} te^{-t}\,dt = [-(t+1)e^{-t}]_a^{\infty} = (a+1)e^{-a}$ .

No. 580

Let $X_1, X_2, X_3, X_4$ be independent observations of a random variable with mean 10 and standard deviation 3.

Calculate the variance of $X_1 - X_2 - X_3 - 3X_4$ .

a. $27$
b. $36$
c. $54$
d. $81$
e. $108$

≡Jawaban No. 580›

(e). $108$

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 independen dengan $\text{Var}(X_i) = \sigma^2 = 9$ :

\text{Var}(a_1X_1 + \ldots + a_nX_n) = \sum a_i^2 \text{Var}(X_i)

Diketahui:

$X_i$ i.i.d., $\sigma = 3$ , $\sigma^2 = 9$ ; target: $\text{Var}(X_1 - X_2 - X_3 - 3X_4)$

▸Langkah Pengerjaan›

Langkah 1: Identifikasi koefisien

W = 1 \cdot X_1 + (-1) \cdot X_2 + (-1) \cdot X_3 + (-3) \cdot X_4

Langkah 2: Hitung variansi

\text{Var}(W) = 1^2(9) + (-1)^2(9) + (-1)^2(9) + (-3)^2(9) = 9 + 9 + 9 + 81 = 108

Hasil Akhir: (e). $108$

◈Jebakan Umum›

⬡Kesalahan Konseptual›

Mengira $\text{Var}(-X_i) < \text{Var}(X_i)$ ; variansi selalu non-negatif dan $\text{Var}(-X) = \text{Var}(X)$ .
Lupa mengkuadratkan koefisien $-3$ : $(-3)^2 = 9$ , bukan $-3$ .

▲Red Flags›

$\text{Var}(aX) = a^2\text{Var}(X)$ ; tanda koefisien tidak mempengaruhi variansi.

No. 581

The following table gives the joint probability function of two random variables $X$ and $Y$ :

	$X=0$	$X=1$	$X=2$
$Y=0$	$1/6$	$0$	$1/2$
$Y=1$	$0$	$1/3$	$0$

Calculate $\text{Cov}(X, Y)$ .

a. $-1/9$
b. $0$
c. $1/18$
d. $1/9$
e. $2/9$

≡Jawaban No. 581›

(a). $-1/9$

Field	Isi
Topik CF2	Topik 3 — Variabel Acak Multivariat
Sub-topik	3.5 Independensi dan Korelasi, 3.6 Matriks Variansi-Kovariansi
Difficulty	Medium
Prerequisite	3.1 Distribusi Gabungan
Connected Topics	3.2 Distribusi Marginal
Referensi	Hogg-Tanis-Zimm Bab 4.1; Miller Bab 4

ℹRumus›

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

Diketahui:

Tabel joint $p(x,y)$ diberikan

▸Langkah Pengerjaan›

Langkah 1: Distribusi marginal

p_X(0) = 1/6, \quad p_X(1) = 1/3, \quad p_X(2) = 1/2

p_Y(0) = 1/6 + 1/2 = 2/3, \quad p_Y(1) = 1/3

Langkah 2: Hitung $E[X]$ , $E[Y]$ , dan $E[XY]$

E[X] = 0(1/6) + 1(1/3) + 2(1/2) = 0 + 1/3 + 1 = 4/3

E[Y] = 0(2/3) + 1(1/3) = 1/3

E[XY] = 0 \cdot 0 \cdot (1/6) + 0 \cdot 1 \cdot 0 + 1 \cdot 0 \cdot 0 + 1 \cdot 1 \cdot (1/3) + 2 \cdot 0 \cdot (1/2) + 2 \cdot 1 \cdot 0

= 0 + 0 + 0 + 1/3 + 0 + 0 = 1/3

Langkah 3: Hitung Kovarians

\text{Cov}(X,Y) = E[XY] - E[X]E[Y] = \frac{1}{3} - \frac{4}{3} \cdot \frac{1}{3} = \frac{1}{3} - \frac{4}{9} = \frac{3 - 4}{9} = -\frac{1}{9}

Hasil Akhir: (a). $-\dfrac{1}{9}$

◈Jebakan Umum›

⬡Kesalahan Konseptual›

Mengira $E[XY] = E[X] \cdot E[Y]$ (independensi); tabel menunjukkan $X$ dan $Y$ tidak independen karena $p(0,1) = 0 \neq p_X(0) \cdot p_Y(1) = (1/6)(1/3)$ .
Menjawab $\text{Cov} = 0$ ; meski banyak sel bernilai 0, kovarians tidak nol.

▲Red Flags›

Hitung $E[XY]$ langsung dari tabel joint — hanya sel yang bernilai $xy \neq 0$ yang berkontribusi.

No. 582

The annual amount a travel insurance company pays for claims has standard deviation 3. The 52 weekly amounts the company pays for claims are independent and have the same standard deviation.

Calculate the standard deviation of each of these weekly amounts.

a. $\dfrac{2\sqrt{52}}{3}$
b. $\dfrac{\sqrt{52}}{3}$
c. $\dfrac{\sqrt{52}}{3}$
d. $\dfrac{3}{\sqrt{52}}$
e. $3\sqrt{52}$

≡Jawaban No. 582›

(d). $\dfrac{3}{\sqrt{52}}$

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	4.2 Distribusi Sampel
Referensi	Miller Bab 6; Hogg-Tanis-Zimm Bab 3

ℹRumus›

Jika $A = \sum_{i=1}^{52} W_i$ (independen, variansi sama $\sigma_W^2$ ):

\text{Var}(A) = 52\,\sigma_W^2 = 9 \implies \sigma_W = \frac{3}{\sqrt{52}}

Diketahui:

$\sigma_A = 3$ (tahunan); $A = \sum_{i=1}^{52} W_i$ ; $W_i$ i.i.d.

▸Langkah Pengerjaan›

Langkah 1: Hubungkan variansi mingguan dan tahunan

\text{Var}(A) = 52\,\sigma_W^2 = 3^2 = 9

Langkah 2: Selesaikan untuk $\sigma_W$

\sigma_W^2 = \frac{9}{52} \implies \sigma_W = \frac{3}{\sqrt{52}}

Hasil Akhir: (d). $\dfrac{3}{\sqrt{52}}$

◈Jebakan Umum›

⬡Kesalahan Konseptual›

Membagi standar deviasi langsung: $\sigma_W \neq \sigma_A/52 = 3/52$ ; variansi yang dibagi, bukan standar deviasi.
$\sigma_W = \sigma_A / \sqrt{52}$ , bukan $\sigma_A / 52$ .

