are 9 and 3 respectively.
Based on the above data, answer the given subquestions.
Find the value of
. Enter the answer correct to two decimal place.
Published solution
**1. Identify the support of $X \sim \text{Uniform}(6, 12)$:**
From Question 2, $X$ is uniformly distributed on the interval $[a, b] = [6, 12]$, with PDF:
$$f_X(x) = \frac{1}{12 - 6} = \frac{1}{6} \quad \text{for } 6 \le x \le 12$$
and $f_X(x) = 0$ otherwise.
**2. Compute $P(5 < X < 7)$:**
Since $f_X(x) = 0$ on $(5, 6)$, the integral is only over $[6, 7]$:
$$P(5 < X < 7) = P(6 \le X < 7) = \frac{7 - 6}{12 - 6} = \frac{1}{6} \approx 0.1667 \approx 0.17$$
**Answer: 0.17**
The number of customer arriving at a bank counter follows a Poisson distribution with an average
of 4 arrivals per hour. What is the probability that at least 2 customers arrives in 30 minutes? Enter
the answer correct to two decimal places.
Published solution
**1. Find the Poisson parameter $\lambda$ for a 30-minute interval:**
The average arrival rate is $4$ customers per hour ($60$ minutes).
For a $30$-minute ($0.5$-hour) interval, the mean number of arrivals is:
$$\lambda = 4 \times 0.5 = 2$$
Let $X \sim \text{Poisson}(2)$.
**2. Compute $P(X \ge 2)$ using the complement rule:**
$$P(X \ge 2) = 1 - P(X = 0) - P(X = 1) = 1 - e^{-2} - 2e^{-2} = 1 - 3e^{-2}$$
Using $e^{-2} \approx 0.1353$:
$$P(X \ge 2) \approx 1 - 3(0.1353) = 1 - 0.4060 = 0.5940 \approx 0.59$$
**Answer: 0.59**
The probability mass function of a discrete random variable X is given by
Find the value of
such that
.
Enter the answer correct to one decimal place.
Published solution
**1. Set up two linear equations for $a$ and $b$:**
From the PMF table ($X \in \{1, 2, 3, 4\}$ with probabilities $0.2, a, 0.3, b$):
1. Total probability equals $1$:
$$0.2 + a + 0.3 + b = 1 \implies a + b = 0.5$$
2. Expected value $E(X) = 2.7$:
$$1(0.2) + 2a + 3(0.3) + 4b = 2.7 \implies 0.2 + 2a + 0.9 + 4b = 2.7 \implies 2a + 4b = 1.6 \implies a + 2b = 0.8$$
**2. Solve for $b$:**
Subtracting the first equation ($a + b = 0.5$) from the second equation ($a + 2b = 0.8$) gives:
$$b = 0.8 - 0.5 = 0.3$$
**Answer: 0.3**
**1. Express the event $Y = k$ in terms of $X$:**
Given $X \sim \text{Binomial}\left(n = 10,\; p = \frac{1}{4}\right)$ with support $X \in \{0, 1, 2, \dots, 10\}$, the random variable $Y = X + 10$ takes values in:
$$k \in \{10, 11, 12, \dots, 20\}$$
Moreover, $Y = k \iff X = k - 10$.
**2. Substitute $x = k - 10$ into the Binomial PMF:**
Since $P(X = x) = {}^{10}C_x \left(\frac{1}{4}\right)^x \left(\frac{3}{4}\right)^{10 - x}$, replacing $x$ by $k - 10$ (and noting $10 - (k - 10) = 20 - k$) gives:
$$P(Y = k) = P(X = k - 10) = {}^{10}C_{k - 10} \left(\frac{1}{4}\right)^{k - 10} \left(\frac{3}{4}\right)^{20 - k}, \quad \text{for } k = 10, 11, 12, \dots, 20$$
**Answer: Option A**
A: **Correct:** Substituting $X = k - 10$ into the $\text{Binomial}(10, 1/4)$ PMF gives ${}^{10}C_{k-10}(1/4)^{k-10}(3/4)^{20-k}$ for $k = 10, \dots, 20$.
B: **Incorrect:** The exponent of $1/4$ must be $k - 10$, not $k$.
C: **Incorrect:** This is the PMF of $X$ itself on $k = 0, \dots, 10$, not $Y = X + 10$.
D: **Incorrect:** The number of trials is $n = 10$ (${}^{10}C_{k-10}$), not $20$ (${}^{20}C_{k-10}$).
**1. Calculate $P(X = 0)$ and $P(X = 3)$ using the hypergeometric distribution:**
There are $3$ red balls and $3$ blue balls ($6$ balls total), and $3$ balls are drawn without replacement. The total number of ways to pick $3$ balls out of $6$ is $\binom{6}{3} = 20$.
- $P(X = 0)$ (picking $0$ red and $3$ blue balls):
$$P(X = 0) = \frac{\binom{3}{0}\binom{3}{3}}{\binom{6}{3}} = \frac{1 \times 1}{20} = \frac{1}{20}$$
- $P(X = 3)$ (picking $3$ red and $0$ blue balls):
$$P(X = 3) = \frac{\binom{3}{3}\binom{3}{0}}{\binom{6}{3}} = \frac{1 \times 1}{20} = \frac{1}{20}$$
- $P(X = 1) = \frac{\binom{3}{1}\binom{3}{2}}{\binom{6}{3}} = \frac{9}{20}$.
**2. Compare the probabilities:**
We see that **$P(X = 0) = P(X = 3) = \frac{1}{20}$**.
**Answer: Option D**
A: **Incorrect:** $P(X = 0) = 1/20$, whereas $P(X = 1) = 9/20$.
