Graded Assignment 8

❓ Question 1

Problem: Let the moment generating function (MGF) of a random variable $X$ be given by:

Find the distribution of $X$.

Concepts: The Moment Generating Function (MGF) for a discrete random variable $X$ is a unique signature of its probability distribution. It’s defined as:

If $X$ can take values $x_1, x_2, \ldots, x_k$ with probabilities $p_1, p_2, \ldots, p_k$ (where $p_i = P(X=x_i)$), the MGF is:

To find the distribution, we just need to match the terms in the given MGF to this standard form. Each term $(p_i)e^{\lambda x_i}$ tells us that $P(X = x_i) = p_i$.

Solution:

1. Write down the given MGF: $$M_X(\lambda) = \left(\frac{1}{4}\right)e^{-2\lambda} + \left(\frac{1}{40}\right) + \left(\frac{3}{10}\right)e^{-\lambda} + \left(\frac{3}{40}\right)e^{2\lambda} + \left(\frac{7}{20}\right)e^{\lambda}$$
2. Rewrite the MGF by ordering the terms by the exponent (the value of $x_i$) from smallest to largest. Remember that a constant term like $\frac{1}{40}$ is just $\frac{1}{40}e^{0\lambda}$.
   • $e^{-2\lambda}$ term: $\left(\frac{1}{4}\right)e^{\lambda(-2)}$
   • $e^{-\lambda}$ term: $\left(\frac{3}{10}\right)e^{\lambda(-1)}$
   • Constant term: $\left(\frac{1}{40}\right)e^{\lambda(0)}$
   • $e^{\lambda}$ term: $\left(\frac{7}{20}\right)e^{\lambda(1)}$
   • $e^{2\lambda}$ term: $\left(\frac{3}{40}\right)e^{\lambda(2)}$
3. Match coefficients ($p_i$) to values ($x_i$):
   • From $\left(\frac{1}{4}\right)e^{\lambda(-2)}$, we see $P(X = -2) = \frac{1}{4}$.
   • From $\left(\frac{3}{10}\right)e^{\lambda(-1)}$, we see $P(X = -1) = \frac{3}{10}$.
   • From $\left(\frac{1}{40}\right)e^{\lambda(0)}$, we see $P(X = 0) = \frac{1}{40}$.
   • From $\left(\frac{7}{20}\right)e^{\lambda(1)}$, we see $P(X = 1) = \frac{7}{20}$.
   • From $\left(\frac{3}{40}\right)e^{\lambda(2)}$, we see $P(X = 2) = \frac{3}{40}$.
4. Construct the probability distribution table (PMF):

   $X$	$-2$	$-1$	$0$	$1$	$2$
   $P(X=x)$	$\frac{1}{4}$	$\frac{3}{10}$	$\frac{1}{40}$	$\frac{7}{20}$	$\frac{3}{40}$

5. Compare this table to the options provided in the image. This table matches the third option.

Answer: The correct distribution is given by the third option:

❓ Question 2

Problem: A fair coin is tossed 1700 times. Use CLT to compute the probability that head appears at most 560 times. Enter the answer correct to 2 decimal places.

Concepts:

1. Binomial Distribution: The number of heads (successes) $S_n$ in $n$ independent coin tosses (trials) follows a Binomial distribution, $S_n \sim \text{Bin}(n, p)$. Here $n = 1700$ and $p = 0.5$ (since the coin is fair).
2. Central Limit Theorem (CLT): For a large $n$, the Binomial distribution $\text{Bin}(n, p)$ can be approximated by a Normal distribution $N(\mu, \sigma^2)$ with:
   • Mean: $\mu = np$
   • Variance: $\sigma^2 = np(1-p)$
3. Continuity Correction: When approximating a discrete distribution (Binomial, which uses integers) with a continuous one (Normal, which uses all real numbers), we must use a continuity correction. To find the probability of “at most 560” (which is $P(S_n \le 560)$), we must include the entire integer “bar” for 560. We do this by extending the range to 560.5. The rule is: $P(S_n \le k) \approx P(Y \le k+0.5)$.