▲Red Flags›

Variansi bersifat aditif untuk variabel independen; standar deviasi tidak. Selalu kerjakan di level variansi, lalu ambil akar.

No. 583

Claim amounts from accidents in an amusement park are uniformly distributed on an interval. The probability that a randomly selected claim amount is less than 12 is 0.50. The probability that a randomly selected claim amount exceeds 6 is 0.875.

Calculate the probability that a randomly selected claim amount is less than 10, given that it exceeds 6.

a. $2/9$
b. $1/4$
c. $2/7$
d. $3/8$
e. $3/7$

≡Jawaban No. 583›

(c). $2/7$

Field	Isi
Topik CF2	Topik 2 — Variabel Acak Univariat
Sub-topik	2.6 Distribusi Kontinu Umum
Difficulty	Medium
Prerequisite	1.4 Probabilitas Bersyarat
Connected Topics	2.2 Variabel Acak Kontinu
Referensi	Miller Bab 5; Hogg-Tanis-Zimm Bab 3

ℹRumus›

$X \sim U(a, b)$ : $P(X < x) = (x-a)/(b-a)$ .

Diketahui:

$P(X < 12) = 0{,}50$ dan $P(X > 6) = 0{,}875$

▸Langkah Pengerjaan›

Langkah 1: Tentukan $a$ dan $b$

P(X < 12) = \frac{12 - a}{b - a} = 0{,}50 \implies 12 - a = 0{,}50(b - a) \quad \cdots (1)

P(X > 6) = \frac{b - 6}{b - a} = 0{,}875 \implies P(X < 6) = \frac{6 - a}{b - a} = 0{,}125 \quad \cdots (2)

Dari (2): $6 - a = 0{,}125(b-a)$ .

Kurangkan (2) dari (1): $(12-a) - (6-a) = (0{,}5 - 0{,}125)(b-a)$

6 = 0{,}375(b-a) \implies b - a = 16

Dari (2): $6 - a = 0{,}125 \times 16 = 2 \implies a = 4$ .

Jadi $b = 4 + 16 = 20$ .

Langkah 2: Hitung probabilitas bersyarat

P(X < 10 \mid X > 6) = \frac{P(6 < X < 10)}{P(X > 6)} = \frac{(10-6)/16}{(20-6)/16} = \frac{4}{14} = \frac{2}{7}

Hasil Akhir: (c). $2/7$

◈Jebakan Umum›

⬡Kesalahan Konseptual›

Mengasumsikan $a = 0$ ; batas bawah $a$ harus dicari dari dua kondisi yang diberikan.
Menggunakan $b - a = 40$ (tidak berdasar); selalu tentukan interval dari kondisi yang ada.

▲Red Flags›

Dua probabilitas Uniform → dua persamaan linear untuk menentukan $a$ dan $b$ .

No. 584

The number of thefts in any year a warehouse owner experiences is modeled by a Poisson distribution. In each year, the probability of at least one theft is 0.10, and the yearly number of thefts are independent.

Calculate the variance of the total number of thefts the owner experiences in the next 15 years.

a. $0{,}11$
b. $0{,}41$
c. $1{,}35$
d. $1{,}50$
e. $1{,}58$

≡Jawaban No. 584›

(e). $1{,}58$

Field	Isi
Topik CF2	Topik 2 — Variabel Acak Univariat
Sub-topik	2.5 Distribusi Diskrit Umum
Difficulty	Medium
Prerequisite	2.1 Variabel Acak Diskrit, 3.5 Independensi dan Korelasi
Connected Topics	4.2 Distribusi Sampel
Referensi	Miller Bab 5; Hogg-Tanis-Zimm Bab 2

ℹRumus›

P(\geq 1 \text{ pencurian}) = 1 - e^{-\lambda} = 0{,}10 \implies e^{-\lambda} = 0{,}9

Jumlah 15 tahun Poisson independen $\sim$ Poisson $(15\lambda)$ ; variansi $= 15\lambda$ .

Diketahui:

$1 - e^{-\lambda} = 0{,}10 \implies \lambda = -\ln(0{,}9)$ ; 15 tahun independen

▸Langkah Pengerjaan›

Langkah 1: Tentukan $\lambda$

e^{-\lambda} = 0{,}90 \implies \lambda = -\ln(0{,}90) \approx 0{,}10536

Langkah 2: Variansi total 15 tahun

\text{Var}(S_{15}) = 15\lambda = 15 \times 0{,}10536 \approx 1{,}5804 \approx 1{,}58

Hasil Akhir: (e). $1{,}58$

◈Jebakan Umum›

⬡Kesalahan Konseptual›

Mengira $\lambda = 0{,}10$ langsung dari $P(\geq 1) = 0{,}10$ ; hubungannya adalah $\lambda = -\ln(1 - 0{,}10) = -\ln(0{,}9)$ , bukan $\lambda = 0{,}10$ .
Mengira variansi total $= 15 \times P(\geq 1) = 1{,}5$ ; harus menggunakan $15\lambda$ .

▲Red Flags›

“Probability at least one” untuk Poisson: $P(\geq 1) = 1 - e^{-\lambda}$ ; selalu balikkan untuk mendapat $\lambda = -\ln(1 - P)$ .

No. 585

The random variable $Y$ has density function

f(y) = \begin{cases} \dfrac{y}{6} - \dfrac{y^2}{36}, & 0 < y < 6 \\ 0, & \text{selainnya} \end{cases}

Calculate $P[1 < Y < 3 \mid 2 < Y < 4]$ .

a. $0{,}111$
b. $0{,}241$
c. $0{,}481$
d. $0{,}500$
e. $0{,}885$

≡Jawaban No. 585›

(d). $0{,}500$

Field	Isi
Topik CF2	Topik 2 — Variabel Acak Univariat
Sub-topik	2.2 Variabel Acak Kontinu
Difficulty	Medium
Prerequisite	1.4 Probabilitas Bersyarat
Connected Topics	2.4 Transformasi Variabel Acak Univariat
Referensi	Miller Bab 4; Hogg-Tanis-Zimm Bab 2

ℹRumus›

P(1 < Y < 3 \mid 2 < Y < 4) = \frac{P(\{1<Y<3\} \cap \{2<Y<4\})}{P(2<Y<4)} = \frac{P(2<Y<3)}{P(2<Y<4)}

Antiderivatif $F(y) = \frac{y^2}{12} - \frac{y^3}{108}$ .

