Graded Assignment 5

❓ Question 1: Battery Life

A person randomly chooses a battery from a store which has 90 batteries of type A and 260 batteries of type B. Battery life of type A and type B batteries are exponentially distributed with average life of 9.0 years and 13.0 years, respectively. If the chosen battery lasts for 5 years, what is the probability that the battery is of type A? Enter your answer correct to two decimals accuracy.

Core Concepts

1. Bayes’ Theorem: This theorem is used to update the probability of an event based on new evidence. The formula is: $$P(A|E) = \frac{P(E|A)P(A)}{P(E)}$$ Where $P(E) = P(E|A)P(A) + P(E|B)P(B)$ (the Law of Total Probability).
2. Exponential Distribution: This distribution models the time until an event (like battery failure).
   • Mean (Average Life): $E[T] = \frac{1}{\lambda}$, where $\lambda$ is the rate parameter.
   • Survival Function $P(T > t)$: The probability that the item lasts longer than time $t$ is $P(T > t) = e^{-\lambda t}$.

Solution

1. Define Events:

   • $A$: The chosen battery is Type A.
   • $B$: The chosen battery is Type B.
   • $E$: The battery lasts for more than 5 years ($T > 5$).
   • Goal: We need to find $P(A|E)$, the probability it was Type A given that it lasted more than 5 years.

2. Find Prior Probabilities ($P(A)$ and $P(B)$):

   • Total batteries = $90 + 260 = 350$.
   • $P(A) = \frac{90}{350} = \frac{9}{35}$
   • $P(B) = \frac{260}{350} = \frac{26}{35}$

3. Find Likelihoods ($P(E|A)$ and $P(E|B)$): This is the probability of survival $P(T > 5)$ for each battery type.

   • For Type A:
     • Average life $E[T_A] = 9 = \frac{1}{\lambda_A} \implies \lambda_A = \frac{1}{9}$.
     • $P(E|A) = P(T_A > 5) = e^{-\lambda_A \cdot 5} = e^{-5/9}$.
   • For Type B:
     • Average life $E[T_B] = 13 = \frac{1}{\lambda_B} \implies \lambda_B = \frac{1}{13}$.
     • $P(E|B) = P(T_B > 5) = e^{-\lambda_B \cdot 5} = e^{-5/13}$.

4. Apply Bayes’ Theorem:

   $$P(A|E) = \frac{P(E|A)P(A)}{P(E|A)P(A) + P(E|B)P(B)}$$$$P(A|E) = \frac{e^{-5/9} \cdot \left(\frac{9}{35}\right)}{e^{-5/9} \cdot \left(\frac{9}{35}\right) + e^{-5/13} \cdot \left(\frac{26}{35}\right)}$$
   • We can cancel the $\frac{1}{35}$ from the numerator and denominator: $$P(A|E) = \frac{9e^{-5/9}}{9e^{-5/9} + 26e^{-5/13}}$$

5. Calculate the Value:

   • $e^{-5/9} \approx e^{-0.555…} \approx 0.5738$
   • $e^{-5/13} \approx e^{-0.3846…} \approx 0.6807$
   • $P(A|E) \approx \frac{9 \cdot 0.5738}{9 \cdot 0.5738 + 26 \cdot 0.6807}$
   • $P(A|E) \approx \frac{5.1642}{5.1642 + 17.6982} = \frac{5.1642}{22.8624} \approx 0.22588$

6. Final Answer: Rounded to two decimal places, the probability is 0.23.

   (Note: The multiple-choice options provided in the image are inconsistent with the problem’s numbers. For example, Option 1, $\frac{1}{1 + 2e^{4000/23400}}$, implies a prior probability ratio $P(B)/P(A) = 2$, but the problem gives $260/90 \approx 2.89$. The calculation above correctly uses the numbers from the problem statement.)

❓ Question 2: $Y = XZ$ Transformation

Let $Y = XZ$, where $X \sim \text{Uniform}{1, 2, 3}$ and $Z \sim \text{Normal}(1, 4)$ are independent. Find the value of $f_{X,Y|Y=2}(2)$.

Core Concepts

1. Bayes’ Theorem for Mixed Variables: We are finding the probability of a discrete state ($X=k$) given a continuous observation ($Y=y$). $$P(X=k | Y=y) = \frac{f_{Y|X=k}(y) P(X=k)}{f_Y(y)}$$
2. Law of Total Probability (Density): The denominator $f_Y(y)$ is the marginal density of $Y$, found by summing over all possibilities of $X$: $$f_Y(y) = \sum_{i} f_{Y|X=i}(y) P(X=i)$$
3. Transformation of Normal Variables: If $Z \sim N(\mu, \sigma^2)$, then $Y = kZ \sim N(k\mu, k^2\sigma^2)$.

Solution

1. Interpret the Notation: The notation $f_{X,Y|Y=2}(2)$ is ambiguous. The most standard interpretation is that it is a typo for $P(X=2 | Y=2)$. We will solve for this value.

2. Identify Priors $P(X=k)$:

   • $X \sim \text{Uniform}{1, 2, 3}$, so $P(X=1) = P(X=2) = P(X=3) = \frac{1}{3}$.

3. Find Conditional Likelihoods $f_{Y|X=k}(y)$:

   • We are given $Z \sim N(\mu=1, \sigma^2=4)$.
   • If $X=1$: $Y = 1 \cdot Z = Z$. So $Y|X=1 \sim N(1, 4)$.
     • $f_{Y|X=1}(y) = \frac{1}{\sqrt{2\pi \cdot 4}} e^{-\frac{(y-1)^2}{2 \cdot 4}} = \frac{1}{\sqrt{8\pi}} e^{-\frac{(y-1)^2}{8}}$.
   • If $X=2$: $Y = 2Z$. The new mean is $2 \cdot 1 = 2$. The new variance is $2^2 \cdot 4 = 16$.
     • $Y|X=2 \sim N(2, 16)$.
     • $f_{Y|X=2}(y) = \frac{1}{\sqrt{2\pi \cdot 16}} e^{-\frac{(y-2)^2}{2 \cdot 16}} = \frac{1}{\sqrt{32\pi}} e^{-\frac{(y-2)^2}{32}}$.
   • If $X=3$: $Y = 3Z$. The new mean is $3 \cdot 1 = 3$. The new variance is $3^2 \cdot 4 = 36$.
     • $Y|X=3 \sim N(3, 36)$.
     • $f_{Y|X=3}(y) = \frac{1}{\sqrt{2\pi \cdot 36}} e^{-\frac{(y-3)^2}{2 \cdot 36}} = \frac{1}{\sqrt{72\pi}} e^{-\frac{(y-3)^2}{72}}$.