B: **Incorrect:** $P(X = 0)$ and $P(X = 3)$ are both equal to $1/20$.
C: **Incorrect:** $P(X = 0)$ is not less than $P(X = 3)$; they are equal by symmetry.
D: **Correct:** By symmetry between the 3 red and 3 blue balls, $P(X = 0) = P(X = 3) = 1/20$.
Consider the dataset 4, 7, 5, 3, 9. If 3 is added to all observations, what is the
percentile of the
new dataset?
Published solution
**1. Sort the original dataset and add $3$ to every observation:**
- Original dataset sorted in ascending order: $3, 4, 5, 7, 9$ ($n = 5$ observations).
- Adding $3$ to each observation gives the new sorted dataset:
$$6, 7, 8, 10, 12$$
**2. Find the $75^{\text{th}}$ percentile of the new dataset:**
For $n = 5$ sorted observations $x_{(1)}, x_{(2)}, x_{(3)}, x_{(4)}, x_{(5)}$, the $75^{\text{th}}$ percentile (third quartile) corresponds to the $4^{\text{th}}$ ordered observation (since $\lceil 0.75 \times 5 \rceil = 4$, or using $1 + 0.75(n - 1) = 1 + 0.75(4) = 4$):
$$x_{(4)} = 10$$
**Answer: 10**
The variable height of students is measured on an interval scale.
6406536030492.
The average temperature recorded each day for one week is time-series data.
6406536030493.
Cross-sectional data are collected at a single point in time across several
individuals, firms, or regions etc.
6406536030494.
The brand of smartphone a person uses is a qualitative variable.
Published solution
**1. Evaluate each statement:**
- **Option A:** Height has a true, meaningful zero point (0 cm means no height) and ratios of heights are meaningful, so height is measured on a **ratio scale**, not an interval scale. (**Incorrect**)
- **Option B:** Recording the average temperature each day over a sequence of time points (one week) produces **time-series data**. (**Correct**)
- **Option C:** Data collected at a single point in time across multiple subjects (individuals, firms, or regions) is the exact definition of **cross-sectional data**. (**Correct**)
- **Option D:** Smartphone brand (e.g., Apple, Samsung, Google) is a non-numeric categorical attribute, so it is a **qualitative (categorical) variable**. (**Correct**)
**Answer: Options B, C, D**
A: **Incorrect:** Height is measured on a ratio scale because it has a true zero point.
B: **Correct:** Observations recorded sequentially over time (daily for a week) constitute time-series data.
C: **Correct:** Cross-sectional data are observations collected from multiple entities at a single point in time.
D: **Correct:** Smartphone brand is a non-numerical category, making it a qualitative variable.
At a seminar, there are 30 participants, consisting of 20 men and 10 women.
• Each man shakes hands only with other men.
• Each woman shakes hands only with other women.
If every allowed pair of people has exactly one handshake, find the total number of handshakes
that occur at the seminar.
Published solution
**1. Count the handshakes among the 20 men:**
Since each man shakes hands with every other man exactly once, the number of male-male handshakes is:
$$\binom{20}{2} = \frac{20 \times 19}{2} = 190$$
**2. Count the handshakes among the 10 women:**
Since each woman shakes hands with every other woman exactly once, the number of female-female handshakes is:
$$\binom{10}{2} = \frac{10 \times 9}{2} = 45$$
**3. Add the two totals:**
$$\text{Total handshakes} = 190 + 45 = 235$$
**Answer: 235**
There are two bags, Bag A and Bag B. Bag A contains 2 red balls and 1 blue ball, while Bag B
contains 1 red ball and 2 blue balls. The probability of choosing Bag A is
, and the probability
of choosing Bag B is
. After selecting a bag, one ball is drawn from it. What is the probability
that the ball drawn is red?
**1. Identify the prior and conditional probabilities:**
- Probability of choosing Bag A: $P(A) = \frac{1}{3}$; Bag A has $2$ red and $1$ blue ball ($3$ total), so $P(\text{Red} \mid A) = \frac{2}{3}$.
- Probability of choosing Bag B: $P(B) = \frac{2}{3}$; Bag B has $1$ red and $2$ blue balls ($3$ total), so $P(\text{Red} \mid B) = \frac{1}{3}$.
**2. Apply the Law of Total Probability:**
$$P(\text{Red}) = P(A)P(\text{Red} \mid A) + P(B)P(\text{Red} \mid B) = \left(\frac{1}{3}\right)\left(\frac{2}{3}\right) + \left(\frac{2}{3}\right)\left(\frac{1}{3}\right) = \frac{2}{9} + \frac{2}{9} = \frac{4}{9}$$
**Answer: Option B**
A: **Incorrect:** $1/3 = 3/9$, whereas the total probability is $2/9 + 2/9 = 4/9$.
B: **Correct:** By the law of total probability, $P(\text{Red}) = (1/3)(2/3) + (2/3)(1/3) = 4/9$.
C: **Incorrect:** $5/9$ is the probability of drawing a blue ball ($1 - 4/9 = 5/9$).
D: **Incorrect:** $2/3$ is $P(\text{Red} \mid A)$, not the overall probability $P(\text{Red})$.
**1. Use the property of the median under a linear transformation:**
If a dataset $y_1, y_2, \dots, y_n$ has median $M = 20$, and each observation is transformed by the strictly increasing linear function $z_i = 5y_i - 3$, the order of the observations is preserved.
**2. Compute the new median:**
$$\text{New Median} = 5M - 3 = 5(20) - 3 = 100 - 3 = 97$$
**Answer: 97**