Diketahui:

$f(y) = y/6 - y^2/36$ pada $(0,6)$

▸Langkah Pengerjaan›

Langkah 1: Tentukan irisan

\{1<Y<3\} \cap \{2<Y<4\} = \{2<Y<3\}

Langkah 2: Hitung integral menggunakan $F(y) = y^2/12 - y^3/108$

P(2<Y<3) = F(3) - F(2) = \left(\frac{9}{12} - \frac{27}{108}\right) - \left(\frac{4}{12} - \frac{8}{108}\right)

= \left(\frac{3}{4} - \frac{1}{4}\right) - \left(\frac{1}{3} - \frac{2}{27}\right) = \frac{1}{2} - \frac{7}{27} = \frac{27 - 14}{54} = \frac{13}{54}

P(2<Y<4) = F(4) - F(2) = \left(\frac{16}{12} - \frac{64}{108}\right) - \left(\frac{4}{12} - \frac{8}{108}\right)

= \left(\frac{4}{3} - \frac{16}{27}\right) - \left(\frac{1}{3} - \frac{2}{27}\right) = 1 - \frac{14}{27} = \frac{13}{27}

Langkah 3: Hitung probabilitas bersyarat

P(1<Y<3 \mid 2<Y<4) = \frac{13/54}{13/27} = \frac{13}{54} \times \frac{27}{13} = \frac{27}{54} = \frac{1}{2} = 0{,}500

Hasil Akhir: (d). $0{,}500$

◈Jebakan Umum›

⬡Kesalahan Konseptual›

Mengira irisan $\{1<Y<3\} \cap \{2<Y<4\} = \{1<Y<4\}$ (gabungan, bukan irisan); irisan yang benar adalah $\{2<Y<3\}$ .
Menghitung $P(2<Y<3)/P(1<Y<3)$ alih-alih dibagi $P(2<Y<4)$ .

▲Red Flags›

$P(A \mid B) = P(A \cap B)/P(B)$ ; identifikasi irisan $A \cap B$ secara grafis pada garis bilangan sebelum menghitung.

No. 586

An insurance company sells an automobile policy that pays 80% of any losses that exceed the deductible of 500. The loss amounts (in units of 1000) follow an exponential distribution with mean 6. Calculate the probability that the policy payment on a randomly selected loss exceeds 5000.

a. $0{,}024$
b. $0{,}318$
c. $0{,}325$
d. $0{,}675$
e. $0{,}682$

≡Jawaban No. 586›

(c). $0{,}325$

Field	Isi
Topik CF2	Topik 2 — Variabel Acak Univariat
Sub-topik	2.6 Distribusi Kontinu Umum
Difficulty	Hard
Prerequisite	2.2 Variabel Acak Kontinu, 1.4 Probabilitas Bersyarat
Connected Topics	3.7 Distribusi Majemuk
Referensi	Hogg-Tanis-Zimm Bab 3.2; Miller Bab 5

ℹRumus›

Pembayaran $Y = 0{,}8(L - 0{,}5)$ jika $L > 0{,}5$ , dan $Y = 0$ jika $L \leq 0{,}5$ (satuan ribuan).

P(Y > 5) = P(0{,}8(L - 0{,}5) > 5) = P(L > 5/0{,}8 + 0{,}5) = P(L > 6{,}75)

Diketahui:

$L \sim \text{Exp}(\beta = 6)$ (dalam ribuan); target: $P(Y > 5)$ (5000 = $5$ dalam satuan ribuan)

▸Langkah Pengerjaan›

Langkah 1: Terjemahkan kondisi pembayaran

Pembayaran $Y = 0{,}8(L - 0{,}5)$ untuk $L > 0{,}5$ .

P(Y > 5) = P(0{,}8(L - 0{,}5) > 5) = P\!\left(L - 0{,}5 > \frac{5}{0{,}8}\right) = P(L > 6{,}25 + 0{,}5) = P(L > 6{,}75)

Langkah 2: Hitung probabilitas

P(L > 6{,}75) = e^{-6{,}75/6} = e^{-1{,}125} \approx 0{,}3247 \approx 0{,}325

Hasil Akhir: (c). $0{,}325$

◈Jebakan Umum›

⬡Kesalahan Konseptual›

Mengira $P(Y > 5) = P(L > 5/0{,}8) = P(L > 6{,}25)$ ; lupa menambahkan deductible $0{,}5$ ke ambang batas kerugian.
Salah konversi satuan: 5000 dalam satuan ribuan $= 5$ , bukan 5000.

▲Red Flags›

Perhatikan deductible: $P(Y > c) = P(L > c/r + d)$ di mana $r$ = persentase coverage dan $d$ = deductible.

No. 587

Once each morning and once each afternoon, the driver of a delivery truck is assigned to a route. The morning route may be 5, 10 or 40 miles. The afternoon route may be 0, 5 or 10 miles. The probabilities are:

	Siang = 0 mi	Siang = 5 mi	Siang = 10 mi
Pagi = 5 mi	0	$2p$	$3p$
Pagi = 10 mi	0	$2p$	0
Pagi = 40 mi	$q$	0	0

The lengths of the assigned afternoon routes have a mean of 6 miles.

Calculate the variance of the length of an afternoon route.

a. $11{,}0$
b. $12{,}0$
c. $16{,}5$
d. $26{,}5$
e. $42{,}0$

≡Jawaban No. 587›

(b). $12{,}0$

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›

Total probabilitas $= 7p + q = 1$ ; $E[\text{Siang}] = 5(4p) + 10(3p) = 50p = 6$ .

Diketahui:

$E[\text{Siang}] = 6$ ; target: $\text{Var}(\text{Siang})$

▸Langkah Pengerjaan›

Langkah 1: Cari $p$ dari $E[\text{Siang}]$

E[\text{Siang}] = 0 \cdot q + 5(2p + 2p) + 10(3p) = 0 + 20p + 30p = 50p = 6 \implies p = 0{,}12

Langkah 2: Cari $q$

Total: $(2p + 3p) + (2p) + q = 7p + q = 1 \implies q = 1 - 7(0{,}12) = 1 - 0{,}84 = 0{,}16$

Langkah 3: Distribusi marginal siang

P(\text{Siang} = 0) = q = 0{,}16

P(\text{Siang} = 5) = 2p + 2p = 4p = 0{,}48

P(\text{Siang} = 10) = 3p = 0{,}36

Langkah 4: Hitung variansi

E[\text{Siang}^2] = 0^2(0{,}16) + 5^2(0{,}48) + 10^2(0{,}36) = 0 + 12 + 36 = 48

\text{Var}(\text{Siang}) = 48 - 6^2 = 48 - 36 = 12

Hasil Akhir: (b). $12{,}0$

◈Jebakan Umum›

⬡Kesalahan Konseptual›

Mengira $P(\text{Siang} = 5) = 2p$ (hanya satu sel); ada dua sel dengan nilai 5 (pagi=5 dan pagi=10), sehingga $P = 2p + 2p = 4p$ .
Lupa menghitung $q$ dari total prob $= 1$ .