4. Apply Bayes’ Theorem for $P(X=2 | Y=2)$:

   $$P(X=2 | Y=2) = \frac{f_{Y|X=2}(2) P(X=2)}{\sum_{i=1}^3 f_{Y|X=i}(2) P(X=i)}$$
   • Since $P(X=i) = \frac{1}{3}$ for all $i$, we can cancel it from the numerator and denominator: $$P(X=2 | Y=2) = \frac{f_{Y|X=2}(2)}{f_{Y|X=1}(2) + f_{Y|X=2}(2) + f_{Y|X=3}(2)}$$

5. Evaluate each density at $y=2$:

   • $f_{Y|X=1}(2) = \frac{1}{\sqrt{8\pi}} e^{-\frac{(2-1)^2}{8}} = \frac{1}{\sqrt{8\pi}} e^{-1/8}$
   • $f_{Y|X=2}(2) = \frac{1}{\sqrt{32\pi}} e^{-\frac{(2-2)^2}{32}} = \frac{1}{\sqrt{32\pi}} e^{0} = \frac{1}{\sqrt{32\pi}}$
   • $f_{Y|X=3}(2) = \frac{1}{\sqrt{72\pi}} e^{-\frac{(2-3)^2}{72}} = \frac{1}{\sqrt{72\pi}} e^{-1/72}$

6. Substitute and Simplify:

   • First, simplify the root terms: $\sqrt{32\pi} = \sqrt{4 \cdot 8\pi} = 2\sqrt{8\pi}$ and $\sqrt{72\pi} = \sqrt{9 \cdot 8\pi} = 3\sqrt{8\pi}$. $$P(X=2 | Y=2) = \frac{\frac{1}{2\sqrt{8\pi}}}{\frac{1}{\sqrt{8\pi}} e^{-1/8} + \frac{1}{2\sqrt{8\pi}} + \frac{1}{3\sqrt{8\pi}} e^{-1/72}}$$
   • Cancel $\frac{1}{\sqrt{8\pi}}$ from all terms: $$P(X=2 | Y=2) = \frac{1/2}{e^{-1/8} + 1/2 + \frac{1}{3}e^{-1/72}}$$
   • To make it cleaner, multiply the numerator and denominator by 6: $$P(X=2 | Y=2) = \frac{3}{6e^{-1/8} + 3 + 2e^{-1/72}}$$

7. Final Answer: The derived answer is $\frac{3}{6e^{-1/8} + 3 + 2e^{-1/72}}$. None of the provided options match this correct derivation. The options given are fundamentally inconsistent with the problem statement $Z \sim N(1, 4)$. For example, the term $e^{-9/8}$ in the options cannot be derived from the given parameters.

❓ Question 3: $Y = (1-X)^9$ Transformation

Let $X$ be a continuous random variable with the following PDF: $f_X(x) = \begin{cases} (6)(1-x)^5 & 0 < x < 1 \ 0 & \text{otherwise} \end{cases}$ Define $Y = (1-X)^9$. Find the PDF of the random variable $Y$.

Core Concepts

1. Transformation of Variables (Formula Method): If $Y=g(X)$ and $g(x)$ is a monotonic function, we can find the new PDF using the formula: $$f_Y(y) = f_X(g^{-1}(y)) \left| \frac{dx}{dy} \right|$$ where $x = g^{-1}(y)$ is the inverse function.

Solution

1. Check Monotonicity and Find Range of Y:

   • The function $g(x) = (1-x)^9$ is monotonic decreasing on the interval $0 < x < 1$.
   • Find the bounds for $y$:
     • When $x \to 0$, $y \to (1-0)^9 = 1$.
     • When $x \to 1$, $y \to (1-1)^9 = 0$.
   • Therefore, the range for $Y$ is $0 < y < 1$.

2. Find the Inverse Function ($x = g^{-1}(y)$):

   • $y = (1-x)^9$
   • $y^{1/9} = 1-x$
   • $x = 1 - y^{1/9}$

3. Find the Derivative of the Inverse ($|\frac{dx}{dy}|$):

   • $\frac{dx}{dy} = \frac{d}{dy} (1 - y^{1/9}) = 0 - \frac{1}{9}y^{(1/9 - 1)} = -\frac{1}{9}y^{-8/9}$
   • The absolute value is $\left| \frac{dx}{dy} \right| = \frac{1}{9}y^{-8/9}$.

4. Apply the Transformation Formula:

   • $f_Y(y) = f_X(x) \left| \frac{dx}{dy} \right|$
   • Substitute $x = 1 - y^{1/9}$ into $f_X(x)$:
     • $f_X(1 - y^{1/9}) = 6(1 - (1 - y^{1/9}))^5 = 6(y^{1/9})^5 = 6y^{5/9}$
   • Now, multiply by the derivative:
     • $f_Y(y) = (6y^{5/9}) \cdot (\frac{1}{9}y^{-8/9})$
     • $f_Y(y) = \frac{6}{9} y^{(5/9 - 8/9)} = \frac{6}{9} y^{-3/9}$

5. Final Answer: The PDF for $Y$ is $f_Y(y) = \frac{6}{9}y^{-3/9}$ for $0 < y < 1$. This matches Option 3, which is $f_Y(y) = \begin{cases} \frac{6}{9}y^{-3/9} & 0 < y < 1 \ 0 & \text{otherwise} \end{cases}$. (This simplifies to $\frac{2}{3}y^{-1/3}$, but is left un-simplified in the option).

❓ Question 4: $Y = \frac{1}{3}(27-X)$ Transformation

Let $X$ be a continuous random variable with the following PDF: $f_X(x) = \begin{cases} \frac{x^2}{81} & -6 < x < 3 \ 0 & \text{otherwise} \end{cases}$ Define $Y = \frac{1}{3}(27-X)$. Find the PDF of the random variable $Y$.

Core Concepts

1. Transformation of Variables (Formula Method): We use the same formula as in the previous question: $$f_Y(y) = f_X(g^{-1}(y)) \left| \frac{dx}{dy} \right|$$

Solution

1. Check Monotonicity and Find Range of Y:

   • The function $g(x) = \frac{1}{3}(27-X) = 9 - \frac{x}{3}$ is a monotonic decreasing linear function.
   • Find the bounds for $y$ by plugging in the bounds for $x$:
     • When $x = -6$, $y = \frac{1}{3}(27 - (-6)) = \frac{33}{3} = 11$.
     • When $x = 3$, $y = \frac{1}{3}(27 - 3) = \frac{24}{3} = 8$.
   • Since the function is decreasing, the bounds flip. The new range is $8 < y < 11$, or $\frac{24}{3} < y < \frac{33}{3}$. This immediately eliminates options 2 and 4.

2. Find the Inverse Function ($x = g^{-1}(y)$):

   • $y = \frac{1}{3}(27 - x)$
   • $3y = 27 - x$
   • $x = 27 - 3y$

3. Find the Derivative of the Inverse ($|\frac{dx}{dy}|$):

   • $\frac{dx}{dy} = \frac{d}{dy} (27 - 3y) = -3$
   • The absolute value is $\left| \frac{dx}{dy} \right| = 3$.

4. Apply the Transformation Formula:

   • $f_Y(y) = f_X(x) \left| \frac{dx}{dy} \right|$
   • Substitute $x = 27 - 3y$ into $f_X(x)$:
     • $f_X(27 - 3y) = \frac{(27 - 3y)^2}{81}$
   • Now, multiply by the derivative:
     • $f_Y(y) = \frac{(27 - 3y)^2}{81} \cdot 3 = \frac{3(27 - 3y)^2}{81}$

5. Final Answer: The PDF for $Y$ is $f_Y(y) = \frac{3(27 - 3y)^2}{81}$ for $\frac{24}{3} < y < \frac{33}{3}$. This matches Option 3.

❓ Question 5: $6E[X]$

Let $X$ be a continuous random variable with the following PDF: $f_X(x) = \begin{cases} x^{14}(17x^2 + 16x - 15) & 0 < x < 1 \ 0 & \text{otherwise} \end{cases}$ Find the value of $6E[X]$. Enter your answer correct to two decimals accuracy.

Core Concepts

1. Expected Value of $X$: For a continuous variable, the expected value (or mean) is found by integrating $x$ times the PDF over its support: $$E[X] = \int_{-\infty}^{\infty} x \cdot f_X(x) \,dx$$
2. Polynomial Integration: The integral of $x^n$ is $\frac{x^{n+1}}{n+1}$.

