🗓️ Mock for Quiz 2

❓ Question 1

Question: Suppose $X_1, X_2, \dots, X_{10} \sim iid \text{ Exp}(5)$. Compute the expected value and variance of sample mean, $\bar{X} = \frac{X_1 + \dots + X_{10}}{10}$.

• $E[\bar{X}] = 0.2 \text{ and } \text{Var}[\bar{X}] = \frac{4}{900}$
• $E[\bar{X}] = 0.2 \text{ and } \text{Var}[\bar{X}] = 0.004$
• $E[\bar{X}] = 5 \text{ and } \text{Var}[\bar{X}] = 2.5$
• $E[\bar{X}] = 5 \text{ and } \text{Var}[\bar{X}] = 0.004$

💡 Concepts

1. Exponential Distribution (Rate Parameter): The notation $\text{Exp}(\lambda)$ refers to an exponential distribution with a rate parameter $\lambda$.

   • The mean (expected value) is $E[X_i] = \mu = \frac{1}{\lambda}$.
   • The variance is $\text{Var}[X_i] = \sigma^2 = \frac{1}{\lambda^2}$.

2. Properties of the Sample Mean ($\bar{X}$): For any $iid$ (independent and identically distributed) sample of size $n$ from a population with mean $\mu$ and variance $\sigma^2$:

   • Expected Value of $\bar{X}$: $E[\bar{X}] = \mu$. The expected value of the sample mean is simply the population mean.
   • Variance of $\bar{X}$: $\text{Var}[\bar{X}] = \frac{\sigma^2}{n}$. The variance of the sample mean is the population variance divided by the sample size.

Solution

1. Identify Parameters:

   • Sample size: $n = 10$
   • Rate parameter: $\lambda = 5$

2. Find Population Mean ($\mu$) and Variance ($\sigma^2$):

   • $\mu = E[X_i] = \frac{1}{\lambda} = \frac{1}{5} = \mathbf{0.2}$
   • $\sigma^2 = \text{Var}[X_i] = \frac{1}{\lambda^2} = \frac{1}{5^2} = \frac{1}{25} = \mathbf{0.04}$

3. Calculate $E[\bar{X}]$:

   • $E[\bar{X}] = \mu = \mathbf{0.2}$

4. Calculate $\text{Var}[\bar{X}]$:

   • $\text{Var}[\bar{X}] = \frac{\sigma^2}{n} = \frac{0.04}{10} = \mathbf{0.004}$

Answer: This matches the second option: $E[\bar{X}] = 0.2 \text{ and } \text{Var}[\bar{X}] = 0.004$.

❓ Question 2

(Please use the below information for the questions 2 & 3) Question: Suppose the random variables $X$ and $Y$ are uniformly distributed over the region $D = [0, 1] \times [0, 1] \cup [2, 3] \times [0, 1]$. Determine $f_{XY}(x, y)$.

💡 Concept: Joint Uniform Distribution

For a continuous distribution to be uniform over a specific region $D$, its probability density function (PDF) $f_{XY}(x, y)$ must be a constant value, $c$, for every point inside that region.

To find this constant $c$, we use the fact that the total probability (the volume under the PDF) must equal 1.

• $\text{Volume} = \text{Area}(D) \times \text{Height}(c) = 1$
• Therefore, the height $c = \frac{1}{\text{Area}(D)}$

Solution

1. Sketch the Region $D$: The region $D$ consists of two separate rectangles.

   • Rectangle 1: $[0, 1] \times [0, 1]$ $\implies$ $0 \le x \le 1$ and $0 \le y \le 1$.
   • Rectangle 2: $[2, 3] \times [0, 1]$ $\implies$ $2 \le x \le 3$ and $0 \le y \le 1$.