▲Red Flags›

Distribusi marginal siang: jumlahkan kolom-kolom sesuai nilai siang dari semua baris.

No. 588

The lifetimes of light bulbs are independent and exponentially distributed with common mean 1000 hours. A lamp has a single bulb and is continuously on. Each light bulb is replaced immediately after it burns out.

Calculate the standard deviation of the time, in hours, until the failure of the fifth bulb.

a. $71$
b. $158$
c. $1000$
d. $2236$
e. $5000$

≡Jawaban No. 588›

(d). $2236$

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	4.2 Distribusi Sampel
Referensi	Miller Bab 6; Hogg-Tanis-Zimm Bab 5.5

ℹRumus›

Total waktu $= T_1 + T_2 + T_3 + T_4 + T_5$ (i.i.d. Eksponensial); $\text{Var}(T_i) = \beta^2 = 10^6$ .

\text{Var}(S) = 5 \times 10^6 \implies \sigma_S = \sqrt{5} \times 1000 = 2236

Diketahui:

$T_i \sim \text{Exp}(\beta = 1000)$ i.i.d.; target: $\sigma_{S_5}$

▸Langkah Pengerjaan›

Langkah 1: Variansi total 5 bola lampu

\text{Var}(S_5) = 5 \times (1000)^2 = 5{,}000{,}000

Langkah 2: Standar deviasi

\sigma_{S_5} = \sqrt{5{,}000{,}000} = 1000\sqrt{5} \approx 2236

Hasil Akhir: (d). $2236$

◈Jebakan Umum›

⬡Kesalahan Konseptual›

Menjawab $\sigma = 5 \times 1000 = 5000$ ; standar deviasi total $= \sqrt{n} \times \sigma$ , bukan $n \times \sigma$ .
Mengira $\text{Var}(\text{Eksponensial}) = \beta$ (mean); untuk Eksponensial: $\text{Var} = \beta^2$ .

▲Red Flags›

$n$ Eksponensial i.i.d. dijumlahkan: $\sigma_S = \sqrt{n}\,\beta$ ; maka $\sigma_{S_5} = \sqrt{5} \times 1000$ .

No. 589

In a one-year period, the number of fires that occur in buildings insured by ABC insurance is Poisson distributed with mean 12. The resulting losses are independent random variables that are also independent of the number of losses, each having density function

f(x) = \begin{cases} \dfrac{8-x}{32}, & 0 < x < 8 \\ 0, & \text{selainnya} \end{cases}

The payment for any loss is equal to the amount of the loss in excess of a deductible of 2.

Calculate the expected total payment that ABC must make for losses due to fire that occur in a one-year period.

a. $8{,}0$
b. $13{,}5$
c. $16{,}5$
d. $30{,}0$
e. $32{,}0$

≡Jawaban No. 589›

(b). $13{,}5$

Field	Isi
Topik CF2	Topik 3 — Variabel Acak Multivariat
Sub-topik	3.7 Distribusi Majemuk
Difficulty	Hard
Prerequisite	2.2 Variabel Acak Kontinu, 2.5 Distribusi Diskrit Umum
Connected Topics	3.4 Nilai Harapan dan Variansi Bersyarat
Referensi	Hogg-Tanis-Zimm Bab 4.4; Miller Bab 5

ℹRumus›

$E[\text{Total Pembayaran}] = E[N] \times E[\text{Pembayaran per klaim}]$ (model majemuk)

Pembayaran per klaim: $Y = (X - 2)^+ = \max(X-2, 0)$

E[Y] = \int_2^8 (x-2) \cdot \frac{8-x}{32}\,dx

Diketahui:

$N \sim \text{Poisson}(12)$ ; $X \sim f(x) = (8-x)/32$ pada $(0,8)$ ; deductible $= 2$

▸Langkah Pengerjaan›

Langkah 1: Hitung $E[Y]$ (pembayaran per klaim)

E[Y] = \int_2^8 (x-2) \cdot \frac{8-x}{32}\,dx

Substitusi $u = x - 2$ : ketika $x=2$ , $u=0$ ; ketika $x=8$ , $u=6$ ; $8-x = 6-u$ :

E[Y] = \frac{1}{32}\int_0^6 u(6-u)\,du = \frac{1}{32}\int_0^6 (6u - u^2)\,du = \frac{1}{32}\left[3u^2 - \frac{u^3}{3}\right]_0^6

= \frac{1}{32}\left(108 - 72\right) = \frac{36}{32} = \frac{9}{8} = 1{,}125

Langkah 2: Hitung $E[\text{Total Pembayaran}]$

E[\text{Total}] = E[N] \times E[Y] = 12 \times \frac{9}{8} = \frac{108}{8} = 13{,}5

Hasil Akhir: (b). $13{,}5$

◈Jebakan Umum›

⬡Kesalahan Konseptual›

Mengira pembayaran $= E[X] \times E[N]$ tanpa memperhitungkan deductible; harus menggunakan $E[(X-2)^+]$ bukan $E[X]$ .
Mengintegralkan dari 0 alih-alih dari 2; deductible $= 2$ berarti pembayaran dimulai dari $X > 2$ .

▲Red Flags›

Model majemuk Poisson: $E[\text{Total}] = \lambda \times E[Y]$ di mana $Y$ adalah pembayaran per klaim (setelah deductible).

No. 590

Each of three people is holding a biased coin that has probability 0.25 of landing heads. They toss the coins simultaneously. After each round of tosses, if the outcomes are all heads or all tails, they will continue to another round; otherwise, they will stop.

Calculate the probability that there will be exactly five rounds of tosses.

a. $0{,}0003$
b. $0{,}0029$
c. $0{,}0206$
d. $0{,}0438$
e. $0{,}1030$

≡Jawaban No. 590›

(c). $0{,}0206$

Field	Isi
Topik CF2	Topik 1 — Dasar-Dasar Probabilitas
Sub-topik	1.5 Kejadian Independen, 1.2 Aksioma dan Perhitungan Probabilitas
Difficulty	Medium
Prerequisite	1.3 Metode Enumerasi
Connected Topics	2.5 Distribusi Diskrit Umum
Referensi	Miller Bab 2; Hogg-Tanis-Zimm Bab 1

ℹRumus›

P(\text{lanjut}) = P(\text{HHH}) + P(\text{TTT}) = (0{,}25)^3 + (0{,}75)^3 = 0{,}015625 + 0{,}421875 = 0{,}4375

P(\text{berhenti}) = 1 - 0{,}4375 = 0{,}5625

Diketahui:

Tepat 5 putaran = 4 putaran pertama lanjut, putaran ke-5 berhenti

▸Langkah Pengerjaan›

Langkah 1: Probabilitas permainan berlanjut per putaran

P(\text{HHH}) = (0{,}25)^3 = 0{,}015625

P(\text{TTT}) = (0{,}75)^3 = 0{,}421875

P(\text{lanjut}) = 0{,}015625 + 0{,}421875 = 0{,}4375

Langkah 2: Probabilitas tepat 5 putaran

P(\text{5 putaran}) = (0{,}4375)^4 \times (1 - 0{,}4375) = (0{,}4375)^4 \times 0{,}5625

(0{,}4375)^4 = (0{,}4375)^2 \times (0{,}4375)^2 = 0{,}19141 \times 0{,}19141 \approx 0{,}03664

P = 0{,}03664 \times 0{,}5625 \approx 0{,}02061 \approx 0{,}0206

Hasil Akhir: (c). $0{,}0206$

◈Jebakan Umum›

⬡Kesalahan Konseptual›

Mengira $P(\text{lanjut}) = (0{,}25)^3 = 0{,}016$ (hanya HHH); perlu menambahkan TTT.
“Tepat 5 putaran” berarti tidak berhenti di putaran 1–4 dan berhenti di putaran 5; ini seperti distribusi Geometrik.

▲Red Flags›

Tepat $n$ putaran = $(P_{\text{lanjut}})^{n-1} \times P_{\text{berhenti}}$ ; pola Geometrik.

No. 591

In a given year, a manufacturer’s profit from selling warranties is normally distributed with mean 20 and variance 16.

Calculate the interval, centered at the mean, containing 25% of the probability associated with a single yearly profit.

a. $(12{,}2,\; 27{,}8)$
b. $(14{,}9,\; 25{,}1)$
c. $(16{,}0,\; 24{,}0)$
d. $(17{,}3,\; 22{,}7)$
e. $(18{,}7,\; 21{,}3)$

≡Jawaban No. 591›

(e). $(18{,}7,\; 21{,}3)$

Field	Isi
Topik CF2	Topik 2 — Variabel Acak Univariat
Sub-topik	2.6 Distribusi Kontinu Umum
Difficulty	Medium
Prerequisite	2.2 Variabel Acak Kontinu
Connected Topics	4.7 Selang Kepercayaan
Referensi	Miller Bab 6; Hogg-Tanis-Zimm Bab 5.5

ℹRumus›

Interval simetris di sekitar mean yang mengandung $p\%$ prob: $(\mu - z_{(1+p)/2}\sigma,\; \mu + z_{(1+p)/2}\sigma)$ .

Untuk $p = 25\%$ : $\Phi(c) = 0{,}625 \implies c \approx 0{,}3186$ .

Diketahui:

$X \sim N(20, 16)$ ; $\sigma = 4$ ; target: interval simetris dengan $P(X \in I) = 0{,}25$

▸Langkah Pengerjaan›

Langkah 1: Tentukan batas $z$

P(\mu - c\sigma < X < \mu + c\sigma) = 0{,}25

2\Phi(c) - 1 = 0{,}25 \implies \Phi(c) = 0{,}625 \implies c \approx 0{,}3186

Langkah 2: Hitung batas interval

\text{Batas bawah} = 20 - 0{,}3186 \times 4 = 20 - 1{,}274 \approx 18{,}73

\text{Batas atas} = 20 + 0{,}3186 \times 4 = 20 + 1{,}274 \approx 21{,}27

Hasil Akhir: (e). $(18{,}7,\; 21{,}3)$

◈Jebakan Umum›

⬡Kesalahan Konseptual›

Mengira $c = z_{0{,}25} = 0{,}674$ (persentil ke-25 dari bawah); untuk interval simetris yang mengandung 25% total, perlu $\Phi(c) = 0{,}625$ , bukan $0{,}75$ .
Mengira $\text{Var} = 16$ berarti $\sigma = 16$ ; $\sigma = \sqrt{16} = 4$ .

▲Red Flags›

Interval simetris mengandung $p$ : setengah di kanan, setengah di kiri → $\Phi(c) = 0{,}5 + p/2$ .

No. 592

A group of 18 patients is tested for diabetes. Each patient independently has a 0.15 probability of testing positive.

Calculate the mode of the number of patients in the group who test positive.

a. $0$
b. $1$
c. $2$
d. $3$
e. $4$

≡Jawaban No. 592›

(c). $2$

Field	Isi
Topik CF2	Topik 2 — Variabel Acak Univariat
Sub-topik	2.5 Distribusi Diskrit Umum
Difficulty	Easy
Prerequisite	2.1 Variabel Acak Diskrit
Connected Topics	4.5 Estimasi Parameter
Referensi	Miller Bab 5; Hogg-Tanis-Zimm Bab 2

ℹRumus›

$X \sim B(18, 0{,}15)$ ; mean $= np = 2{,}7$ .

Modus Binomial: $\lfloor (n+1)p \rfloor$ atau $\lceil (n+1)p \rceil - 1$ .

Diketahui:

$n = 18$ , $p = 0{,}15$ ; mean $= 2{,}7$

▸Langkah Pengerjaan›

Langkah 1: Hitung probabilitas di sekitar mean

$(n+1)p = 19 \times 0{,}15 = 2{,}85$ → modus $= \lfloor 2{,}85 \rfloor = 2$ .

Verifikasi:

P(X=1) \approx 0{,}1704, \quad P(X=2) \approx 0{,}2556, \quad P(X=3) \approx 0{,}2406

$P(X=2) > P(X=3) > P(X=1)$ ; modus $= 2$ .

Hasil Akhir: (c). $2$

◈Jebakan Umum›

⬡Kesalahan Konseptual›

Mengira modus $= \text{round}(np) = \text{round}(2{,}7) = 3$ ; modus Binomial $\approx \lfloor (n+1)p \rfloor = 2$ .
Modus dan mean Binomial berbeda kecuali untuk kasus tertentu.

▲Red Flags›

Modus Binomial $B(n,p)$ : jika $(n+1)p$ bukan bilangan bulat → modus $= \lfloor (n+1)p \rfloor$ ; hitung beberapa nilai PMF untuk konfirmasi.

No. 593

An insurance company’s annual profit is normally distributed with mean 39. The 28th percentile of the annual profit is 21.