Solution:

1. Identify parameters:
   • Number of trials, $n = 1700$
   • Probability of success (heads), $p = 0.5$
2. Find the mean and variance of the approximating Normal distribution:
   • Mean: $\mu = np = 1700 \times 0.5 = 850$.
   • Variance: $\sigma^2 = np(1-p) = 1700 \times 0.5 \times 0.5 = 425$.
   • Standard Deviation: $\sigma = \sqrt{425} \approx 20.616$.
3. State the problem: We want to find $P(S_n \le 560)$.
4. Apply continuity correction:
   • We approximate $P(S_n \le 560)$ with $P(Y \le 560.5)$, where $Y \sim N(850, 425)$.
5. Standardize the value 560.5 by calculating its Z-score: $$Z = \frac{Y - \mu}{\sigma}$$ $$z = \frac{560.5 - 850}{\sqrt{425}} = \frac{-289.5}{20.616} \approx -14.04$$
6. Find the probability:
   • We need to find $P(Z \le -14.04)$.
   • A Z-score of -14.04 is extremely far to the left of the mean (0). Standard Z-tables typically only go up to -3.5 or -4.0, at which point the probability is already $0.0002$ or $0.00003$.
   • A Z-score of -14.04 means the value is 14 standard deviations below the mean. The probability of this is virtually zero.

Answer: The probability is $P(Z \le -14.04) \approx 0$. Correct to 2 decimal places, the answer is 0.00.

❓ Question 3

Problem: Let $X_1, \ldots, X_{500} \sim i.i.d \text{ Normal}(0, 1)$. Evaluate $P(X_1^2 + \ldots + X_{500}^2 > 520)$ using Central Limit theorem. Enter the answer correct to 2 decimal places. Hint: $(X_1^2 + \ldots + X_{500}^2) \sim \text{Gamma}(250, 0.5)$.

Concepts:

1. Central Limit Theorem (CLT): The CLT states that the sum of a large number $n$ of independent and identically distributed (i.i.d.) random variables will be approximately normally distributed, regardless of the original distribution.
2. Chi-Squared Distribution: The square of a standard normal variable, $X_i^2$, follows a Chi-squared distribution with 1 degree of freedom, $\chi^2(1)$.
3. Sum of variables: We are asked to find the probability of a sum, $S = Y_1 + \ldots + Y_{500}$, where each $Y_i = X_i^2$.
4. Mean and Variance of $\chi^2(1)$: A $\chi^2(k)$ distribution has mean $k$ and variance $2k$.
   • Therefore, our $Y_i \sim \chi^2(1)$ has:
   • $E[Y_i] = 1$
   • $Var(Y_i) = 2$

Solution:

1. Define the sum: We are interested in the sum $S = \sum_{i=1}^{500} Y_i$, where $Y_i = X_i^2$. We have 500 i.i.d. variables $Y_i$.
2. Apply CLT: According to the CLT, the sum $S$ will be approximately normally distributed: $S \approx N(\mu_S, \sigma_S^2)$.
3. Find the mean and variance of the sum $S$:
   • Mean: $\mu_S = n \times E[Y_i] = 500 \times 1 = 500$.
   • Variance: $\sigma_S^2 = n \times Var(Y_i) = 500 \times 2 = 1000$.
   • Standard Deviation: $\sigma_S = \sqrt{1000} \approx 31.62$.
   • (This matches the hint: $\text{Gamma}(\alpha=250, \beta=0.5 \text{ rate})$ has mean $\alpha/\beta = 250/0.5 = 500$ and variance $\alpha/\beta^2 = 250/(0.5^2) = 1000$.)
4. State the problem: We need to find $P(S > 520)$. We will use our normal approximation $N(500, 1000)$.
5. Standardize the value 520: $$Z = \frac{S - \mu_S}{\sigma_S}$$ $$z = \frac{520 - 500}{\sqrt{1000}} = \frac{20}{31.62} \approx 0.632$$
6. Find the probability:
   • We need to find $P(Z > 0.632)$.
   • Using a standard Z-table, $P(Z \le 0.63) = 0.7357$.
   • $P(Z > 0.63) = 1 - P(Z \le 0.63) = 1 - 0.7357 = 0.2643$.
7. Format the answer: The question asks for the answer correct to 2 decimal places. $0.2643 \approx 0.26$.

Answer: The probability is 0.26.