2. Calculate the Total Area of $D$:

   • $\text{Area}(\text{Rect 1}) = \text{width} \times \text{height} = (1 - 0) \times (1 - 0) = 1$
   • $\text{Area}(\text{Rect 2}) = \text{width} \times \text{height} = (3 - 2) \times (1 - 0) = 1$
   • $\text{Total Area}(D) = \text{Area}(\text{Rect 1}) + \text{Area}(\text{Rect 2}) = 1 + 1 = \mathbf{2}$

3. Find the PDF Height ($c$):

   • $c = \frac{1}{\text{Total Area}(D)} = \frac{1}{2}$

4. Write the PDF: The PDF $f_{XY}(x, y)$ is $\frac{1}{2}$ for any point $(x, y)$ inside $D$, and 0 otherwise.

Answer: This matches the third option: $f_{XY}(x, y) = \begin{cases} \frac{1}{2} & \text{for } (x, y) \in D \ 0 & \text{otherwise} \end{cases}$ (Note: The option text “$x \in D$” is likely a typo and should mean $(x, y) \in D$.)

❓ Question 3

Question: Are $X$ and $Y$ independent?

• Yes
• No

💡 Concept: Independence of Continuous Variables

Two random variables $X$ and $Y$ are independent if and only if their joint PDF $f_{XY}(x, y)$ can be factored into the product of their marginal PDFs, $f_X(x)$ and $f_Y(y)$. $f_{XY}(x, y) = f_X(x) \cdot f_Y(y) \quad \text{for all } x, y$

A quick test is to check if the support (the region $D$ where the PDF is non-zero) is a “product space.” A product space $D$ can be written as $A \times B$, where $A$ is a set of $x$-values and $B$ is a set of $y$-values.

Solution

1. Check the Support $D$:

   • Our support is $D = ([0, 1] \cup [2, 3]) \times [0, 1]$.
   • This is a product space, where:
     • The set of $x$-values is $A = [0, 1] \cup [2, 3]$.
     • The set of $y$-values is $B = [0, 1]$.
   • Because the support is a product space and the PDF ($f_{XY} = \frac{1}{2}$) is constant, $X$ and $Y$ are independent.

2. Formal Proof (by finding marginals):

   • Find $f_X(x)$: Integrate $f_{XY}$ over all $y$. The range of $y$ is $[0, 1]$. $f_X(x) = \int_{0}^{1} f_{XY}(x, y) ,dy = \int_{0}^{1} \frac{1}{2} ,dy = \frac{1}{2} [y]_0^1 = \frac{1}{2}(1 - 0) = \frac{1}{2}$ (This is valid for $x \in [0, 1] \cup [2, 3]$)
   • Find $f_Y(y)$: Integrate $f_{XY}$ over all $x$. The range of $x$ is $[0, 1] \cup [2, 3]$. $f_Y(y) = \int_{-\infty}^{\infty} f_{XY}(x, y) ,dx = \int_{0}^{1} \frac{1}{2} ,dx + \int_{2}^{3} \frac{1}{2} ,dx$ $f_Y(y) = \left[ \frac{1}{2}x \right]_0^1 + \left[ \frac{1}{2}x \right]_2^3 = \left(\frac{1}{2} - 0\right) + \left(\frac{3}{2} - \frac{2}{2}\right) = \frac{1}{2} + \frac{1}{2} = 1$ (This is valid for $y \in [0, 1]$)
   • Check the Product: $f_X(x) \cdot f_Y(y) = \frac{1}{2} \cdot 1 = \frac{1}{2}$
   • This product, $\frac{1}{2}$, is equal to the joint PDF $f_{XY}(x, y)$ over the entire support $D$. Therefore, the variables are independent.

Answer: Yes

❓ Question 4

(Please use the below information for the questions 4 & 5) Question: Let $X_1, X_2 \sim iid X$ and $X$ be a discrete random variable with following probability mass function $P(X = k) = \begin{cases} 0.3 & \text{for } k = -4 \ 0.2 & \text{for } k = 1 \ 0.5 & \text{for } k = 2 \ 0 & \text{otherwise} \end{cases}$

Find the moment generating function (MGF) of $X$.

💡 Concept: Moment Generating Function (MGF)

The MGF of a random variable $X$ is defined as $M_X(t) = E[e^{tX}]$.

