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Question 258 MCQ · 0.0 marks
Discrete random variables:
Distribution PMF (fx (k)) CDF (Fx()) E[X] Var(X)
0
x <0
x=k
Uniform(AА) k-a+1 k<x<k+1 a+b n2-1
72
n=b-a+1 12
A= {a,a+1,..., 6} k=a,a+1,..., 6-1,b
2
k= a, a+1,...,b
1 xn
0 x <0
x=1
Bernoulli(p( 1-p 0≤ x <1 d p(1-p)
1-p=0
1 x≥1
0 x<0
"Скpk(1-p)n-k, "Cip (1-p)n- k<x<k+1
Binomial(n, p) -0 up np(1-p)
k =0,1,...,n
k =0,1,...,n
1 xn
0
x <0
1ㄧ2 p2
(1-p)k-1p, 1-p
Geometric(p) 1− (1-p) k≤x<k+1
k =1,..., 00
k = 1,..., 00
이 x<0
rrーa
A 시
Poisson(A) k! k<x<k+1
e2i!i=0
k = 0, 1,...,0∞
k = 0, 1,...,∞
6406535934300.
Useful Data has been mentioned above.
6406535934301.
This data attachment is just for a reference & not for an evaluation.
Published solution
**1. Review the reference sheet:**
Question 258 provides the official formula and table reference sheet ("Useful data") for the examination. Selecting **"Useful Data has been mentioned above."** acknowledges the reference sheet.
**Answer: Option A**
A: **Correct:** Confirms and acknowledges the useful data and formulas provided above.
B: **Incorrect:** Standard exam instructions require selecting the first option to acknowledge the reference sheet.
**1. Find the normalization constant $k$:**
The joint PDF is $f_{XY}(x, y) = kxy$ for $1 < x < 3$ and $0 < y < x$ (since $X$ and $Y$ represent weights of apples in quintals, $y > 0$).
Set the double integral over the support equal to $1$:
$$\int_1^3 \int_0^x kxy\,dy\,dx = k \int_1^3 x \left[\frac{y^2}{2}\right]_0^x dx = \frac{k}{2} \int_1^3 x^3\,dx = \frac{k}{2}\left[\frac{x^4}{4}\right]_1^3 = \frac{k}{8}(81 - 1) = 10k = 1$$
Thus, $k = \frac{1}{10}$.
**2. Compute $P(2 < X < 3,\; 0 < Y < 1)$:**
Note that for $2 < x < 3$ and $0 < y < 1$, the condition $y < x$ is automatically satisfied:
$$P(2 < X < 3,\; 0 < Y < 1) = \int_2^3 \int_0^1 \frac{1}{10}xy\,dy\,dx = \frac{1}{10} \left(\int_2^3 x\,dx\right)\left(\int_0^1 y\,dy\right)$$
$$= \frac{1}{10} \cdot \left[\frac{x^2}{2}\right]_2^3 \cdot \left[\frac{y^2}{2}\right]_0^1 = \frac{1}{10} \cdot \frac{9 - 4}{2} \cdot \frac{1}{2} = \frac{1}{10} \cdot \frac{5}{4} = \frac{1}{8} = 0.125$$
**Answer: 0.125**
**1. Expand the expectation in equation (1) (Blank A):**
Given $E\left[\frac{3X_1 + kX_2 - 5X_3}{4}\right] = 50$, multiply both sides by $4$ and use linearity of expectation:
$$3E[X_1] + kE[X_2] - 5E[X_3] = 200$$
Therefore, Blank A is **$kE[X_2] - 5E[X_3]$**.
**2. Match with the numbered options list:**
In the options list, **$kE[X_2] - 5E[X_3]$** is option **(2)**.
**Answer: 2**
**1. Factor the constant $1/4$ out of the variance (Blank C):**
Using the property $\text{Var}(cW) = c^2\text{Var}(W)$ with $c = \frac{1}{4}$:
$$\text{Var}\left(\frac{3X_1 + 4X_2 - 5X_3}{4}\right) = \left(\frac{1}{4}\right)^2 \text{Var}(3X_1 + 4X_2 - 5X_3) = \frac{1}{16}\text{Var}(3X_1 + 4X_2 - 5X_3)$$
Thus, Blank C is **$\frac{1}{16}$**.
**2. Match with the numbered options list:**
In the options list, **$\frac{1}{16}$** is option **(7)**.
**Answer: 7**
**1. Compute the final value of $\text{Var}(Z)$ (Blank F):**
$$\text{Var}(Z) = \frac{1}{16} \times 1250 = 78.125 \approx 78.13$$
**2. Match with the numbered options list:**
In the options list, **$78.13$** is option **(9)**.
**Answer: 9**
**1. Find the range and inverse transformation for $Y = 70 + 5X^2$:**
- Since $0 \le X \le 4$ is non-negative, the transformation $y = 70 + 5x^2$ is strictly increasing on $[0, 4]$ and maps $x \in [0, 4]$ to $y \in [70,\; 70 + 5(16)] = [70, 150]$.
- Solving for $x > 0$ in terms of $y$:
$$x = \sqrt{\frac{y - 70}{5}}$$
- Differentiating $x$ with respect to $y$ gives the Jacobian factor:
$$\frac{dx}{dy} = \frac{1}{2\sqrt{5(y - 70)}}$$
**2. Substitute into the change-of-variables formula $f_Y(y) = f_X(x)\left|\frac{dx}{dy}\right|$:**
$$f_Y(y) = \frac{1}{8}\left(4 - \sqrt{\frac{y - 70}{5}}\right) \cdot \frac{1}{2\sqrt{5(y - 70)}} = \frac{1}{16}\left(\frac{4}{\sqrt{5(y - 70)}} - \frac{1}{5}\right) = \frac{1}{20}\left(\sqrt{\frac{5}{y - 70}} - \frac{1}{4}\right)$$
for $70 \le y \le 150$, and $0$ otherwise.
