Statistics for Data Science I · End Term · Sep 2025 · FN

**1. Use the mean and variance formulas for $X \sim \text{Uniform}(a, b)$:** - Mean: $$E[X] = \frac{a + b}{2} = 8 \implies a + b = 16$$ - Variance: $$\text{Var}(X) = \frac{(b - a)^2}{12} = \frac{4}{3} \implies (b - a)^2 = 16$$ Since $b > a$, we have: $$b - a = 4$$ **2. Solve for $a$ and $b$, then compute $2a - b$:** Adding and subtracting $a + b = 16$ and $b - a = 4$: $$2b = 20 \implies b = 10, \qquad a = 16 - 10 = 6$$ Therefore: $$2a - b = 2(6) - 10 = 12 - 10 = 2$$ **Answer: 2**

**1. Identify the support of $X \sim \text{Uniform}(6, 10)$:** From Question 2, $a = 6$ and $b = 10$, so $X \sim \text{Uniform}(6, 10)$ with constant density $f_X(x) = \frac{1}{10 - 6} = \frac{1}{4}$ on $[6, 10]$. **2. Compute $P(7 < X < 9)$:** Since $[7, 9] \subseteq [6, 10]$: $$P(7 < X < 9) = \frac{9 - 7}{10 - 6} = \frac{2}{4} = 0.5$$ **Answer: 0.5**

**1. Model the number of complaint calls using a Binomial distribution:** Let $X$ be the number of complaint calls out of $n = 12$ independent calls, where the probability of a complaint call is $p = 0.30$ (and $1 - p = 0.70$). Then $X \sim \text{Binomial}(12, 0.30)$. **2. Compute $P(X < 2) = P(X = 0) + P(X = 1)$:** - $P(X = 0) = \binom{12}{0}(0.3)^0(0.7)^{12} = (0.7)^{12} \approx 0.013841$ - $P(X = 1) = \binom{12}{1}(0.3)^1(0.7)^{11} = 12(0.3)(0.7)^{11} = 3.6(0.7)^{11} \approx 0.071184$ Adding the two probabilities: $$P(X < 2) = (0.7)^{11}(0.7 + 3.6) = 4.3 \times (0.7)^{11} \approx 0.085025 \approx 0.085$$ **Answer: 0.085**

**1. Count the outcomes for tossing an unbiased coin 4 times:** There are $2^4 = 16$ equally likely outcomes: - **4 consecutive heads (`HHHH`):** $1$ outcome, so $P(X = 41) = \frac{1}{16}$. - **Exactly 3 consecutive heads (`HHHT` or `THHH`):** $2$ outcomes, so $P(X = 10) = \frac{2}{16}$. - **Otherwise:** $16 - 1 - 2 = 13$ outcomes, so $P(X = -1) = \frac{13}{16}$. **2. Compute the expected value $E[X]$:** $$E[X] = 41\left(\frac{1}{16}\right) + 10\left(\frac{2}{16}\right) + (-1)\left(\frac{13}{16}\right) = \frac{41 + 20 - 13}{16} = \frac{48}{16} = 3$$ **Answer: 3**

**1. Find the Poisson parameter $\lambda$ for a 30-minute interval:** With an average of $4$ arrivals per hour, the mean number of arrivals in $30$ minutes ($0.5$ hours) is $\lambda = 4 \times 0.5 = 2$. Let $X \sim \text{Poisson}(2)$. **2. Compute $P(X \le 2)$:** $$P(X \le 2) = P(X = 0) + P(X = 1) + P(X = 2) = e^{-2}\left(\frac{2^0}{0!} + \frac{2^1}{1!} + \frac{2^2}{2!}\right) = e^{-2}(1 + 2 + 2) = 5e^{-2}$$ Using $e^{-2} \approx 0.135335$: $$P(X \le 2) \approx 5 \times 0.135335 = 0.6767 \approx 0.68$$ **Answer: 0.68**

**1. Express $P(X > 1 \mid X < 3)$ in terms of the CDF values:** From the piecewise constant CDF $F(x)$, $X$ is a discrete random variable taking values in $\{0, 1, 2, 3\}$ with: - $P(X < 3) = P(X \le 2) = F(2) = k$ - $P(1 < X < 3) = P(X = 2) = F(2) - F(1) = k - 0.45$ **2. Solve for $k$ using $P(X > 1 \mid X < 3) = \frac{5}{14}$:** $$P(X > 1 \mid X < 3) = \frac{P(X = 2)}{P(X < 3)} = \frac{k - 0.45}{k} = \frac{5}{14}$$ Cross-multiplying: $$14(k - 0.45) = 5k \implies 14k - 6.3 = 5k \implies 9k = 6.3 \implies k = 0.7$$ **Answer: 0.7**