• For a discrete random variable, this is calculated by summing $e^{tk} \cdot P(X=k)$ over all possible values of $k$. $M_X(t) = \sum_k e^{tk} P(X=k)$

(Note: The question uses $\lambda$ instead of the more common $t$. We will use $\lambda$ to match the options.)

Solution

1. Apply the MGF definition: $M_X(\lambda) = E[e^{\lambda X}] = \sum_k e^{\lambda k} P(X=k)$

2. Plug in the values from the PMF: $M_X(\lambda) = e^{\lambda(-4)} P(X=-4) + e^{\lambda(1)} P(X=1) + e^{\lambda(2)} P(X=2)$

3. Substitute the probabilities: $M_X(\lambda) = e^{-4\lambda} \cdot (0.3) + e^{\lambda} \cdot (0.2) + e^{2\lambda} \cdot (0.5)$

4. Reorder to match the options: $M_X(\lambda) = 0.3e^{-4\lambda} + 0.2e^{\lambda} + 0.5e^{2\lambda}$

Answer: This matches the second option.

❓ Question 5

Question: Define $Y = X_1 + X_2$. Find the distribution of $Y$.

💡 Concept: Distribution of a Sum of Independent Variables

We need to find the probability mass function (PMF) for $Y = X_1 + X_2$. Since $X_1$ and $X_2$ are independent, the probability of any joint outcome is the product of their individual probabilities: $P(X_1 = a, X_2 = b) = P(X_1 = a)P(X_2 = b)$.

To find $P(Y=y)$, we must find all pairs $(X_1, X_2)$ that sum to $y$ and add their probabilities. This process is called convolution.

Solution

1. List all possible outcomes and their probabilities: The possible values for $X_1$ and $X_2$ are ${-4, 1, 2}$ with probabilities ${0.3, 0.2, 0.5}$.

2. Find all possible values of $Y = X_1 + X_2$:

   $X_1 \downarrow$	$X_2 \to$ -4 (P=0.3)	1 (P=0.2)	2 (P=0.5)
   -4 (P=0.3)	$Y = -8$	$Y = -3$	$Y = -2$
   1 (P=0.2)	$Y = -3$	$Y = 2$	$Y = 3$
   2 (P=0.5)	$Y = -2$	$Y = 3$	$Y = 4$
   The possible values for $Y$ are ${-8, -3, -2, 2, 3, 4}$.

3. Calculate the probability for each value of $Y$:

   • $P(Y = -8)$:
     • $X_1 = -4, X_2 = -4$
     • $P = (0.3) \times (0.3) = \mathbf{0.09}$
   • $P(Y = -3)$:
     • ($X_1 = -4, X_2 = 1$) OR ($X_1 = 1, X_2 = -4$)
     • $P = (0.3)(0.2) + (0.2)(0.3) = 0.06 + 0.06 = \mathbf{0.12}$
   • $P(Y = -2)$:
     • ($X_1 = -4, X_2 = 2$) OR ($X_1 = 2, X_2 = -4$)
     • $P = (0.3)(0.5) + (0.5)(0.3) = 0.15 + 0.15 = \mathbf{0.30}$
   • $P(Y = 2)$:
     • $X_1 = 1, X_2 = 1$
     • $P = (0.2) \times (0.2) = \mathbf{0.04}$
   • $P(Y = 3)$:
     • ($X_1 = 1, X_2 = 2$) OR ($X_1 = 2, X_2 = 1$)
     • $P = (0.2)(0.5) + (0.5)(0.2) = 0.10 + 0.10 = \mathbf{0.20}$
   • $P(Y = 4)$:
     • $X_1 = 2, X_2 = 2$
     • $P = (0.5) \times (0.5) = \mathbf{0.25}$

4. Assemble the PMF Table for $Y$:

   $Y$	$P(Y=y)$
   -8	0.09
   -3	0.12
   -2	0.30
   2	0.04
   3	0.20
   4	0.25

Answer: This matches the first option.

❓ Question 15

Question: From past experience, it is known that the number of tickets purchased by a person standing in line at the ticket window for a movie follows a distribution that has mean $\mu = 2.4$ and standard deviation $\sigma = 2$. Suppose that few hours before the start of a movie there are 100 persons standing in the line to purchase tickets. If only 250 tickets remain, what is the approximate probability that all 100 person will be able to purchase the tickets they desire?