❓ Question 4

Problem: Let $X$ be a random variable having the gamma distribution with $\alpha = 2n$ and $\beta = 1$. Use the Weak Law of Large Numbers (WLLN) to find the value of $n$ such that $P\left(\left|\frac{X}{2n} - 1\right| > 0.01\right) < 0.01$.

Hint: If $Y \sim \text{Gamma}(\alpha, \beta)$, $E[Y] = \alpha/\beta$ and $Var[Y] = \alpha/\beta^2$. Sum of $n$ independent $\text{Gamma}(\alpha, \beta)$ is $\text{Gamma}(n\alpha, \beta)$.

Concepts:

1. Weak Law of Large Numbers (WLLN): This law relates to the convergence of a sample mean $\bar{X}_n$ to the population mean $\mu$. The hint about the sum of Gamma variables suggests we should re-frame $X$ as a sum.
2. Chebyshev’s Inequality: This is the tool used to find a bound for the WLLN. For any random variable $Y$ with mean $\mu_Y$ and variance $\sigma_Y^2$, and any $\epsilon > 0$: $$P(|Y - \mu_Y| > \epsilon) \le \frac{\sigma_Y^2}{\epsilon^2}$$ We are asked to find $n$ such that this probability is less than 0.01. We can do this by finding the $n$ that makes the bound equal to 0.01.

Solution:

1. Re-frame $X$ as a sum:
   • We are given $X \sim \text{Gamma}(\alpha=2n, \beta=1)$.
   • Using the hint, we can express $X$ as the sum of $n$ i.i.d. variables, $X = \sum_{i=1}^{n} X_i$.
   • If $X \sim \text{Gamma}(n\alpha, \beta)$, we must have $n\alpha = 2n$ and $\beta=1$. This implies $\alpha=2$.
   • So, we can model $X$ as the sum of $n$ variables $X_i$, where each $X_i \sim \text{Gamma}(\alpha=2, \beta=1)$.
2. Find Mean and Variance of $X_i$:
   • $E[X_i] = \alpha/\beta = 2/1 = 2$.
   • $Var(X_i) = \alpha/\beta^2 = 2/1^2 = 2$.
3. Rewrite the probability expression:
   • We are interested in the term $\frac{X}{2n}$.
   • $\frac{X}{2n} = \frac{\sum X_i}{2n} = \frac{1}{2} \left( \frac{\sum X_i}{n} \right) = \frac{1}{2}\bar{X}_n$, where $\bar{X}_n$ is the sample mean.
   • The probability becomes: $P\left(\left|\frac{1}{2}\bar{X}_n - 1\right| > 0.01\right)$.
   • The mean of $\bar{X}_n$ is $E[\bar{X}_n] = E[X_i] = 2$. Let’s factor out $\frac{1}{2}$ to get the standard WLLN form:
   • $P\left(\frac{1}{2}\left|\bar{X}_n - 2\right| > 0.01\right) = P\left(\left|\bar{X}_n - 2\right| > 0.02\right)$.
4. Apply Chebyshev’s Inequality:
   • We now have the form $P(|\bar{X}_n - \mu| > \epsilon) < \delta$, where:
     • $\mu = E[X_i] = 2$
     • $\epsilon = 0.02$
     • $\delta = 0.01$
   • Chebyshev’s Inequality states: $P(|\bar{X}_n - \mu| > \epsilon) \le \frac{Var(\bar{X}_n)}{\epsilon^2}$.
   • We need to find $Var(\bar{X}_n)$: $Var(\bar{X}_n) = \frac{Var(X_i)}{n} = \frac{2}{n}$.
   • So, $P(|\bar{X}_n - 2| > 0.02) \le \frac{2/n}{(0.02)^2}$.
5. Solve for $n$:
   • We want to find $n$ such that the probability is $< 0.01$. We set the upper bound to be equal to $0.01$ to find the minimum $n$. (Any $n$ larger than this will also satisfy the inequality).
   • $\frac{2/n}{(0.02)^2} = 0.01$
   • $\frac{2/n}{0.0004} = 0.01$
   • $\frac{2}{0.0004 n} = 0.01$
   • $\frac{5000}{n} = 0.01$
   • $5000 = 0.01 n$
   • $n = \frac{5000}{0.01} = 500,000$.
6. Interpret the result:
   • This $n=500,000$ is the boundary value. To make the probability strictly less than 0.01, we need $n$ to be strictly greater than 500,000.
   • Looking at the options: {505000, 470000, 498000, 482000}.
   • The only option that satisfies $n > 500,000$ is 505000.