**Answer: Option A**
A: **Correct:** Applying the change-of-variables formula yields $f_Y(y) = \frac{1}{20}\left(\sqrt{\frac{5}{y-70}} - \frac{1}{4}\right)$ for $70 \le y \le 150$.
B: **Incorrect:** Missing parentheses around $\sqrt{\frac{5}{y-70}} - \frac{1}{4}$, so $-\frac{1}{4}$ is not multiplied by $\frac{1}{20}$.
C: **Incorrect:** The square root term is inverted ($\sqrt{\frac{y-70}{5}}$ instead of $\sqrt{\frac{5}{y-70}}$).
D: **Incorrect:** Both the outside factor $\frac{1}{2}$ and the inverted square root are incorrect.
**1. Find $E[X]$ from the power series expansion of $M_X(\lambda)$:**
The moment generating function of $X$ is:
$$M_X(\lambda) = 1 + E[X]\lambda + \frac{E[X^2]\lambda^2}{2!} + \frac{E[X^3]\lambda^3}{3!} + \dots = 1 + 6\lambda + \frac{20\lambda^2}{2!} + \frac{70\lambda^3}{3!} + \dots$$
Comparing the coefficient of $\lambda$, we see that $E[X] = 6$.
**2. Compute the MGF of $Y = X - E[X] = X - 6$:**
$$M_Y(\lambda) = E\left[e^{\lambda(X - 6)}\right] = e^{-6\lambda} E\left[e^{\lambda X}\right] = e^{-6\lambda} M_X(\lambda)$$
Substituting $M_X(\lambda)$ gives:
$$M_Y(\lambda) = e^{-6\lambda}\left(1 + 6\lambda + \frac{20\lambda^2}{2!} + \frac{70\lambda^3}{3!} + \dots\right)$$
**Answer: Option D**
A: **Incorrect:** Uses $e^{+6\lambda}$ instead of $e^{-6\lambda}$ for $Y = X - 6$.
B: **Incorrect:** $E[X]$ is $6$, not $-10$.
C: **Incorrect:** Uses $e^{\lambda}$ instead of $e^{-6\lambda}$.
D: **Correct:** Since $E[X] = 6$, $M_Y(\lambda) = E[e^{\lambda(X-6)}] = e^{-6\lambda}M_X(\lambda)$.
**1. Identify the distribution of $X$ from its MGF:**
Given $M_X(\lambda) = \frac{1}{4}e^{\lambda} + \frac{1}{2}e^{2\lambda} + \frac{1}{4}e^{3\lambda}$:
- Since $M_X(\lambda) = \sum_x P(X = x)e^{\lambda x}$, $X$ is a **discrete random variable with range $\{1, 2, 3\}$** and probabilities $P(X = 1) = \frac{1}{4}$, $P(X = 2) = \frac{1}{2}$, $P(X = 3) = \frac{1}{4}$. (**Option A is correct**)
**2. Compute $E[X]$ and $\text{Var}(X)$:**
- Mean:
$$E[X] = 1\left(\frac{1}{4}\right) + 2\left(\frac{1}{2}\right) + 3\left(\frac{1}{4}\right) = 2$$
(**Option B is correct**)
- Second moment and variance:
$$E[X^2] = 1^2\left(\frac{1}{4}\right) + 2^2\left(\frac{1}{2}\right) + 3^2\left(\frac{1}{4}\right) = \frac{1}{4} + 2 + \frac{9}{4} = 4.5$$
$$\text{Var}(X) = E[X^2] - (E[X])^2 = 4.5 - 4 = 0.5$$
**3. Apply the Central Limit Theorem to $Y = \sum_{i=1}^{100} X_i$:**
For $n = 100$ i.i.d. variables, $Y = \sum_{i=1}^{100} X_i$ is approximately normal with:
- Mean: $nE[X] = 100 \times 2 = 200$
- Variance: $n\text{Var}(X) = 100 \times 0.5 = 50$
(**Option C is correct**, and Option D is incorrect).
**Answer: Options A, B, C**
A: **Correct:** The exponents $1, 2, 3$ in $M_X(\lambda)$ show that $X$ takes values in $\{1, 2, 3\}$.
B: **Correct:** $E[X] = \frac{1}{4}(1) + \frac{1}{2}(2) + \frac{1}{4}(3) = 2$.
C: **Correct:** By the CLT, $Y = \sum_{i=1}^{100} X_i$ is approximately $\text{Normal}(100 \times 2,\; 100 \times 0.5) = \text{Normal}(200, 50)$.
D: **Incorrect:** $2$ and $0.5$ are the mean and variance of a single $X_i$, not of the sum $Y = \sum_{i=1}^{100} X_i$.
is the sample
mean.
Based on the above data, answer the given subquestions.
Find the value of
.Enter the answer correct to one decimal place.
Published solution
**1. Find the mean of $X_i \sim \text{Uniform}[0, 1]$:**
For $X_i \sim \text{Uniform}[a, b]$ with $a = 0$ and $b = 1$:
$$E[X_i] = \frac{a + b}{2} = \frac{0 + 1}{2} = 0.5$$
**2. Find the expected value of the sample mean $Y = \frac{1}{6}\sum_{i=1}^6 X_i$:**
$$E[Y] = E[X_1] = 0.5$$
**Answer: 0.5**