• $F_Z(1.5)$
• $1 - F_Z(1.5)$
• $F_Z(0.5)$
• $1 - F_Z(0.5)$

💡 Concept: The Central Limit Theorem (CLT)

This problem involves the sum of a large number of independent random variables (the number of tickets each of the 100 people wants). The Central Limit Theorem (CLT) states that when you sum a large number ($n \ge 30$ is a good rule of thumb) of independent and identically distributed (iid) random variables, their sum $S_n$ is approximately normally distributed, regardless of the original distribution.

• Mean of the Sum: $E[S_n] = n \cdot \mu$
• Variance of the Sum: $\text{Var}[S_n] = n \cdot \sigma^2$
• Standard Deviation of the Sum: $\text{SD}[S_n] = \sqrt{n} \cdot \sigma$

We can then standardize this sum using a Z-score: $Z = \frac{S_n - E[S_n]}{\text{SD}[S_n]}$

Solution

1. Define Variables:

   • Let $X_i$ be the number of tickets desired by person $i$.
   • $E[X_i] = \mu = 2.4$
   • $\text{SD}[X_i] = \sigma = 2$
   • $\text{Var}[X_i] = \sigma^2 = 2^2 = 4$
   • The sample size is $n = 100$.

2. Define the Sum:

   • Let $S_n$ be the total number of tickets desired by all 100 people: $S_n = X_1 + X_2 + \dots + X_{100}$.
   • We want to find the probability that this total is less than or equal to the 250 tickets available: $P(S_n \le 250)$.

3. Apply the Central Limit Theorem:

   • Since $n=100$ is large, $S_n$ is approximately normally distributed.
   • Mean of Sum: $E[S_n] = n \cdot \mu = 100 \times 2.4 = 240$
   • Variance of Sum: $\text{Var}[S_n] = n \cdot \sigma^2 = 100 \times 4 = 400$
   • Standard Deviation of Sum: $\text{SD}[S_n] = \sqrt{400} = 20$
   • So, $S_n \sim \text{Normal}(\mu = 240, \sigma = 20)$ (approximately).

4. Standardize (find the Z-score):

   • We need to find the Z-score for the value 250.
   • $Z = \frac{\text{Value} - \text{Mean}}{\text{Standard Deviation}} = \frac{250 - 240}{20} = \frac{10}{20} = 0.5$

5. Find the Probability:

   • $P(S_n \le 250)$ is approximately equal to $P(Z \le 0.5)$.
   • The probability $P(Z \le z)$ is the definition of the standard normal cumulative distribution function (CDF), which is denoted by $F_Z(z)$.
   • Therefore, $P(Z \le 0.5) = F_Z(0.5)$.

Answer: This matches the third option: $F_Z(0.5)$

❓ Question 16

Question: A random sample of size 50 is collected from a population with a standard deviation of 10. Let $A$ be the event that the sample mean $\bar{X}$ is within 4 of the population mean $\mu$, i.e., $A = |\bar{X} - \mu| < 4$. Find a lower bound on $P(A)$ using the weak law of large numbers. Enter the answers correct to three decimal places.

💡 Concept: Chebyshev’s Inequality

The “weak law of large numbers” can be proved using Chebyshev’s Inequality. When a question asks for a “lower bound” on a probability, it almost always implies using this inequality.

Chebyshev’s Inequality provides a lower bound for the probability that a random variable $Y$ will fall within $k$ standard deviations of its mean. A more general form, and the one we’ll use, is: $P(|Y - E[Y]| \ge k) \le \frac{\text{Var}[Y]}{k^2}$

The complement of this is more useful here, as it gives a lower bound on the probability of being within a certain distance $k$ of the mean: $P(|Y - E[Y]| < k) \ge 1 - \frac{\text{Var}[Y]}{k^2}$

Solution

1. Identify Components:

   • We are interested in the sample mean, so our random variable is $Y = \bar{X}$.
   • The expected value of the sample mean is the population mean: $E[Y] = E[\bar{X}] = \mu$.
   • The event $A$ is $|\bar{X} - \mu| < 4$. This matches our inequality form with $k = 4$.
   • Population standard deviation $\sigma = 10$, so population variance $\sigma^2 = 100$.
   • Sample size $n = 50$.