Calculate the percentile associated with an annual profit of 3 for this insurance company.

a. 4th
b. 6th
c. 8th
d. 10th
e. 12th

≡Jawaban No. 593›

(e). 12th

Field	Isi
Topik CF2	Topik 2 — Variabel Acak Univariat
Sub-topik	2.6 Distribusi Kontinu Umum
Difficulty	Medium
Prerequisite	2.2 Variabel Acak Kontinu
Connected Topics	4.7 Selang Kepercayaan
Referensi	Miller Bab 6; Hogg-Tanis-Zimm Bab 5.5

ℹRumus›

Persentil ke-28 $\iff$ $\Phi\!\left(\frac{21-39}{\sigma}\right) = 0{,}28 \implies \frac{-18}{\sigma} = z_{0{,}28} = -0{,}5828$

Diketahui:

$\mu = 39$ ; persentil ke-28 $= 21$ ; target: persentil untuk $X = 3$

▸Langkah Pengerjaan›

Langkah 1: Tentukan $\sigma$

\Phi\!\left(\frac{21-39}{\sigma}\right) = 0{,}28 \implies \frac{-18}{\sigma} = -0{,}5828 \implies \sigma = \frac{18}{0{,}5828} \approx 30{,}885

Langkah 2: Hitung persentil untuk $X = 3$

z = \frac{3 - 39}{30{,}885} = \frac{-36}{30{,}885} \approx -1{,}166

\Phi(-1{,}166) \approx 0{,}122 \approx 12\%

Hasil Akhir: (e). 12th

◈Jebakan Umum›

⬡Kesalahan Konseptual›

Mengira $z_{0{,}28} = 0{,}58$ (positif); persentil ke-28 berada di bawah mean, sehingga $z$ negatif.
Lupa bahwa nilai 3 sangat jauh di bawah mean 39, sehingga persentilnya rendah.

▲Red Flags›

Cari $\sigma$ dari persentil yang diberikan, lalu gunakan $\sigma$ tersebut untuk mencari persentil nilai baru.

No. 594

From a standard deck of 52 playing cards, five cards are chosen without replacement.

Calculate the probability of obtaining at least two kings, but no aces or jacks.

a. $0{,}0174$
b. $0{,}0228$
c. $0{,}0240$
d. $0{,}0298$
e. $0{,}0417$

≡Jawaban No. 594›

(c). $0{,}0240$

Field	Isi
Topik CF2	Topik 1 — Dasar-Dasar Probabilitas
Sub-topik	1.3 Metode Enumerasi
Difficulty	Hard
Prerequisite	2.5 Distribusi Diskrit Umum
Connected Topics	1.2 Aksioma dan Perhitungan Probabilitas
Referensi	Miller Bab 2; Hogg-Tanis-Zimm Bab 1

ℹRumus›

Deck dikurangi ace dan jack: $52 - 4 - 4 = 44$ kartu, di antaranya 4 raja dan 40 lainnya.

P = \frac{\binom{4}{k}\binom{40}{5-k}}{\binom{52}{5}}, \quad k = 2, 3, 4

Diketahui:

52 kartu; 4 raja, 4 ace, 4 jack, 40 lainnya (termasuk 4 raja)
$\binom{52}{5} = 2{,}598{,}960$

▸Langkah Pengerjaan›

Langkah 1: Identifikasi kelompok kartu yang boleh dipilih

Tidak ada ace (4 kartu) dan tidak ada jack (4 kartu). Sisa: $52 - 4 - 4 = 44$ kartu, terdiri dari 4 raja dan 40 non-raja.

Langkah 2: Hitung probabilitas tiap kasus

$k=2$ raja: $\binom{4}{2}\binom{40}{3} = 6 \times 9880 = 59{,}280$

$k=3$ raja: $\binom{4}{3}\binom{40}{2} = 4 \times 780 = 3{,}120$

$k=4$ raja: $\binom{4}{4}\binom{40}{1} = 1 \times 40 = 40$

Langkah 3: Hitung total probabilitas

P = \frac{59{,}280 + 3{,}120 + 40}{2{,}598{,}960} = \frac{62{,}440}{2{,}598{,}960} \approx 0{,}02402 \approx 0{,}0240

Hasil Akhir: (c). $0{,}0240$

◈Jebakan Umum›

⬡Kesalahan Konseptual›

Menggunakan denominator $\binom{44}{5}$ (hanya dari 44 kartu yang diizinkan) alih-alih $\binom{52}{5}$ ; pengambilan dilakukan dari seluruh deck 52 kartu.
Lupa memasukkan kasus $k=3$ dan $k=4$ raja.

▲Red Flags›

“At least two kings, no aces or jacks” → kondisi gabungan; hitung jumlah cara memilih dari subkelompok yang diizinkan, dibagi total cara dari seluruh deck.

No. 595

The check-out times for customers at a grocery store are modeled by a continuous uniform distribution between 1.1 and 8.6 minutes. The store would like to advertise that “no more than 5% of customers require more than $y$ minutes to check out.”

Calculate the minimum value for $y$ .

a. $1{,}475$
b. $7{,}125$
c. $8{,}170$
d. $8{,}225$
e. $8{,}545$

≡Jawaban No. 595›

(d). $8{,}225$

Field	Isi
Topik CF2	Topik 2 — Variabel Acak Univariat
Sub-topik	2.6 Distribusi Kontinu Umum
Difficulty	Easy
Prerequisite	2.2 Variabel Acak Kontinu
Connected Topics	4.7 Selang Kepercayaan
Referensi	Miller Bab 5; Hogg-Tanis-Zimm Bab 3

ℹRumus›

$T \sim U(1{,}1, 8{,}6)$ ; $P(T > y) \leq 0{,}05 \implies P(T \leq y) \geq 0{,}95$ .

P(T \leq y) = \frac{y - 1{,}1}{8{,}6 - 1{,}1} = \frac{y - 1{,}1}{7{,}5} \geq 0{,}95

Diketahui:

$T \sim U(1{,}1, 8{,}6)$ ; target: nilai minimum $y$ sehingga $P(T > y) \leq 0{,}05$

▸Langkah Pengerjaan›

Langkah 1: Terjemahkan pernyataan iklan

“Tidak lebih dari 5% pelanggan memerlukan lebih dari $y$ menit” $\iff P(T > y) \leq 0{,}05$ .

P(T > y) = \frac{8{,}6 - y}{7{,}5} \leq 0{,}05

Langkah 2: Selesaikan untuk $y$

8{,}6 - y \leq 0{,}05 \times 7{,}5 = 0{,}375

y \geq 8{,}6 - 0{,}375 = 8{,}225

Nilai minimum $y = 8{,}225$ .

Hasil Akhir: (d). $8{,}225$

◈Jebakan Umum›

⬡Kesalahan Konseptual›

Mengira $P(T > y) = 0{,}05 \implies y = 1{,}1 + 0{,}05 \times 7{,}5 = 1{,}475$ (jawaban (A)); ini adalah persentil ke-5, bukan ke-95.
“Lebih dari $y$ menit tidak lebih dari 5%” → cari persentil ke-95, bukan ke-5.