Solution

1. Set up the $E[X]$ Integral:

   $$E[X] = \int_0^1 x \cdot \left[ x^{14}(17x^2 + 16x - 15) \right] \,dx$$$$E[X] = \int_0^1 x^{15}(17x^2 + 16x - 15) \,dx$$

2. Distribute and Simplify the Integrand:

   $$E[X] = \int_0^1 (17x^{17} + 16x^{16} - 15x^{15}) \,dx$$

3. Integrate Term by Term:

   $$E[X] = \left[ 17 \frac{x^{18}}{18} + 16 \frac{x^{17}}{17} - 15 \frac{x^{16}}{16} \right]_0^1$$

4. Evaluate at the Bounds:

   • Evaluating at the upper bound $x=1$: $E[X] = \frac{17(1)^{18}}{18} + \frac{16(1)^{17}}{17} - \frac{15(1)^{16}}{16}$
   • Evaluating at the lower bound $x=0$ gives 0.
   • So, $E[X] = \frac{17}{18} + \frac{16}{17} - \frac{15}{16}$.

5. Calculate the Value of $E[X]$:

   • $\frac{17}{18} \approx 0.94444…$
   • $\frac{16}{17} \approx 0.94117…$
   • $\frac{15}{16} = 0.9375$
   • $E[X] \approx 0.94444 + 0.94117 - 0.9375 \approx 0.94811…$

6. Find the Final Answer $6E[X]$:

   • $6E[X] = 6 \cdot (\frac{17}{18} + \frac{16}{17} - \frac{15}{16})$
   • $6 \times 0.94811… \approx 5.6887…$

7. Final Answer: Rounding to two decimals accuracy, the value is 5.69.

❓ Question 6: Variance of $Y=78X+5$

Let $X$ be a continuous random variable with the following PDF: $f_X(x) = \begin{cases} x & 0 \le x \le 1 \ 2-x & 1 < x \le 2 \ 0 & \text{otherwise} \end{cases}$ Define $Y = 78X + 5$. Find the variance of $Y$.

Core Concepts

1. Properties of Variance: For constants $a$ and $b$: $$\text{Var}(aX + b) = a^2 \text{Var}(X)$$
2. Variance Definition: The variance of $X$ is: $$\text{Var}(X) = E[X^2] - (E[X])^2$$
3. Symmetry: If a PDF is symmetric about a point $c$, then the mean $E[X] = c$.

Solution

1. Find $\text{Var}(Y)$ using its properties:

   • $Y = 78X + 5$, so $a = 78$ and $b = 5$.
   • $\text{Var}(Y) = 78^2 \text{Var}(X)$.
   • Our goal is to find $\text{Var}(X)$ and multiply it by $78^2$.

2. Find $E[X]$:

   • The PDF is a peaking at $x=1$. It is perfectly symmetric around $x=1$.
   • Due to this symmetry, the mean $E[X] = 1$.

3. Find $E[X^2]$:

   $$E[X^2] = \int_{-\infty}^{\infty} x^2 f_X(x) \,dx$$
   • We must split the integral based on the piecewise PDF: $$E[X^2] = \int_0^1 x^2(x) \,dx + \int_1^2 x^2(2-x) \,dx$$
   • First part: $\int_0^1 x^3 ,dx = \left[ \frac{x^4}{4} \right]_0^1 = \frac{1}{4}$.
   • Second part: $\int_1^2 (2x^2 - x^3) ,dx = \left[ \frac{2x^3}{3} - \frac{x^4}{4} \right]_1^2$
     • At $x=2$: $\frac{2(8)}{3} - \frac{16}{4} = \frac{16}{3} - 4 = \frac{4}{3}$
     • At $x=1$: $\frac{2(1)}{3} - \frac{1}{4} = \frac{8-3}{12} = \frac{5}{12}$
     • Value of second part: $\frac{4}{3} - \frac{5}{12} = \frac{16}{12} - \frac{5}{12} = \frac{11}{12}$.
   • Total $E[X^2]$: $\frac{1}{4} + \frac{11}{12} = \frac{3}{12} + \frac{11}{12} = \frac{14}{12} = \frac{7}{6}$.

4. Find $\text{Var}(X)$:

   • $\text{Var}(X) = E[X^2] - (E[X])^2 = \frac{7}{6} - (1)^2 = \frac{7}{6} - 1 = \frac{1}{6}$.

5. Find $\text{Var}(Y)$:

   • $\text{Var}(Y) = 78^2 \text{Var}(X) = 78^2 \cdot \frac{1}{6}$
   • $\text{Var}(Y) = \frac{78 \cdot 78}{6} = 13 \cdot 78 = 1014$.

6. Final Answer: The variance of $Y$ is 1014.

❓ Question 7: Normal Distribution Transformation

Suppose $X \sim \text{Normal}(3, 4)$. Find the PDF of $Y = 2X + 9$.

Core Concepts

1. Linear Transformation of a Normal Variable: A key property of the normal distribution is that it is “closed under linear transformation.” This means a linear function of a normal variable is also a normal variable.
2. Rules for Mean and Variance:
   • If $X \sim N(\mu_X, \sigma_X^2)$ and $Y = aX + b$:
   • New Mean: $\mu_Y = E[aX+b] = aE[X] + b = a\mu_X + b$
   • New Variance: $\sigma_Y^2 = \text{Var}(aX+b) = a^2 \text{Var}(X) = a^2 \sigma_X^2$
3. Normal PDF Formula: A variable $Y \sim N(\mu, \sigma^2)$ has the PDF: $$f_Y(y) = \frac{1}{\sqrt{2\pi \sigma^2}} e^{-\frac{(y-\mu)^2}{2\sigma^2}}$$

Solution

1. Identify Parameters of X:

   • $X \sim \text{Normal}(3, 4)$
   • Mean $\mu_X = 3$
   • Variance $\sigma_X^2 = 4$

2. Identify Transformation $Y = 2X + 9$:

   • $a = 2$
   • $b = 9$

3. Calculate New Mean $\mu_Y$:

   • $\mu_Y = a\mu_X + b = 2(3) + 9 = 6 + 9 = 15$.

4. Calculate New Variance $\sigma_Y^2$:

   • $\sigma_Y^2 = a^2 \sigma_X^2 = (2)^2 \cdot 4 = 4 \cdot 4 = 16$.

5. Determine the New Distribution:

   • $Y$ is also normally distributed: $Y \sim \text{Normal}(\mu_Y=15, \sigma_Y^2=16)$.

6. Write the PDF for Y:

   • We use the general Normal PDF formula with $\mu = 15$ and $\sigma^2 = 16$. $$f_Y(y) = \frac{1}{\sqrt{2\pi \cdot 16}} e^{-\frac{(y-15)^2}{2 \cdot 16}}$$ $$f_Y(y) = \frac{1}{\sqrt{32\pi}} e^{-\frac{(y-15)^2}{32}}$$

7. Final Answer: This matches Option 3.

❓ Question 8: Chebyshev’s Inequality

Suppose $X$ is a continuous random variable with mean 50 and variance 16. Using Chebyshev inequality, find the greatest lower bound of the probability that $X$ takes a value in between 42 and 58. (Enter the answer correct to two decimal places)

Core Concept

Chebyshev’s Inequality provides a lower bound on the probability that a random variable falls within a certain distance from its mean. The formula is:

where:

• $\mu$ is the mean.
• $\sigma$ is the standard deviation.
• $k$ is the number of standard deviations from the mean (and $k > 1$).