2. Calculate the Variance of the Sample Mean ($\text{Var}[\bar{X}]$):

   • $\text{Var}[Y] = \text{Var}[\bar{X}] = \frac{\sigma^2}{n}$
   • $\text{Var}[\bar{X}] = \frac{100}{50} = 2$

3. Apply Chebyshev’s Inequality:

   • We want to find the lower bound for $P(A) = P(|\bar{X} - \mu| < 4)$.
   • $P(|\bar{X} - \mu| < 4) \ge 1 - \frac{\text{Var}[\bar{X}]}{4^2}$
   • $P(A) \ge 1 - \frac{2}{16}$
   • $P(A) \ge 1 - \frac{1}{8}$
   • $P(A) \ge \frac{7}{8}$

4. Format Answer:

   • As a decimal, the lower bound is $7 \div 8 = 0.875$.

Answer: 0.875

❓ Questions 17 & 18

(Please use the below information for the questions 17 & 18) Let $X$ and $Y$ be independent exponentially distributed random variables, each with expected value 1.

❓ Question 17

Find $P(X < 2Y)$. Enter the answer correct to two decimal places.

💡 Concepts

1. Exponential PDF: For an exponential random variable with mean $E[X] = 1/\lambda$, the rate parameter is $\lambda = 1 / E[X]$. The PDF is $f(x) = \lambda e^{-\lambda x}$.
2. Joint PDF for Independent Variables: If $X$ and $Y$ are independent, their joint PDF is the product of their individual (marginal) PDFs: $f_{XY}(x, y) = f_X(x) f_Y(y)$.
3. Probability as Volume: To find the probability of a region (e.g., $X < 2Y$), we must integrate the joint PDF over that region within the support (where the variables are defined).

Solution

1. Find PDFs:
   • $E[X] = 1 \implies \lambda_X = 1/1 = 1$. So, $f_X(x) = e^{-x}$ for $x > 0$.
   • $E[Y] = 1 \implies \lambda_Y = 1/1 = 1$. So, $f_Y(y) = e^{-y}$ for $y > 0$.
2. Find Joint PDF:
   • Since $X$ and $Y$ are independent, $f_{XY}(x, y) = e^{-x} \cdot e^{-y} = e^{-(x+y)}$.
   • The support for this joint PDF is the first quadrant ($x > 0, y > 0$).
3. Define Region of Integration:
   • We want $P(X < 2Y)$, which is the same as $P(Y > X/2)$.
   • This is the region in the first quadrant above the line $Y = X/2$.
   • 4. Set up and Solve the Double Integral:
   • We’ll integrate with respect to $y$ first, then $x$.
   • $P(X < 2Y) = \int_{x=0}^{\infty} \left( \int_{y=x/2}^{\infty} e^{-(x+y)} ,dy \right) ,dx$
   • Inner Integral (w.r.t $y$): $\int_{x/2}^{\infty} e^{-x} e^{-y} ,dy = e^{-x} \int_{x/2}^{\infty} e^{-y} ,dy = e^{-x} \left[ -e^{-y} \right]_{y=x/2}^{y=\infty}$ $= e^{-x} ( (0) - (-e^{-x/2}) ) = e^{-x} \cdot e^{-x/2} = e^{-3x/2}$
   • Outer Integral (w.r.t $x$): $\int_{0}^{\infty} e^{-3x/2} ,dx = \left[ -\frac{1}{3/2} e^{-3x/2} \right]_0^\infty = \left[ -\frac{2}{3} e^{-3x/2} \right]_0^\infty$ $= (0) - (-\frac{2}{3} e^0) = \frac{2}{3}$
4. Format Answer:
   • $2/3 = 0.6666…$
   • Rounded to two decimal places, this is $0.67$.