▲Red Flags›

Iklan berbunyi " $P(\text{lama} > y) \leq 0{,}05$ " → $y$ adalah persentil ke-95 distribusi waktu check-out.

No. 596

An insurance company’s annual profit is normally distributed. The probability that the annual profit exceeds 3.50 is 0.3264 and the probability that it exceeds 3.62 is 0.2743.

Calculate the variance of the insurance company’s annual profit.

a. $0{,}44$
b. $0{,}64$
c. $0{,}80$
d. $1{,}25$
e. $1{,}56$

≡Jawaban No. 596›

(b). $0{,}64$

Field	Isi
Topik CF2	Topik 2 — Variabel Acak Univariat
Sub-topik	2.6 Distribusi Kontinu Umum
Difficulty	Hard
Prerequisite	2.2 Variabel Acak Kontinu
Connected Topics	4.5 Estimasi Parameter
Referensi	Miller Bab 6; Hogg-Tanis-Zimm Bab 5.5

ℹRumus›

Dua probabilitas ekor Normal → dua persamaan: $\frac{x_i - \mu}{\sigma} = z_i$ .

Diketahui:

$P(X > 3{,}50) = 0{,}3264 \implies P(X \leq 3{,}50) = 0{,}6736 \implies z_1 = 0{,}4499 \approx 0{,}45$
$P(X > 3{,}62) = 0{,}2743 \implies P(X \leq 3{,}62) = 0{,}7257 \implies z_2 = 0{,}5999 \approx 0{,}60$

▸Langkah Pengerjaan›

Langkah 1: Tulis sistem persamaan

\frac{3{,}50 - \mu}{\sigma} = 0{,}4499 \quad \cdots (1)

\frac{3{,}62 - \mu}{\sigma} = 0{,}5999 \quad \cdots (2)

Langkah 2: Kurangkan (1) dari (2)

\frac{3{,}62 - 3{,}50}{\sigma} = 0{,}5999 - 0{,}4499 = 0{,}15 \implies \frac{0{,}12}{\sigma} = 0{,}15 \implies \sigma = \frac{0{,}12}{0{,}15} = 0{,}8

Langkah 3: Hitung variansi

\sigma^2 = (0{,}8)^2 = 0{,}64

Hasil Akhir: (b). $0{,}64$

◈Jebakan Umum›

⬡Kesalahan Konseptual›

Salah membaca tabel Normal: $P(X > 3{,}50) = 0{,}3264$ berarti $z_1 = 0{,}45$ (positif karena $x > \mu$ jika $z > 0$ ).
Lupa bahwa mengurangkan dua persamaan menghilangkan $\mu$ secara efisien.

▲Red Flags›

Dua persamaan Normal dengan dua unknown → kurangkan untuk menghilangkan satu unknown sebelum menyelesaikan.

No. 597

A theft insurance company offers two policies, A and B. Under each policy, a loss is uniformly distributed on the interval $[a, 19]$ . For a loss under policy A, the deductible is 0 and the expected value of the claim payment is 10. For a loss under policy B, the deductible is 4.

Calculate the expected value of the claim payment for a loss under policy B.

a. $6{,}00$
b. $6{,}25$
c. $7{,}50$
d. $7{,}89$
e. $8{,}33$

≡Jawaban No. 597›

(b). $6{,}25$

Field	Isi
Topik CF2	Topik 2 — Variabel Acak Univariat
Sub-topik	2.6 Distribusi Kontinu Umum
Difficulty	Medium
Prerequisite	2.2 Variabel Acak Kontinu
Connected Topics	3.3 Distribusi Bersyarat
Referensi	Miller Bab 5; Hogg-Tanis-Zimm Bab 3

ℹRumus›

$L \sim U(a, 19)$ ; tanpa deductible: $E[L] = (a+19)/2 = 10 \implies a = 1$ .

Dengan deductible $d = 4$ : $E[(L-4)^+] = \int_4^{19} (x-4) \cdot \frac{1}{18}\,dx$

Diketahui:

Policy A: tidak ada deductible, $E[\text{pembayaran}] = 10 \implies E[L] = 10$
Policy B: deductible $= 4$

▸Langkah Pengerjaan›

Langkah 1: Tentukan $a$

E[L] = \frac{a + 19}{2} = 10 \implies a + 19 = 20 \implies a = 1

Jadi $L \sim U(1, 19)$ , panjang interval $= 18$ .

Langkah 2: Hitung $E[(L-4)^+]$

E[(L-4)^+] = \int_4^{19} (x-4) \cdot \frac{1}{18}\,dx = \frac{1}{18}\left[\frac{(x-4)^2}{2}\right]_4^{19} = \frac{1}{18} \cdot \frac{(15)^2}{2} = \frac{225}{36} = \frac{25}{4} = 6{,}25

Hasil Akhir: (b). $6{,}25$

◈Jebakan Umum›

⬡Kesalahan Konseptual›

Mengira $E[\text{pembayaran B}] = E[L] - 4 = 6$ ; ini salah karena pembayaran hanya dilakukan jika $L > 4$ .
Lupa bahwa $P(L > 4) = 15/18 \neq 1$ ; klaim hanya dibayar untuk $15/18$ dari total kasus.

▲Red Flags›

Untuk $L \sim U(a,b)$ dengan deductible $d$ : $E[(L-d)^+] = \frac{(b-d)^2}{2(b-a)}$ .

No. 598

The profit $X$ on an insurance policy is normally distributed with $E(X^2) = 74$ and $P[X > 0] = 0{,}92$ .

Calculate the variance of the profit on this policy.

a. $24{,}88$
b. $30{,}77$
c. $34{,}50$
d. $44{,}20$
e. $62{,}86$

≡Jawaban No. 598›

(a). $24{,}88$

Field	Isi
Topik CF2	Topik 2 — Variabel Acak Univariat
Sub-topik	2.6 Distribusi Kontinu Umum
Difficulty	Hard
Prerequisite	2.2 Variabel Acak Kontinu
Connected Topics	4.5 Estimasi Parameter
Referensi	Miller Bab 6; Hogg-Tanis-Zimm Bab 5.5

ℹRumus›

P(X > 0) = 0{,}92 \implies \Phi(\mu/\sigma) = 0{,}92 \implies \mu/\sigma = 1{,}405

E[X^2] = \text{Var}(X) + (E[X])^2 = \sigma^2 + \mu^2 = 74

Diketahui:

$E[X^2] = 74$ ; $P(X > 0) = 0{,}92$

▸Langkah Pengerjaan›

Langkah 1: Tentukan $\mu/\sigma$

P(X > 0) = \Phi\!\left(\frac{\mu}{\sigma}\right) = 0{,}92 \implies \frac{\mu}{\sigma} = 1{,}405

\implies \mu = 1{,}405\,\sigma

Langkah 2: Substitusi ke $E[X^2]$

\sigma^2 + \mu^2 = \sigma^2 + (1{,}405\sigma)^2 = \sigma^2(1 + 1{,}974) = 2{,}974\,\sigma^2 = 74

\sigma^2 = \frac{74}{2{,}974} \approx 24{,}88

Hasil Akhir: (a). $24{,}88$

◈Jebakan Umum›

⬡Kesalahan Konseptual›

Mengira $E[X^2] = \text{Var}(X)$ ; yang benar adalah $E[X^2] = \text{Var}(X) + (E[X])^2$ .
Salah membaca tabel: $P(X > 0) = 0{,}92$ berarti $z = 1{,}405$ , bukan $z = 0{,}92$ .