Solution

1. Identify Parameters:

   • Mean $\mu = 50$.
   • Variance $\sigma^2 = 16$.
   • Standard Deviation $\sigma = \sqrt{16} = 4$.

2. Define the Interval:

   • We want to find the probability $P(42 < X < 58)$.
   • We can rewrite this interval in terms of the mean: $P(50 - 8 < X < 50 + 8)$
   • This is the same as finding the probability that $X$ is within 8 units of the mean: $P(|X - 50| < 8)$

3. Find $k$:

   • The inequality is in the form $P(|X - \mu| < k\sigma)$.
   • We set our distance (8) equal to $k\sigma$: $k\sigma = 8$ $k(4) = 8$ $k = 2$

4. Apply Chebyshev’s Inequality:

   • Now we plug $k = 2$ into the formula: $P(|X - 50| < 8) \ge 1 - \frac{1}{k^2}$ $P(42 < X < 58) \ge 1 - \frac{1}{2^2}$ $P(42 < X < 58) \ge 1 - \frac{1}{4}$ $P(42 < X < 58) \ge 0.75$

5. Final Answer: The greatest lower bound for the probability is 0.75.

❓ Question 9: Expected Value of a Transformed Variable

A passenger train arrives punctually at a station every 20 minutes. Each morning, a passenger walks in to the train station. Let $X$ denote the amount of time (in minutes) the passenger waits for the train from the time he reaches the train station. It is known that the probability density function of $X$ is $f_X(x) = \begin{cases} 1/20 & \text{if } 0 < x < 20 \ 0 & \text{Otherwise} \end{cases}$. Find the expected value of $Y = X^3 + 22$.

Core Concepts

1. Uniform Distribution: The PDF $f_X(x) = 1/(b-a)$ for $a < x < b$ describes a continuous uniform distribution, $X \sim \text{Uniform}(0, 20)$.
2. Linearity of Expectation: $E[aG(X) + b] = aE[G(X)] + b$. In this case, $E[Y] = E[X^3 + 22] = E[X^3] + 22$.
3. LOTUS (Law of the Unconscious Statistician): This rule allows us to find the expected value of a function of $X$ without first finding the PDF of $Y$. $$E[g(X)] = \int_{-\infty}^{\infty} g(x) f_X(x) \,dx$$

Solution

1. Set up the Expectation:

   • We need to find $E[Y]$. Using linearity of expectation: $E[Y] = E[X^3 + 22] = E[X^3] + 22$
   • Our first step is to calculate $E[X^3]$.

2. Calculate $E[X^3]$ using LOTUS:

   • $g(x) = x^3$ and $f_X(x) = 1/20$.
   • $E[X^3] = \int_{0}^{20} x^3 \cdot \left(\frac{1}{20}\right) ,dx$
   • $E[X^3] = \frac{1}{20} \int_{0}^{20} x^3 ,dx$

3. Evaluate the Integral:

   • $E[X^3] = \frac{1}{20} \left[ \frac{x^4}{4} \right]_{0}^{20}$
   • $E[X^3] = \frac{1}{20} \left( \frac{20^4}{4} - \frac{0^4}{4} \right)$
   • $E[X^3] = \frac{1}{20} \cdot \frac{20^4}{4} = \frac{20^3}{4}$
   • $E[X^3] = \frac{8000}{4} = 2000$

4. Find $E[Y]$:

   • $E[Y] = E[X^3] + 22 = 2000 + 22 = 2022$.

5. Final Answer: The expected value of $Y$ is 2022.

❓ Question 10: Bayes’ Theorem (Mixed Variables)

60% of the total people in a city were male and 40% were female. The age of the males is Normal(60,25) and the age of the females is Normal(55,36). If the age of a randomly selected person is 60, what is the probability that the selected candidate is male?

Core Concept

Bayes’ Theorem for mixed variables (discrete events, continuous data) allows us to update our belief about a discrete event (like “Male” or “Female”) given a continuous measurement (like “Age = 60”).

The denominator, $f(\text{Age}=60)$, is found using the Law of Total Probability: $f(\text{Age}=60) = f(\text{Age}=60 | \text{Male})P(\text{Male}) + f(\text{Age}=60 | \text{Female})P(\text{Female})$

Normal PDF: $f(x) = \frac{1}{\sqrt{2\pi \sigma^2}} e^{-\frac{(x-\mu)^2}{2\sigma^2}}$

Solution

1. Define Events and Parameters:

   • $M$: Event the person is Male. $P(M) = 0.60$.
   • $F$: Event the person is Female. $P(F) = 0.40$.
   • $A$: Random variable for age.
   • Age given Male: $A|M \sim N(\mu=60, \sigma^2=25)$.
   • Age given Female: $A|F \sim N(\mu=55, \sigma^2=36)$.
   • Goal: Find $P(M | A=60)$.

2. Calculate Likelihoods (PDF values at A=60):

   • $f(A=60 | M)$: (PDF of $N(60, 25)$ at $x=60$)
     • The mean is 60, so we are at the peak. The exponent is $e^0 = 1$.
     • $f(A=60 | M) = \frac{1}{\sqrt{2\pi \cdot 25}} e^{-\frac{(60-60)^2}{2 \cdot 25}} = \frac{1}{\sqrt{50\pi}} = \frac{1}{5\sqrt{2\pi}}$
   • $f(A=60 | F)$: (PDF of $N(55, 36)$ at $x=60$)
     • $f(A=60 | F) = \frac{1}{\sqrt{2\pi \cdot 36}} e^{-\frac{(60-55)^2}{2 \cdot 36}} = \frac{1}{\sqrt{72\pi}} e^{-\frac{25}{72}} = \frac{1}{6\sqrt{2\pi}} e^{-25/72}$

3. Apply Bayes’ Theorem:

   $$P(M | A=60) = \frac{f(A=60 | M) P(M)}{f(A=60 | M)P(M) + f(A=60 | F)P(F)}$$$$P(M | A=60) = \frac{\left(\frac{1}{5\sqrt{2\pi}}\right) (0.60)}{\left(\frac{1}{5\sqrt{2\pi}}\right) (0.60) + \left(\frac{1}{6\sqrt{2\pi}} e^{-25/72}\right) (0.40)}$$

4. Simplify the Expression:

   • We can cancel the $\frac{1}{\sqrt{2\pi}}$ term from the numerator and all parts of the denominator. $$P(M | A=60) = \frac{\frac{0.60}{5}}{\frac{0.60}{5} + \frac{0.40}{6} e^{-25/72}}$$ $$P(M | A=60) = \frac{0.12}{0.12 + \left(\frac{0.40}{6}\right) e^{-25/72}}$$
   • To match the options, let’s multiply the numerator and denominator by a common multiple, like 100 or 300. Let’s try 75.
   • Or, simpler, let’s divide the numerator and denominator by the numerator ($0.12$): $$P(M | A=60) = \frac{1}{1 + \frac{0.40/6}{0.12} e^{-25/72}}$$
   • Let’s evaluate the fraction: $\frac{0.40/6}{0.12} = \frac{0.40}{6 \cdot 0.12} = \frac{0.40}{0.72} = \frac{40}{72} = \frac{5}{9}$. $$P(M | A=60) = \frac{1}{1 + \frac{5}{9} e^{-25/72}}$$
   • Now, multiply the numerator and denominator by 9 to clear the fraction: $$P(M | A=60) = \frac{9}{9(1 + \frac{5}{9} e^{-25/72})} = \frac{9}{9 + 5 e^{-25/72}}$$

5. Final Answer: The probability is $\frac{9}{9 + 5\exp\left(-\frac{25}{72}\right)}$, which is Option 1.

❓ Question 11: Transformation of Variables

Let $X \sim \text{Uniform}[0, 1]$ be a continuous random variable. Define a new random variable $Y = -\log(X)$. Find the probability density function of $Y$.