Answer (17): 0.67

❓ Question 18

Find $P(X > Y \mid X < 2Y)$. Write your answer correct to two decimal places.

💡 Concept: Conditional Probability

The definition of conditional probability is $P(A \mid B) = \frac{P(A \cap B)}{P(B)}$.

• Let $A$ be the event $X > Y$.
• Let $B$ be the event $X < 2Y$.
• We need to find $\frac{P(X > Y \text{ and } X < 2Y)}{P(X < 2Y)}$.

This simplifies to finding the probability of the intersection $P(Y < X < 2Y)$ and dividing it by the probability we found in Question 17.

Solution

1. Identify Probabilities:
   • Denominator: $P(B) = P(X < 2Y)$. We found this in Q17. $P(B) = 2/3$.
   • Numerator: $P(A \cap B) = P(Y < X < 2Y)$. We must calculate this.
2. Define Region for Numerator:
   • The region is bounded by the lines $Y = X$ and $Y = X/2$ in the first quadrant. It’s a “wedge” between these two lines.
   • 3. Set up and Solve Integral for Numerator:
   • $P(Y < X < 2Y) = \int_{x=0}^{\infty} \left( \int_{y=x/2}^{y=x} e^{-(x+y)} ,dy \right) ,dx$
   • Inner Integral (w.r.t $y$): $\int_{x/2}^{x} e^{-x} e^{-y} ,dy = e^{-x} \int_{x/2}^{x} e^{-y} ,dy = e^{-x} \left[ -e^{-y} \right]_{y=x/2}^{y=x}$ $= e^{-x} ( (-e^{-x}) - (-e^{-x/2}) ) = e^{-x} (e^{-x/2} - e^{-x})$ $= e^{-3x/2} - e^{-2x}$
   • Outer Integral (w.r.t $x$): $\int_{0}^{\infty} (e^{-3x/2} - e^{-2x}) ,dx = \int_{0}^{\infty} e^{-3x/2} ,dx - \int_{0}^{\infty} e^{-2x} ,dx$
     • From Q17, we know $\int_{0}^{\infty} e^{-3x/2} ,dx = 2/3$.
     • $\int_{0}^{\infty} e^{-2x} ,dx = \left[ -\frac{1}{2} e^{-2x} \right]_0^\infty = (0) - (-\frac{1}{2} e^0) = 1/2$.
   • $\text{Numerator} = \frac{2}{3} - \frac{1}{2} = \frac{4}{6} - \frac{3}{6} = \frac{1}{6}$
3. Calculate Final Conditional Probability:
   • $P(A \mid B) = \frac{\text{Numerator}}{\text{Denominator}} = \frac{1/6}{2/3}$
   • $P(A \mid B) = \frac{1}{6} \times \frac{3}{2} = \frac{3}{12} = \frac{1}{4}$
4. Format Answer:
   • $1/4 = 0.25$

Answer (18): 0.25

💡 Concept: The Central Limit Theorem (CLT)

The problem relies on the Central Limit Theorem (CLT). This theorem is fundamental to statistics and states that if you take a sufficiently large sample ($n \ge 30$ is a common rule of thumb) from any population (regardless of its original distribution), the distribution of the sample mean ($\bar{X}$) will be approximately normal.

• Mean of Sample Means: The mean of this new normal distribution is the same as the original population mean, $\mu$. $E[\bar{X}] = \mu$
• Standard Deviation of Sample Means: The standard deviation of this new normal distribution, known as the standard error (SE), is the original population standard deviation, $\sigma$, divided by the square root of the sample size, $n$. $\text{SD}[\bar{X}] = \frac{\sigma}{\sqrt{n}}$

❓ Question 19 (Blank A)

Question: By Central Limit Theorem, the sampling distribution of $\bar{X}$ will approximately be A…

Solution: As explained by the Central Limit Theorem, since the sample size $n = 64$ is large (it’s greater than 30), the sampling distribution of the sample mean ($\bar{X}$) will be approximately Normal.

• Answer: 2 (Normal)

❓ Question 20 (Blank B)

Question: …with mean B…

Solution: The mean of the sampling distribution of $\bar{X}$ is always equal to the population mean ($\mu$).