Answer: The correct value for $n$ from the options is 505000.

❓ Question 5

Problem: Suppose speeds of vehicles on a particular road are normally distributed with mean 36 mph and standard deviation 2 mph. Find the probability that the mean speed $\bar{X}$ of 25 randomly selected vehicles is between 32 and 49 mph.

Concepts: Sampling Distribution of the Sample Mean: If a population is normally distributed as $X \sim N(\mu, \sigma^2)$, then the distribution of the sample mean $\bar{X}$ for a sample of size $n$ is also normally distributed:

The mean of the sample mean ($\mu_{\bar{X}}$) is the same as the population mean ($\mu$). The standard deviation of the sample mean ($\sigma_{\bar{X}}$), also called the standard error, is $\sigma_{\bar{X}} = \frac{\sigma}{\sqrt{n}}$.

Solution:

1. Identify population and sample parameters:
   • Population Mean $\mu = 36$
   • Population Standard Deviation $\sigma = 2$
   • Sample Size $n = 25$
2. Determine the distribution of the sample mean $\bar{X}$:
   • Mean of $\bar{X}$: $\mu_{\bar{X}} = \mu = 36$.
   • Standard Error of $\bar{X}$: $\sigma_{\bar{X}} = \frac{\sigma}{\sqrt{n}} = \frac{2}{\sqrt{25}} = \frac{2}{5}$.
   • So, $\bar{X} \sim N(36, (2/5)^2)$.
3. Set up the probability statement:
   • We want to find $P(32 < \bar{X} < 49)$.
   • This can be written as $P(\bar{X} \le 49) - P(\bar{X} \le 32)$.
4. Standardize the values:
   • To find $P(\bar{X} \le x)$, we calculate the Z-score: $Z = \frac{x - \mu_{\bar{X}}}{\sigma_{\bar{X}}} = \frac{x - 36}{2 / \sqrt{25}}$.
   • The probability is $P(Z \le z)$, which is denoted by the CDF $F_Z(z)$.
5. Apply to our two boundaries:
   • For $x = 49$: $P(\bar{X} \le 49) = F_Z\left( \frac{49 - 36}{2 / \sqrt{25}} \right)$
   • For $x = 32$: $P(\bar{X} \le 32) = F_Z\left( \frac{32 - 36}{2 / \sqrt{25}} \right)$
6. Combine the terms: $$P(32 < \bar{X} < 49) = F_Z\left( \frac{49 - 36}{2 / \sqrt{25}} \right) - F_Z\left( \frac{32 - 36}{2 / \sqrt{25}} \right)$$
7. Compare this expression to the given options. It matches the last option.

Answer: The correct expression is the last one:

❓ Question 6

Problem: Let the MGF of $X$ be given as $M_X(\lambda) = e^{5\lambda + 7\lambda^2}$. Define a new random variable $Z = 3X + c$. Which of the following are true for the MGF of $Z$, $M_Z(\lambda)$?

• $M_Z(\lambda)$ depends on the constant $c$.
• $M_Z(\lambda)$ does not depend on the constant $c$.
• $M_Z(\lambda) = e^{c\lambda + 15\lambda + 21\lambda^2}$
• $M_Z(\lambda) = e^{15\lambda + 21\lambda^2}$
• $M_Z(\lambda) = e^{c\lambda + 15\lambda + 63\lambda^2}$

Concepts: Properties of MGFs (Linear Transformation): A key property of MGFs relates to linear transformations. If $Y = aX + b$, its MGF $M_Y(\lambda)$ can be found from $M_X(\lambda)$ as follows:

Since $e^{b\lambda}$ is a constant with respect to $X$, we can pull it out:

By the definition of MGF, $E[e^{(a\lambda)X}] = M_X(a\lambda)$. Therefore, the property is: $M_{aX+b}(\lambda) = e^{b\lambda} M_X(a\lambda)$

Solution:

1. Identify the transformation: We have $Z = 3X + c$.
   • This matches $aX+b$ with $a = 3$ and $b = c$.
2. Apply the MGF property:
   • $M_Z(\lambda) = M_{3X+c}(\lambda) = e^{c\lambda} M_X(3\lambda)$.
3. Find the term $M_X(3\lambda)$:
   • We are given $M_X(\lambda) = e^{5\lambda + 7\lambda^2}$.
   • To find $M_X(3\lambda)$, we substitute $(3\lambda)$ everywhere we see $\lambda$:
   • $M_X(3\lambda) = e^{5(3\lambda) + 7(3\lambda)^2}$
   • $M_X(3\lambda) = e^{15\lambda + 7(9\lambda^2)}$
   • $M_X(3\lambda) = e^{15\lambda + 63\lambda^2}$
4. Combine the parts to find $M_Z(\lambda)$:
   • $M_Z(\lambda) = e^{c\lambda} \times M_X(3\lambda)$
   • $M_Z(\lambda) = e^{c\lambda} \times e^{15\lambda + 63\lambda^2}$
   • When multiplying exponentials, we add the exponents:
   • $M_Z(\lambda) = e^{c\lambda + 15\lambda + 63\lambda^2}$
5. Evaluate the options:
   • $M_Z(\lambda)$ depends on the constant $c$: True. The term $c\lambda$ is clearly in the final expression.
   • $M_Z(\lambda)$ does not depend on the constant $c$: False.
   • $M_Z(\lambda) = e^{c\lambda + 15\lambda + 21\lambda^2}$: False. The coefficient of $\lambda^2$ is 63, not 21. (This error comes from $7 \times 3$ instead of $7 \times 3^2$).
   • $M_Z(\lambda) = e^{15\lambda + 21\lambda^2}$: False.
   • $M_Z(\lambda) = e^{c\lambda + 15\lambda + 63\lambda^2}$: True. This exactly matches our result.

Answer: There are two true statements:

• $M_Z(\lambda)$ depends on the constant $c$.
• $M_Z(\lambda) = e^{c\lambda + 15\lambda + 63\lambda^2}$

❓ Question 7

Problem: Let $X$ be a random variable with PMF as:

Concepts: Linearity of Expectation: This is a fundamental property of expectation. It states that the expectation of a sum of random variables is equal to the sum of their individual expectations.

This property holds whether or not the variables are independent.

Expected Value of a Discrete Variable: The expected value (or mean) of a discrete random variable $X$ is calculated by summing the product of each possible value $x$ and its probability $f_X(x)$.

Solution:

1. Goal: We need to find $E[Y] = E[X_1 + X_2]$.
2. Apply Linearity of Expectation:
   • $E[Y] = E[X_1] + E[X_2]$.
3. Use i.i.d. property:
   • Since $X_1$ and $X_2$ are independent and identically distributed (i.i.d.) from $X$, they both have the same expected value as $X$.
   • $E[X_1] = E[X]$ and $E[X_2] = E[X]$.
   • Therefore, $E[Y] = E[X] + E[X] = 2E[X]$.
4. Calculate $E[X]$ from the given PMF table:
   • $E[X] = (0 \times \frac{1}{5}) + (2 \times \frac{3}{5}) + (3 \times \frac{1}{5})$
   • $E[X] = 0 + \frac{6}{5} + \frac{3}{5}$
   • $E[X] = \frac{9}{5}$
   • $E[X] = 1.8$.
5. Calculate $E[Y]$:
   • $E[Y] = 2 \times E[X] = 2 \times 1.8 = 3.6$.
6. Format the answer: The answer 3.6 is already correct to one decimal place.

Answer: The value of $E[Y]$ is 3.6.

❓ Question 8

Problem: The number of online orders received by a bookstore follows a Poisson distribution with a mean rate of 20 orders per hour. The store operates for 10 hours daily. Using the Central Limit Theorem, estimate the probability that the total number of orders received in a 10-hour period is between 180 and 220. Enter the answer correct to two decimal places. {Hint: Use $F_Z(0.1) = 0.5398, F_Z(1.41) = 0.92$}

Concepts:

1. Sum of Poisson Variables: If $X_i \sim \text{Poisson}(\lambda_i)$ are independent Poisson variables, their sum $S_n = \sum X_i$ is also a Poisson variable with a parameter equal to the sum of the individual parameters, $S_n \sim \text{Poisson}(\sum \lambda_i)$.
2. Central Limit Theorem (CLT) for Poisson: For a large parameter $\lambda$, a Poisson distribution, $\text{Poisson}(\lambda)$, can be approximated by a Normal distribution $Y \sim N(\mu, \sigma^2)$ where both the mean and the variance are equal to $\lambda$.
   • Mean: $\mu = \lambda$
   • Variance: $\sigma^2 = \lambda$