Core Concept

Transformation of Variables (CDF Method): This is often the most reliable way to find the PDF of a new variable $Y = g(X)$.

1. Find the range of $Y$.
2. Find the CDF of $Y$: $F_Y(y) = P(Y \le y)$.
3. Substitute $Y$ with $g(X)$: $P(g(X) \le y)$.
4. Solve the inequality for $X$ (be careful when multiplying/dividing by negatives).
5. Express $P(X \le \dots)$ using the known CDF of $X$, $F_X(x)$.
6. Differentiate the resulting $F_Y(y)$ to get the PDF: $f_Y(y) = \frac{d}{dy} F_Y(y)$.

Solution

1. Find the Range of $Y$:

   • The range of $X$ is $[0, 1]$.
   • As $x \to 0^+$, $Y = -\log(x) \to \infty$.
   • As $x \to 1^-$, $Y = -\log(1) \to 0$.
   • Therefore, the range of $Y$ is $(0, \infty)$, or $y > 0$.

2. Find the CDF of $Y$:

   • $F_Y(y) = P(Y \le y)$
   • $F_Y(y) = P(-\log(X) \le y)$
   • Multiply by -1 (this flips the inequality sign): $F_Y(y) = P(\log(X) \ge -y)$
   • Exponentiate both sides (since $e^x$ is a monotonic increasing function, the sign stays the same): $F_Y(y) = P(e^{\log(X)} \ge e^{-y})$ $F_Y(y) = P(X \ge e^{-y})$
   • Since $X \sim \text{Uniform}[0, 1]$, its CDF is $F_X(x) = x$ and $P(X \ge k) = 1 - P(X < k) = 1 - F_X(k) = 1 - k$.
   • Therefore, $P(X \ge e^{-y}) = 1 - e^{-y}$.
   • So, the CDF of $Y$ is $F_Y(y) = 1 - e^{-y}$, for $y > 0$.

3. Differentiate to find the PDF $f_Y(y)$:

   • $f_Y(y) = \frac{d}{dy} F_Y(y) = \frac{d}{dy} (1 - e^{-y})$
   • $f_Y(y) = 0 - (-e^{-y}) = e^{-y}$

4. Final Answer: The PDF is $f_Y(y) = e^{-y}$, for $y > 0$. This is the Exponential(1) distribution and matches Option 1.

❓ Question 12: Expected Value

(Use the following information to answer Q11 and Q12) Let $X \sim \text{Uniform}[0, 1]$ be a continuous random variable. Define a new random variable $Y = -\log(X)$. Find $E[Y]$.

Core Concept

Expected Value of a Known Distribution: If you can identify the distribution of $Y$, you can use its known expected value formula.

• From Question 11, we found that $Y \sim \text{Exponential}(\lambda)$ with $f_Y(y) = e^{-y}$.
• The general form of the exponential PDF is $f(y) = \lambda e^{-\lambda y}$.
• By comparing $e^{-y}$ to $\lambda e^{-\lambda y}$, we can see that $\lambda = 1$.
• The expected value of an $\text{Exponential}(\lambda)$ variable is $E[Y] = \frac{1}{\lambda}$.

Solution

1. Identify Distribution: From the solution to Question 11, $Y \sim \text{Exponential}(1)$.
2. Use $E[Y]$ Formula: The mean (expected value) of an exponential distribution with rate $\lambda$ is $1/\lambda$.
3. Calculate:
   • $\lambda = 1$
   • $E[Y] = \frac{1}{1} = 1$

(Alternatively, using LOTUS: $E[Y] = \int_0^1 (-\log(x)) \cdot 1 ,dx = -[x\log(x) - x]0^1 = -[(1\log(1)-1) - \lim{x\to 0}(x\log(x)-x)] = -[(0-1) - (0-0)] = 1$)

4. Final Answer: The expected value of $Y$ is 1.

❓ Question 13: Minimizing an Expected Value

Let $X$ be a continuous uniform random variable on $[0, 100]$. Define $f(x)$ as $f(x) = \begin{cases} 2(a-x) & \text{if } x \le a \ x-a & \text{if } x > a \end{cases}$ where $a$ is a real number between 0 and 100. What value should you choose for $a$ to minimize the expected value of $f(X)$? Enter the answer correct to two decimal places.

Core Concepts

1. LOTUS: $E[f(X)] = \int_{-\infty}^{\infty} f(x) f_X(x) ,dx$.
2. Uniform PDF: Since $X \sim \text{Uniform}[0, 100]$, its PDF is $f_X(x) = \frac{1}{100-0} = \frac{1}{100}$ for $x \in [0, 100]$.
3. Calculus Optimization: To minimize a function $H(a)$, find its derivative $H’(a)$, set it to zero, and solve for $a$.

Solution

1. Set up the Expectation $E[f(X)]$:

   • Let $H(a) = E[f(X)]$.
   • $H(a) = \int_{0}^{100} f(x) \cdot \left(\frac{1}{100}\right) ,dx$
   • We must split the integral at $x=a$ because the definition of $f(x)$ changes at $a$.
   • $H(a) = \frac{1}{100} \left[ \int_{0}^{a} 2(a-x) ,dx + \int_{a}^{100} (x-a) ,dx \right]$

2. Evaluate the Integrals:

   • First Integral: $\int_{0}^{a} (2a - 2x) ,dx = \left[ 2ax - x^2 \right]_{0}^{a} = (2a^2 - a^2) - (0 - 0) = a^2$
   • Second Integral: $\int_{a}^{100} (x - a) ,dx = \left[ \frac{x^2}{2} - ax \right]_{a}^{100}$ $= \left( \frac{100^2}{2} - 100a \right) - \left( \frac{a^2}{2} - a^2 \right)$ $= (5000 - 100a) - \left( -\frac{a^2}{2} \right) = 5000 - 100a + \frac{a^2}{2}$

3. Create the Function $H(a)$:

   • $H(a) = \frac{1}{100} \left[ (a^2) + \left( 5000 - 100a + \frac{a^2}{2} \right) \right]$
   • $H(a) = \frac{1}{100} \left[ \frac{3}{2}a^2 - 100a + 5000 \right]$

4. Minimize $H(a)$:

   • Find the derivative with respect to $a$: $H’(a) = \frac{1}{100} \frac{d}{da} \left[ \frac{3}{2}a^2 - 100a + 5000 \right]$ $H’(a) = \frac{1}{100} \left[ \left(\frac{3}{2} \cdot 2a\right) - 100 + 0 \right]$ $H’(a) = \frac{1}{100} [ 3a - 100 ]$
   • Set the derivative to zero: $\frac{1}{100} [ 3a - 100 ] = 0$ $3a - 100 = 0$ $a = \frac{100}{3}$
   • (The second derivative $H’’(a) = 3/100 > 0$, so this is a minimum).

5. Final Answer:

   • $a = \frac{100}{3} = 33.333…$
   • Correct to two decimal places, the value is 33.33.

❓ Question 14: Transformation of Variables

The probability density function of a continuous random variable $X$ is given by $f_X(x) = \begin{cases} ax + \frac{1}{10} & 1 \le x \le 2 \ 0 & \text{otherwise} \end{cases}$. Find the distribution of $Y = 3X + 2$.