• Given: Population mean $\mu = 800$.

• Therefore, $E[\bar{X}] = \mu = 800$.

• Answer: 3 (800)

❓ Question 21 (Blank C)

Question: …and standard deviation c…

Solution: The standard deviation of the sampling distribution of $\bar{X}$ (the standard error) is $\frac{\sigma}{\sqrt{n}}$.

• Given: Population standard deviation $\sigma = 120$.

• Given: Sample size $n = 64$.

• Calculation: $\text{SD}[\bar{X}] = \frac{120}{\sqrt{64}} = \frac{120}{8} = 15$.

• Answer: 6 (15)

❓ Question 22 (Blank D)

Question: Now, to verify the company’s claim, we have to calculate the value of $P(\text{D})$.

Solution: The company’s claim is about the “average lifetime” ($\bar{X}$) being “less than 780 hours”. We need to find the probability of this event occurring.

• Therefore, we must calculate $P(\bar{X} < 780)$.

• Answer: 13 ($\bar{X} < 780$)

❓ Question 23 (Blank E)

Question: By using CLT we can write, $P(\text{D}) = P(Z < \text{E})$

Solution: To find the probability, we must convert our value $\bar{X} = 780$ into a standard normal Z-score using the mean and standard deviation of the sampling distribution we found in blanks B and C.

• Z-score formula: $Z = \frac{\bar{X} - E[\bar{X}]}{\text{SD}[\bar{X}]}$

• Our values:

  • $\bar{X} = 780$
  • $E[\bar{X}] = 800$ (from Blank B)
  • $\text{SD}[\bar{X}] = 15$ (from Blank C)

• Calculation: $Z = \frac{780 - 800}{15} = \frac{-20}{15} = -\frac{4}{3} \approx -1.33$

• So, $P(\bar{X} < 780) = P(Z < -1.33)$.

• Answer: 8 (–1.33)

❓ Question 24 (Blank F)

Question: $= 1 - P(Z < \text{F})$

Solution: This step in the solution is calculating the probability $P(Z < -1.33)$ using a standard normal table, which often only provides probabilities for positive Z-values. We use the symmetry of the normal distribution:

1. The probability of being in the left tail $P(Z < -1.33)$ is, by symmetry, equal to the probability of being in the right tail $P(Z > 1.33)$. 2. The probability of being in the right tail $P(Z > 1.33)$ is equal to 1 minus the probability of not being in that tail (i.e., 1 minus the left/main body). $P(Z > 1.33) = 1 - P(Z < 1.33)$
2. Combining these, we get: $P(Z < -1.33) = 1 - P(Z < 1.33)$.

• Comparing this to the template $= 1 - P(Z < \text{F})$, we see that F = 1.33.

• Answer: 10 (1.33)

❓ Question 25 (Blank G)

Question: $= \text{G}$

Solution: This is the final calculation step from the previous question.

• We need to calculate $1 - P(Z < 1.33)$.

• The problem gives us the hint: $F_Z(1.33) = 0.908$. (Note: $F_Z(z)$ is the standard notation for $P(Z < z)$).

• Calculation: $G = 1 - 0.908 = 0.092$.

• So, the actual probability $P(\bar{X} < 780)$ is 0.092, or 9.2%.

• Answer: 14 (0.092)

❓ Question 26 (Blank H)

Question: Hence, the company’s claim is H…

Solution: We must now compare our finding with the company’s claim.

• Our Calculation: The actual probability is $P(\bar{X} < 780) = 0.092$, or 9.2%.

• Company’s Claim: The probability is “only about 5%”.

• Since 9.2% is not “about 5%” (it’s nearly double), the company’s claim is not true.

• Answer: 17 (not true)

❓ Question 27 (Blank I)

Question: …as the value of $P(\bar{X} < 780)$ is I 5%.

Solution: This is the final comparison that justifies our answer for H.

• Our calculated value is 9.2%.

• The company’s claim is 5%.

• The number 9.2% is greater than 5%.

• Answer: 19 (greater than)