Solution:

1. Find the distribution for the 10-hour period:

   • The rate per hour is $\lambda_{\text{hour}} = 20$.
   • The total period is $t = 10$ hours.
   • Let $T$ be the total number of orders in 10 hours. $T$ follows a Poisson distribution with a total mean $\lambda_T$.
   • $\lambda_T = \lambda_{\text{hour}} \times t = 20 \text{ orders/hour} \times 10 \text{ hours} = 200$.
   • So, $T \sim \text{Poisson}(200)$.

2. Apply the Central Limit Theorem:

   • Since $\lambda_T = 200$ is large, we can approximate this Poisson distribution with a Normal distribution $Y \sim N(\mu, \sigma^2)$.
   • Mean: $\mu = \lambda_T = 200$.
   • Variance: $\sigma^2 = \lambda_T = 200$.
   • Standard Deviation: $\sigma = \sqrt{200} = \sqrt{100 \times 2} = 10\sqrt{2} \approx 14.142$.

3. Calculate the probability:

   • We need to estimate $P(180 \le T \le 220)$.
   • We approximate this using our Normal distribution: $P(180 \le Y \le 220)$.
   • To find this, we standardize the endpoints 180 and 220 to get Z-scores.
   • $Z = \frac{Y - \mu}{\sigma}$

4. Standardize the endpoints:

   • $z_{\text{lower}} = \frac{180 - 200}{\sqrt{200}} = \frac{-20}{\sqrt{200}} \approx -1.414$
   • $z_{\text{upper}} = \frac{220 - 200}{\sqrt{200}} = \frac{20}{\sqrt{200}} \approx 1.414$

5. Find the probability using Z-scores:

   • We need to find $P(-1.414 \le Z \le 1.414)$.
   • This is $F_Z(1.414) - F_Z(-1.414)$.
   • Using the hint, we’ll use $z = 1.41$.
   • $P \approx F_Z(1.41) - F_Z(-1.41)$
   • By the symmetry of the standard normal distribution, $F_Z(-1.41) = 1 - F_Z(1.41)$.
   • $P \approx F_Z(1.41) - (1 - F_Z(1.41))$
   • $P \approx 2 \times F_Z(1.41) - 1$

6. Use the hint value:

   • The hint provides $F_Z(1.41) = 0.92$.
   • $P \approx 2 \times 0.92 - 1 = 1.84 - 1 = 0.84$.
   • (The hint $F_Z(0.1) = 0.5398$ appears to be extra information not needed for this calculation).

Answer: The probability is 0.84.

❓ Question 9

Problem: Let $X_1, X_2 \sim i.i.d. X$ and $X$ be a discrete random variable with the following probability mass function (PMF):

Define a new random variable $Z = X_1 + X_2$. Calculate the value of $E[Z]$. Enter the answer correct to two decimal places.

Concepts:

1. Expected Value of a Discrete Variable: The expected value (or mean) of a discrete random variable $X$ is the sum of each possible value $x$ multiplied by its probability $P(X=x)$. $$E[X] = \sum x \cdot P(X=x)$$
2. Linearity of Expectation: This powerful property states that the expected value of a sum of random variables is the sum of their individual expected values. $$E[Z] = E[X_1 + X_2] = E[X_1] + E[X_2]$$
3. i.i.d. Variables: Since $X_1$ and $X_2$ are independent and identically distributed (i.i.d.) from $X$, they have the same expected value as $X$. $E[X_1] = E[X]$ and $E[X_2] = E[X]$.