Core Concepts

1. Valid PDF: A PDF must integrate to 1 over its domain. $\int_{-\infty}^{\infty} f_X(x) ,dx = 1$. This is used to find the constant $a$.
2. Transformation of Variables (Formula Method): For a monotonic $Y=g(X)$: $$f_Y(y) = f_X(g^{-1}(y)) \left| \frac{dx}{dy} \right|$$ where $x = g^{-1}(y)$ is the inverse function.

Solution

1. Find the constant $a$:

   • $\int_{1}^{2} (ax + \frac{1}{10}) ,dx = 1$
   • $\left[ \frac{ax^2}{2} + \frac{x}{10} \right]_{1}^{2} = 1$
   • $\left( \frac{a(2)^2}{2} + \frac{2}{10} \right) - \left( \frac{a(1)^2}{2} + \frac{1}{10} \right) = 1$
   • $\left( 2a + \frac{2}{10} \right) - \left( \frac{a}{2} + \frac{1}{10} \right) = 1$
   • $2a - \frac{a}{2} + \frac{2}{10} - \frac{1}{10} = 1$
   • $\frac{3a}{2} + \frac{1}{10} = 1$
   • $\frac{3a}{2} = \frac{9}{10}$
   • $a = \frac{9}{10} \cdot \frac{2}{3} = \frac{18}{30} = \frac{3}{5}$
   • So, the PDF is $f_X(x) = \frac{3}{5}x + \frac{1}{10}$ for $1 \le x \le 2$.

2. Apply the Transformation $Y = 3X + 2$:

   • Find Range of $Y$: The function is increasing.
     • When $x=1$, $y = 3(1) + 2 = 5$.
     • When $x=2$, $y = 3(2) + 2 = 8$.
     • The new range is $5 \le y \le 8$. (This eliminates option 3).
   • Find Inverse $x = g^{-1}(y)$: $y = 3x + 2 \implies y - 2 = 3x \implies x = \frac{y-2}{3}$
   • Find Derivative $|\frac{dx}{dy}|$: $\frac{dx}{dy} = \frac{1}{3}$. So $|\frac{dx}{dy}| = \frac{1}{3}$.
   • Apply Formula: $f_Y(y) = f_X(x) \left| \frac{dx}{dy} \right|$ $f_Y(y) = f_X\left(\frac{y-2}{3}\right) \cdot \frac{1}{3}$
     • Substitute $x = \frac{y-2}{3}$ into $f_X(x) = \frac{3}{5}x + \frac{1}{10}$: $f_Y(y) = \left[ \frac{3}{5}\left(\frac{y-2}{3}\right) + \frac{1}{10} \right] \cdot \frac{1}{3}$
     • The 3s cancel in the first term: $f_Y(y) = \left[ \frac{y-2}{5} + \frac{1}{10} \right] \cdot \frac{1}{3}$
     • Find a common denominator inside the bracket: $f_Y(y) = \left[ \frac{2(y-2)}{10} + \frac{1}{10} \right] \cdot \frac{1}{3}$ $f_Y(y) = \left[ \frac{2y - 4 + 1}{10} \right] \cdot \frac{1}{3}$ $f_Y(y) = \left[ \frac{2y - 3}{10} \right] \cdot \frac{1}{3} = \frac{2y - 3}{30}$

3. Final Answer: The PDF is $f_Y(y) = \begin{cases} \frac{2y-3}{30} & 5 \le y \le 8 \ 0 & \text{otherwise} \end{cases}$, which is Option 2.

❓ Question 15: Expected Value

Find the value of $E(Y)$. Enter the answer up to two decimal places.

(Note: This question almost certainly refers to the variable $Y$ from Question 14.)

Core Concept

Linearity of Expectation: This is the easiest way to solve this.

In our case, $E[Y] = E[3X + 2] = 3E[X] + 2$. We just need to find $E[X]$.

Solution

1. Find $E[X]$:

   • $E[X] = \int_{-\infty}^{\infty} x \cdot f_X(x) ,dx$
   • $E[X] = \int_{1}^{2} x \left( \frac{3}{5}x + \frac{1}{10} \right) ,dx$
   • $E[X] = \int_{1}^{2} \left( \frac{3}{5}x^2 + \frac{1}{10}x \right) ,dx$

2. Evaluate the Integral:

   • $E[X] = \left[ \frac{3}{5}\frac{x^3}{3} + \frac{1}{10}\frac{x^2}{2} \right]_{1}^{2}$
   • $E[X] = \left[ \frac{x^3}{5} + \frac{x^2}{20} \right]_{1}^{2}$
   • At $x=2$: $\frac{2^3}{5} + \frac{2^2}{20} = \frac{8}{5} + \frac{4}{20} = \frac{32}{20} + \frac{4}{20} = \frac{36}{20} = \frac{9}{5} = 1.8$
   • At $x=1$: $\frac{1^3}{5} + \frac{1^2}{20} = \frac{1}{5} + \frac{1}{20} = \frac{4}{20} + \frac{1}{20} = \frac{5}{20} = \frac{1}{4} = 0.25$
   • $E[X] = 1.8 - 0.25 = 1.55$

3. Find $E[Y]$:

   • $E[Y] = 3E[X] + 2$
   • $E[Y] = 3(1.55) + 2 = 4.65 + 2 = 6.65$

4. Final Answer: The value of $E(Y)$ is 6.65.

❓ Question 16: Bayes’ Theorem (Mixed Variables)

In a university, 30% of the students enrolled in a math course are from the science stream, while 70% are from the commerce stream. The exam scores for science stream students follows $N(55, 36)$, and the exam scores for commerce stream students follows $N(60, 25)$. If a randomly selected student scored 60 in the exam, what is the probability that the student is from the commerce stream? Enter the answer up to two decimal places.

Core Concept

This is another Bayes’ Theorem problem, just like Question 10. We want to find $P(\text{Commerce} | \text{Score}=60)$.

Solution

1. Define Events and Parameters:

   • $S$: Event the student is from Science. $P(S) = 0.30$.
   • $C$: Event the student is from Commerce. $P(C) = 0.70$.
   • $X$: Random variable for score.
   • Score given Science: $X|S \sim N(\mu=55, \sigma^2=36)$.
   • Score given Commerce: $X|C \sim N(\mu=60, \sigma^2=25)$.
   • Goal: Find $P(C | X=60)$.