▲Red Flags›

Dua persamaan ( $\mu/\sigma = c$ dan $\sigma^2 + \mu^2 = k$ ) → substitusi $\mu = c\sigma$ ke persamaan kedua untuk mendapat $\sigma^2$ .

No. 599

A random variable is exponentially distributed with mean $\mu$ .

Determine the difference between the 75th percentile and the 25th percentile of this distribution.

a. $\mu / \ln 3$
b. $\mu$
c. $\mu \ln 2$
d. $\mu \ln 3$
e. $2\mu$

≡Jawaban No. 599›

(d). $\mu \ln 3$

Field	Isi
Topik CF2	Topik 2 — Variabel Acak Univariat
Sub-topik	2.6 Distribusi Kontinu Umum
Difficulty	Medium
Prerequisite	2.2 Variabel Acak Kontinu
Connected Topics	2.4 Transformasi Variabel Acak Univariat
Referensi	Hogg-Tanis-Zimm Bab 3.2; Miller Bab 5

ℹRumus›

Persentil ke- $p$ dari $\text{Exp}(\beta = \mu)$ : $x_p = -\mu \ln(1-p)$

Diketahui:

$X \sim \text{Exp}(\mu)$ ; target: $x_{0{,}75} - x_{0{,}25}$

▸Langkah Pengerjaan›

Langkah 1: Hitung persentil ke-75 ( $a$ )

P(X \leq a) = 1 - e^{-a/\mu} = 0{,}75 \implies e^{-a/\mu} = 0{,}25 \implies a = -\mu \ln(0{,}25) = \mu \ln 4

Langkah 2: Hitung persentil ke-25 ( $b$ )

1 - e^{-b/\mu} = 0{,}25 \implies e^{-b/\mu} = 0{,}75 \implies b = -\mu \ln(0{,}75) = \mu \ln(4/3)

Langkah 3: Hitung selisih

a - b = \mu \ln 4 - \mu \ln(4/3) = \mu \ln\!\left(\frac{4}{4/3}\right) = \mu \ln 3

Hasil Akhir: (d). $\mu \ln 3$

◈Jebakan Umum›

⬡Kesalahan Konseptual›

Mengira IQR Eksponensial $= \mu \ln 2$ (median dikali 2); IQR $= \mu \ln 4 - \mu \ln(4/3) = \mu \ln 3$ .
Salah menghitung $\ln(4) - \ln(4/3) = \ln(4 \times 3/4) = \ln 3$ ; periksa kembali hukum logaritma.

▲Red Flags›

IQR Eksponensial $= -\mu[\ln(0{,}25) - \ln(0{,}75)] = \mu \ln 3 \approx 1{,}099\mu$ .

No. 600

Hospitalization losses under a certain policy are mutually independent and each has cumulative distribution function

F(x) = \begin{cases} 0, & x \leq 0 \\ \dfrac{15x^2 + x}{16}, & 0 < x \leq 1 \\ 1, & x > 1 \end{cases}

For each hospitalization loss, the insurer reimburses up to a maximum of 0.80.

Calculate the probability that a policyholder’s first partially unreimbursed hospitalization loss occurs on the third hospitalization.

a. $0{,}027$
b. $0{,}080$
c. $0{,}128$
d. $0{,}148$
e. $0{,}350$

≡Jawaban No. 600›

(d). $0{,}148$

Field	Isi
Topik CF2	Topik 1 — Dasar-Dasar Probabilitas
Sub-topik	1.5 Kejadian Independen, 1.2 Aksioma dan Perhitungan Probabilitas
Difficulty	Hard
Prerequisite	2.2 Variabel Acak Kontinu, 2.5 Distribusi Diskrit Umum
Connected Topics	3.3 Distribusi Bersyarat
Referensi	Hogg-Tanis-Zimm Bab 2; Miller Bab 5

ℹRumus›

Kerugian terganti penuh ( $\leq 0{,}80$ ) jika $X \leq 0{,}80$ ; terganti sebagian jika $X > 0{,}80$ .

p = P(\text{terganti penuh}) = F(0{,}80) = \frac{15(0{,}64) + 0{,}80}{16} = \frac{9{,}6 + 0{,}8}{16} = \frac{10{,}4}{16} = 0{,}65

“Kerugian terganti sebagian pertama terjadi di rawat inap ke-3” = dua pertama terganti penuh, ketiga terganti sebagian.

Diketahui:

CDF diberikan; cap $= 0{,}80$ ; target: $P(\text{pertama kali sebagian di rawat inap ke-3})$

▸Langkah Pengerjaan›

Langkah 1: Hitung $p = P(\text{terganti penuh}) = P(X \leq 0{,}80)$

p = F(0{,}80) = \frac{15(0{,}80)^2 + 0{,}80}{16} = \frac{15(0{,}64) + 0{,}80}{16} = \frac{9{,}6 + 0{,}8}{16} = \frac{10{,}4}{16} = 0{,}65

Langkah 2: Probabilitas pertama kali sebagian di ke-3

Distribusi Geometrik-like: dua pertama sukses (terganti penuh), ketiga gagal (terganti sebagian):

P = p^2 \times (1-p) = (0{,}65)^2 \times (0{,}35) = 0{,}4225 \times 0{,}35 = 0{,}147875 \approx 0{,}148

Hasil Akhir: (d). $0{,}148$

◈Jebakan Umum›

⬡Kesalahan Konseptual›

Mengira $p = 0{,}80$ (nilai cap) langsung; $p$ harus dihitung dari CDF: $p = F(0{,}80) = 0{,}65$ .
“Terganti sebagian” berarti $X > 0{,}80$ ; probabilitasnya $= 1 - F(0{,}80) = 0{,}35$ .

▲Red Flags›

“Pertama kali terjadi pada rawat inap ke- $n$ ” → $(p)^{n-1}(1-p)$ ; pola Geometrik.