Solution:

1. Goal: We need to find $E[Z]$, where $Z = X_1 + X_2$.
2. Apply Linearity of Expectation: $$E[Z] = E[X_1] + E[X_2]$$
3. Use the i.i.d. property:
   • Since $X_1$ and $X_2$ have the same distribution as $X$, $E[X_1] = E[X]$ and $E[X_2] = E[X]$.
   • Therefore, $E[Z] = E[X] + E[X] = 2E[X]$.
4. Calculate $E[X]$ using the given PMF:
   • $E[X] = (1 \times P(X=1)) + (2 \times P(X=2)) + (3 \times P(X=3))$
   • $E[X] = \left(1 \times \frac{1}{3}\right) + \left(2 \times \frac{1}{2}\right) + \left(3 \times \frac{1}{6}\right)$
   • $E[X] = \frac{1}{3} + 1 + \frac{3}{6}$
   • $E[X] = \frac{1}{3} + 1 + \frac{1}{2}$
   • Find a common denominator (6):
   • $E[X] = \frac{2}{6} + \frac{6}{6} + \frac{3}{6} = \frac{11}{6}$
5. Calculate $E[Z]$:
   • $E[Z] = 2 \times E[X] = 2 \times \left(\frac{11}{6}\right) = \frac{22}{6} = \frac{11}{3}$
6. Convert to decimal:
   • $E[Z] = 11 \div 3 = 3.6666…$
7. Format the answer: Rounding to two decimal places gives 3.67.

Answer: The value of $E[Z]$ is 3.67.

❓ Question 10

Problem: A pharmaceutical company is conducting a clinical trial where historical data shows that 20% of patients experience side effects from a new medication. A research team randomly selects 400 patients for their study. Let a random variable $X$ model the total number of patients who do not experience side effects in the selected sample. Which of the following is true?

• $X \sim \text{Binomial}(400, 0.20)$
• $X \sim \text{Binomial}(400, 0.80)$
• $X \sim \text{Binomial}(400, 0.5)$
• $X \sim \text{Binomial}(400, 0.02)$

Concepts: Binomial Distribution: A random variable $X$ follows a Binomial distribution, $X \sim \text{Binomial}(n, p)$, if it meets these criteria:

1. $n$ (fixed trials): There is a fixed number of trials ($n$).
2. Two Outcomes: Each trial results in one of two outcomes, typically called “success” and “failure”.
3. $p$ (constant probability): The probability of “success” ($p$) is the same for every trial.
4. Independence: The trials are independent of each other.

The variable $X$ then counts the total number of “successes” in the $n$ trials.

Solution:

1. Identify $n$ (number of trials): The team selects 400 patients. Each patient is a trial.
   • $n = 400$
2. Identify “success”: The variable $X$ is defined as the “total number of patients who do not experience side effects.” Therefore, a “success” for this problem is defined as:
   • Success = A patient does not experience side effects.
3. Identify $p$ (probability of success):
   • The problem states that 20% of patients do experience side effects.
   • $P(\text{side effect}) = 0.20$
   • The probability of our defined “success” (no side effect) is the complement of this:
   • $p = P(\text{no side effect}) = 1 - P(\text{side effect})$
   • $p = 1 - 0.20 = \mathbf{0.80}$
4. Conclusion:
   • The random variable $X$ counts the number of successes ($n=400$) where the probability of success is $p=0.80$.
   • Therefore, $X$ follows a Binomial distribution with parameters $n=400$ and $p=0.80$.

Answer: The correct statement is $X \sim \text{Binomial}(400, 0.80)$.

❓ Questions 11-19: Chebyshev’s Inequality Problem

This is a single, multi-part problem that requires filling in 9 blanks (A-I) in a provided solution. Below is the complete, step-by-step derivation, followed by the specific answers for each blank.

🎯 Overall Problem

You are given the probability distribution for a random variable $X$:

You need to use Chebyshev’s inequality to find the minimum sample size $n$ such that the probability of the sample mean ($\bar{X}$) being within 1 unit of the population mean ($\mu$) is at least 0.95. The goal is to find $n$ for:

🧠 Core Concepts

1. Expected Value (Mean, $\mu$): For a discrete variable $X$, the expected value is the weighted average of its possible values. $$\mu = E[X] = \sum x \cdot P(X=x)$$
2. Variance ($\sigma^2$, $Var(X)$): This measures the spread of the data. It’s calculated as: $$Var(X) = E[X^2] - (E[X])^2$$ where $E[X^2] = \sum x^2 \cdot P(X=x)$.
3. Variance of the Sample Mean ($Var(\bar{X})$): The variance of the sample mean of $n$ independent and identically distributed (i.i.d.) variables is the population variance divided by $n$. $$Var(\bar{X}) = \frac{Var(X)}{n}$$
4. Chebyshev’s Inequality: This inequality provides a bound on the probability that a random variable is a certain distance from its mean. For any random variable $Y$ with mean $\mu_Y$ and variance $\sigma_Y^2$, and any $k > 0$:
   • Standard Form: $P(|Y - \mu_Y| \ge k) \le \frac{\sigma_Y^2}{k^2}$
   • Complement Form: $P(|Y - \mu_Y| < k) \ge 1 - \frac{\sigma_Y^2}{k^2}$ In our problem, $Y = \bar{X}$, $\mu_Y = \mu$, $\sigma_Y^2 = Var(\bar{X})$, and $k=1$.

📝 Step-by-Step Solution

Here is the complete calculation to find the minimum value of $n$.

Step 1: Calculate the Population Mean ($\mu$)

Step 2: Calculate $E[X^2]$

Step 3: Calculate the Population Variance ($Var(X)$)

Step 4: State the Variance of the Sample Mean ($Var(\bar{X})$)

Step 5: Apply Chebyshev’s Inequality We want to find $n$ such that $P(|\bar{X} - \mu| < 1) \ge 0.95$. We use the complement form of Chebyshev’s inequality with $k=1$:

Step 6: Solve for $n$ To guarantee our probability is at least 0.95, we set the bound from our formula to be greater than or equal to our desired probability:

Now, we just solve for $n$:

The minimum value of $n$ is 980.

📋 Answers to the Blanks

Now we can fill in each blank from your images based on the detailed solution above.

11) Blank A:

• Question: P(|\bar{X} - \mu| ___ A ___) \ge F
• Context: This is the complement form of Chebyshev’s, $P(|\bar{X} - \mu| < k) \ge \ldots$. The problem text states we want the mean to be “less than 1” from the population mean.
• Answer: < 1

12) Blank B:

• Question: P(|\bar{X} - \mu| < 1) \ge ___ B ___
• Context: This is equation (1), which represents the goal of the problem. The text says we want the probability to be “at least 0.95”.
• Answer: 0.95

13) Blank C:

• Question: P(|\bar{X} - \mu| ___ C ___) \le D
• Context: This is equation (2), the standard (non-complement) form of Chebyshev’s inequality, $P(|\bar{X} - \mu| \ge k) \le \ldots$. Here $k=1$.
• Answer: \ge 1

14) Blank D:

• Question: P(|\bar{X} - \mu| \ge 1) \le ___ D ___
• Context: This is the right side of the standard Chebyshev’s inequality, $\frac{Var(\bar{X})}{k^2}$. We have $k=1$ and $Var(\bar{X}) = 49/n$.
• Answer: 49/n (or $Var(\bar{X})$)

15) Blank E:

• Question: Var(\bar{X}) = \ldots = ___ E ___ / n
• Context: This blank asks for the value of the population variance, $Var(X)$. From Step 3 of our solution, $Var(X) = 49$.
• Answer: 49

16) Blank F:

• Question: P(|\bar{X} - \mu| < 1) \ge ___ F ___
• Context: This is equation (3), the result of applying the complement form of Chebyshev’s inequality. From Step 5, this is $1 - \frac{Var(\bar{X})}{1^2}$.
• Answer: 1 - 49/n

17) Blank G:

• Question: \Rightarrow \frac{49}{n} \le \text{___ G ___}
• Context: This line in the derivation comes from setting $F \ge B$. $1 - \frac{49}{n} \ge 0.95$ $1 - 0.95 \ge \frac{49}{n}$ $0.05 \ge \frac{49}{n}$, which is the same as $\frac{49}{n} \le 0.05$.
• Answer: 0.05

18) Blank H:

• Question: \Rightarrow n \ge \frac{\text{___ H ___}}{0.05}
• Context: This is the next step in solving for $n$ from $\frac{49}{n} \le 0.05$. Multiplying by $n$ and dividing by $0.05$ gives $n \ge \frac{49}{0.05}$.
• Answer: 49

19) Blank I:

• Question: Therefore, minimum value of n is ___ I ___ (This blank was likely mistyped as ‘H’ in your image, but corresponds to question 19 for ‘I’).
• Context: This is the final calculated value for $n$. $n \ge \frac{49}{0.05}$
• Answer: 980