2. Calculate Likelihoods (PDF values at X=60):

   • $f(X=60 | C)$: (PDF of $N(60, 25)$ at $x=60$)
     • We are at the mean. The exponent is $e^0 = 1$.
     • $f(X=60 | C) = \frac{1}{\sqrt{2\pi \cdot 25}} e^{-\frac{(60-60)^2}{2 \cdot 25}} = \frac{1}{\sqrt{50\pi}} = \frac{1}{5\sqrt{2\pi}}$
   • $f(X=60 | S)$: (PDF of $N(55, 36)$ at $x=60$)
     • $f(X=60 | S) = \frac{1}{\sqrt{2\pi \cdot 36}} e^{-\frac{(60-55)^2}{2 \cdot 36}} = \frac{1}{\sqrt{72\pi}} e^{-\frac{25}{72}} = \frac{1}{6\sqrt{2\pi}} e^{-25/72}$

3. Apply Bayes’ Theorem:

   $$P(C | X=60) = \frac{f(X=60 | C) P(C)}{f(X=60 | C)P(C) + f(X=60 | S)P(S)}$$$$P(C | X=60) = \frac{\left(\frac{1}{5\sqrt{2\pi}}\right) (0.70)}{\left(\frac{1}{5\sqrt{2\pi}}\right) (0.70) + \left(\frac{1}{6\sqrt{2\pi}} e^{-25/72}\right) (0.30)}$$

4. Simplify and Calculate:

   • Cancel $\frac{1}{\sqrt{2\pi}}$ from all terms: $$P(C | X=60) = \frac{\frac{0.70}{5}}{\frac{0.70}{5} + \frac{0.30}{6} e^{-25/72}}$$ $$P(C | X=60) = \frac{0.14}{0.14 + 0.05 e^{-25/72}}$$
   • Now, we need to calculate the value.
     • $e^{-25/72} \approx e^{-0.34722…} \approx 0.70668$
   • $P(C | X=60) \approx \frac{0.14}{0.14 + 0.05 \cdot (0.70668)}$
   • $P(C | X=60) \approx \frac{0.14}{0.14 + 0.035334}$
   • $P(C | X=60) \approx \frac{0.14}{0.175334} \approx 0.79848…$

5. Final Answer: Rounded to two decimal places, the probability is 0.80.

❓ Question 17: Widget Production Cost

A company produces three types of widgets: Type A, Type B and Type C. The production cost of a widget depends on two independent random variables: machine efficiency ($M$) and raw material ($Q$). The machine efficiency $M \sim \text{Uniform}{1, 2, 3}$, where $M=1$ represents low efficiency, $M=2$ represents moderate efficiency, and $M=3$ represents high efficiency. The raw material quality $Q \sim N(4, 1)$. The production cost $C$ (in dollars) is given by the formula: $C = M.Q + M$ If a random widget has a production cost of 12, what is the probability that the machine efficiency $M$ was 2?

Core Concepts

Bayes’ Theorem for Mixed Variables: This is the core concept. We’re finding the probability of a discrete event ($M=2$) given a continuous measurement ($C=12$). The formula is:

The denominator $f(C=c)$ is the Law of Total Probability:

Transformation of Normal Variables: If $X \sim N(\mu, \sigma^2)$, then a linear transformation $Y = aX + b$ results in a new normal variable $Y \sim N(a\mu + b, a^2\sigma^2)$.

Solution

1. Goal: We need to find $P(M=2 | C=12)$.

2. Priors $P(M=i)$:

   • $M \sim \text{Uniform}{1, 2, 3}$, so $P(M=1) = P(M=2) = P(M=3) = \frac{1}{3}$.
   • Because the priors are all equal, they will cancel out in the Bayes’ formula: $$P(M=2 | C=12) = \frac{f(C=12 | M=2)}{f(C=12 | M=1) + f(C=12 | M=2) + f(C=12 | M=3)}$$

3. Find Conditional Likelihoods $f(C=12 | M=i)$:

   • First, simplify the cost formula: $C = M.Q + M = M(Q+1)$.
   • Let’s find the distribution of $Q’ = Q+1$. Since $Q \sim N(\mu=4, \sigma^2=1)$, $Q’ = Q+1 \sim N(\mu=4+1, \sigma^2=1) \implies Q’ \sim N(5, 1)$.
   • Now we find the distribution of $C = M \cdot Q’$ for each $M$:
     • If $M=1$: $C = 1 \cdot Q’$. So, $C|M=1 \sim N(5, 1)$. The PDF at $C=12$ is: $f_1 = f(C=12|M=1) = \frac{1}{\sqrt{2\pi \cdot 1}} e^{-\frac{(12-5)^2}{2 \cdot 1}} = \frac{1}{\sqrt{2\pi}} e^{-49/2}$.
     • If $M=2$: $C = 2 \cdot Q’$. The new mean is $2 \cdot 5 = 10$ and the new variance is $2^2 \cdot 1 = 4$. So, $C|M=2 \sim N(10, 4)$. The PDF at $C=12$ is: $f_2 = f(C=12|M=2) = \frac{1}{\sqrt{2\pi \cdot 4}} e^{-\frac{(12-10)^2}{2 \cdot 4}} = \frac{1}{\sqrt{8\pi}} e^{-\frac{4}{8}} = \frac{1}{\sqrt{8\pi}} e^{-1/2}$.
     • If $M=3$: $C = 3 \cdot Q’$. The new mean is $3 \cdot 5 = 15$ and the new variance is $3^2 \cdot 1 = 9$. So, $C|M=3 \sim N(15, 9)$. The PDF at $C=12$ is: $f_3 = f(C=12|M=3) = \frac{1}{\sqrt{2\pi \cdot 9}} e^{-\frac{(12-15)^2}{2 \cdot 9}} = \frac{1}{\sqrt{18\pi}} e^{-\frac{9}{18}} = \frac{1}{\sqrt{18\pi}} e^{-1/2}$.

4. Substitute into Bayes’ Formula:

   $$P(M=2 | C=12) = \frac{f_2}{f_1 + f_2 + f_3} = \frac{\frac{1}{\sqrt{8\pi}} e^{-1/2}}{\frac{1}{\sqrt{2\pi}} e^{-49/2} + \frac{1}{\sqrt{8\pi}} e^{-1/2} + \frac{1}{\sqrt{18\pi}} e^{-1/2}}$$

5. Simplify:

   • Let’s simplify the square roots: $\sqrt{8\pi} = 2\sqrt{2\pi}$ $\sqrt{18\pi} = 3\sqrt{2\pi}$
   • Substitute these in: $$P(M=2 | C=12) = \frac{\frac{1}{2\sqrt{2\pi}} e^{-1/2}}{\frac{1}{\sqrt{2\pi}} e^{-49/2} + \frac{1}{2\sqrt{2\pi}} e^{-1/2} + \frac{1}{3\sqrt{2\pi}} e^{-1/2}}$$
   • Cancel $\frac{1}{\sqrt{2\pi}}$ from all terms: $$P(M=2 | C=12) = \frac{\frac{1}{2} e^{-1/2}}{e^{-49/2} + \frac{1}{2} e^{-1/2} + \frac{1}{3} e^{-1/2}}$$
   • Multiply the numerator and denominator by 6 to clear the fractions: $$P(M=2 | C=12) = \frac{3 e^{-1/2}}{6 e^{-49/2} + 3 e^{-1/2} + 2 e^{-1/2}}$$

6. Match to Options:

   • Our derived answer is: $\frac{3 e^{-1/2}}{6 e^{-49/2} + 3 e^{-1/2} + 2 e^{-1/2}}$
   • Option (a) is: $\frac{3 \exp(-1/8)}{3 \exp(-1/8) + 6 \exp(-49/2) + 2 \exp(-1/9)}$
   • The coefficients (3 in the numerator; 6, 3, and 2 in the denominator) match exactly if we reorder the denominator.
   • There are typos in the exponents in the provided options. $e^{-1/8}$ and $e^{-1/9}$ should both be $e^{-1/2}$.
   • Based on the identical coefficient structure, Option (a) is the correct choice.

Final Answer

The correct option is (a).

❓ Question 18: Chocolate and Vanilla Cakes

A bakery produces two types of cakes: Chocolate and Vanilla. Suppose the type of cake is chosen at random with equal probability. The weight $W$ of cake depends on its type:

• For chocolates cakes, the weight $W \sim \text{Normal}(4, 100)$.
• For vanilla cakes, the weight $W \sim \text{Uniform}[5, 15]$. What is the conditional probability that a cake is a chocolate cake if its weight is $W = 4$?

Core Concepts

Bayes’ Theorem (Mixed Variables):

Probability Density Function (PDF): A PDF $f(x)$ gives the relative likelihood of a continuous variable taking a specific value. If a value is impossible, its PDF is 0.

Solution

1. Priors:

   • $P(\text{Chocolate}) = 0.5$
   • $P(\text{Vanilla}) = 0.5$

2. Likelihoods $f(W=4 | \text{Type})$:

   • Chocolate: $W \sim N(\mu=4, \sigma^2=100)$. The PDF at $W=4$ is: $f(W=4 | \text{Chocolate}) = \frac{1}{\sqrt{2\pi \cdot 100}} e^{-\frac{(4-4)^2}{2 \cdot 100}} = \frac{1}{\sqrt{200\pi}} > 0$. (The exact value doesn’t matter, just that it’s positive).
   • Vanilla: $W \sim \text{Uniform}[5, 15]$. The PDF is $f(w) = \frac{1}{15-5} = \frac{1}{10}$ for $5 \le w \le 15$, and $f(w)=0$ otherwise.
     • We need the density at $W=4$. Since 4 is outside the support interval $[5, 15]$, the density is 0.
     • $f(W=4 | \text{Vanilla}) = 0$.

3. Substitute into Bayes’ Theorem:

   $$P(\text{Chocolate} | W=4) = \frac{f(W=4 | \text{Chocolate}) \cdot (0.5)}{f(W=4 | \text{Chocolate})\cdot(0.5) + (0)\cdot(0.5)}$$$$P(\text{Chocolate} | W=4) = \frac{f(W=4 | \text{Chocolate}) \cdot 0.5}{f(W=4 | \text{Chocolate}) \cdot 0.5 + 0}$$$$P(\text{Chocolate} | W=4) = \frac{f(W=4 | \text{Chocolate}) \cdot 0.5}{f(W=4 | \text{Chocolate}) \cdot 0.5} = 1$$

4. Logical Check: This makes perfect sense. If a vanilla cake must weigh between 5 and 15, and we observe a cake that weighs 4, it is impossible for it to be vanilla. Therefore, it must be a chocolate cake.

Final Answer

The probability is 1.

❓ Question 19: User Engagement Patterns

An online education platform conducted a study to analyze the user engagement patterns. It observed that 70% of its users are regular learners, while the remaining 30% are occasional learners. The duration of the time spent per session (in minutes) follows a Uniform [40, 90] distribution for regular learners and a Uniform [20, 60] distribution for occasional learners. If a randomly selected user is observed to have spent 50 minutes in a session, what is the probability that this user is a regular learner? Enter the answer correct to three decimal places.

Core Concepts

Bayes’ Theorem (Mixed Variables): This is the same setup as the previous two problems.

Uniform PDF: For $X \sim \text{Uniform}[a, b]$, the PDF is $f(x) = \frac{1}{b-a}$ for $x$ in the interval, and $0$ otherwise.

Solution

1. Priors:

   • $P(\text{Regular}) = 0.70$
   • $P(\text{Occasional}) = 0.30$

2. Likelihoods $f(T=50 | \text{Type})$:

   • Regular: $T \sim \text{Uniform}[40, 90]$. The value $T=50$ is inside this interval.
     • $f(T=50 | \text{Regular}) = \frac{1}{90 - 40} = \frac{1}{50}$.
   • Occasional: $T \sim \text{Uniform}[20, 60]$. The value $T=50$ is inside this interval.
     • $f(T=50 | \text{Occasional}) = \frac{1}{60 - 20} = \frac{1}{40}$.

3. Substitute into Bayes’ Theorem:

   $$P(\text{Regular} | T=50) = \frac{(\frac{1}{50}) \cdot (0.70)}{(\frac{1}{50}) \cdot (0.70) + (\frac{1}{40}) \cdot (0.30)}$$

4. Calculate:

   • Numerator: $\frac{0.70}{50} = \frac{7}{500} = 0.014$
   • Denominator: $\frac{0.70}{50} + \frac{0.30}{40} = \frac{7}{500} + \frac{3}{400}$
   • Find a common denominator (2000): $\frac{7 \cdot 4}{2000} + \frac{3 \cdot 5}{2000} = \frac{28}{2000} + \frac{15}{2000} = \frac{43}{2000}$
   • Divide the numerator by the denominator: $P(\text{Regular} | T=50) = \frac{7/500}{43/2000} = \frac{7}{500} \cdot \frac{2000}{43} = \frac{7 \cdot 4}{43} = \frac{28}{43}$

5. Convert to Decimal:

   • $28 \div 43 \approx 0.65116…$

Final Answer

Correct to three decimal places, the probability is 0.651.

❓ Question 20: Delivery Cost

A logistics company monitors the delivery time $X$ (in hours) of same-day intra-city shipments. The delivery time follows the probability density function: $f_X(x) = \begin{cases} \frac{2}{9}(3-x), & 0 \le x \le 3 \ 0, & \text{otherwise} \end{cases}$ The delivery cost (in ₹) for a shipment delivered in time $x$ is modeled as: $C(x) = 100x^2 + 100(1-x)$ Find the probability that the delivery cost exceeds ₹100. Enter the answer correct to two decimal places.

Core Concepts

Transformation of Variables (Finding Probability): We don’t need to find the PDF of $C$. We just need to find the probability $P(C > 100)$.

1. Find the values of $X$ that make $C(x) > 100$.
2. Integrate the PDF of $X$, $f_X(x)$, over that range of $X$.

Solution

1. Find the region of $X$ for $C > 100$:

   • First, simplify $C(x)$: $C(x) = 100x^2 + 100 - 100x = 100(x^2 - x + 1)$.
   • Set up the inequality: $C(x) > 100$ $100(x^2 - x + 1) > 100$
   • Divide by 100: $x^2 - x + 1 > 1$
   • Subtract 1: $x^2 - x > 0$
   • Factor: $x(x-1) > 0$
   • This inequality is true when $x < 0$ or $x > 1$.

2. Constrain with the domain of $X$:

   • The PDF $f_X(x)$ is defined only on the interval $0 \le x \le 3$.
   • We need to find where the two regions overlap:
     • (Region 1: $x < 0$) and (Domain: $0 \le x \le 3$) $\to$ No overlap.
     • (Region 2: $x > 1$) and (Domain: $0 \le x \le 3$) $\to$ The overlap is $1 < x \le 3$.
   • Therefore, $P(C > 100)$ is the same as $P(X > 1)$.

3. Integrate $f_X(x)$ from 1 to 3:

   $$P(X > 1) = \int_1^3 f_X(x) \,dx = \int_1^3 \frac{2}{9}(3-x) \,dx$$$$P(X > 1) = \frac{2}{9} \int_1^3 (3-x) \,dx$$$$P(X > 1) = \frac{2}{9} \left[ 3x - \frac{x^2}{2} \right]_1^3$$

4. Evaluate the integral:

   • Plug in upper bound ($x=3$): $3(3) - \frac{3^2}{2} = 9 - \frac{9}{2} = 4.5$
   • Plug in lower bound ($x=1$): $3(1) - \frac{1^2}{2} = 3 - \frac{1}{2} = 2.5$
   • Subtract: $P(X > 1) = \frac{2}{9} [4.5 - 2.5] = \frac{2}{9} [2] = \frac{4}{9}$

5. Convert to Decimal:

   • $4 \div 9 = 0.4444…$

Final Answer

Correct to two decimal places, the probability is 0.44.