**1. Write out the probabilities for $X \in \{0, 1, 2, 3, 4\}$ where $P(X = x) = \frac{x + 2}{20}$:** - $P(X = 0) = \frac{2}{20} = 0.10$ - $P(X = 1) = \frac{3}{20} = 0.15$ - $P(X = 2) = \frac{4}{20} = 0.20$ - $P(X = 3) = \frac{5}{20} = 0.25$ - $P(X = 4) = \frac{6}{20} = 0.30$ **2. Compute $P(X > 1)$:** $$P(X > 1) = 1 - P(X = 0) - P(X = 1) = 1 - 0.10 - 0.15 = 0.75$$ **Answer: 0.75**

**1. Compute $E(X^2)$ using $P(X = x) = \frac{x + 2}{20}$ for $x = 0, 1, 2, 3, 4$:** $$E(X^2) = 0^2\left(\frac{2}{20}\right) + 1^2\left(\frac{3}{20}\right) + 2^2\left(\frac{4}{20}\right) + 3^2\left(\frac{5}{20}\right) + 4^2\left(\frac{6}{20}\right)$$ $$E(X^2) = \frac{0 + 3 + 16 + 45 + 96}{20} = \frac{160}{20} = 8$$ **2. Compute $E(3X^2 + 5)$ by linearity of expectation:** $$E(3X^2 + 5) = 3E(X^2) + 5 = 3(8) + 5 = 24 + 5 = 29$$ **Answer: Option A** A: **Correct:** Since $E(X^2) = 160/20 = 8$, $E(3X^2 + 5) = 3(8) + 5 = 29$. B: **Incorrect:** $24$ is $3E(X^2)$, forgetting to add $+5$. C: **Incorrect:** Confuses $E(X^2)$ with $3E(X^2)$ before multiplying by $3$. D: **Incorrect:** $3(24) + 5 = 77$ multiplies by $3$ twice.

**1. Read all student scores from the stem-and-leaf plot:** - Stem $2$: $24, 27$ ($2$ students) - Stem $3$: $31, 31, 35, 38, 39$ ($5$ students) - Stem $4$: $41, 43, 46, 48$ ($4$ students) - Stem $5$: $50, 50$ ($2$ students) Total number of students $= 2 + 5 + 4 + 2 = 13$. **2. Count how many students passed (score $\ge 40$):** Students with scores in stems $4$ and $5$ ($41, 43, 46, 48, 50, 50$) passed: $$\text{Number of passing students} = 4 + 2 = 6$$ Therefore, the proportion of students who passed is **$\frac{6}{13}$**. **Answer: Option B** A: **Incorrect:** The total number of students in the plot is $13$, not $11$. B: **Correct:** Out of $13$ students total, $6$ scored $40$ or above, giving a proportion of $6/13$. C: **Incorrect:** Both the numerator ($5$ instead of $6$) and denominator ($11$ instead of $13$) are wrong. D: **Incorrect:** There are $6$ students with scores $\ge 40$ ($41, 43, 46, 48, 50, 50$), not $5$.

**1. List the sorted scores after adding the missing score of $34$:** Originally there were $13$ scores. Adding $34$ gives $n = 14$ sorted scores: $$24,\; 27,\; 31,\; 31,\; 34,\; 35,\; 38,\; 39,\; 41,\; 43,\; 46,\; 48,\; 50,\; 50$$ **2. Compute the median of the $14$ scores:** For $n = 14$ (even), the median is the average of the $7^{\text{th}}$ and $8^{\text{th}}$ ordered scores: - $7^{\text{th}}$ score $= 38$ - $8^{\text{th}}$ score $= 39$ $$\text{Median} = \frac{38 + 39}{2} = 38.5$$ **Answer: 38.5**

**1. Find the total number of students $N$ from the Football row:** In Table 1, Football has $12$ students and a relative frequency of $0.15$: $$\frac{12}{N} = 0.15 \implies N = \frac{12}{0.15} = 80$$ (Check: Badminton has $18$ students, and $18 / 80 = 0.225$, which matches the table!) **2. Find the total relative frequency of Basketball ($c$) and Cricket ($d$):** Since the sum of all relative frequencies across the 5 sports must equal $1$: - Relative frequency of Athletics: $e = \frac{10}{80} = 0.125$ - Therefore: $$c + d = 1 - (0.15 + 0.225 + 0.125) = 1 - 0.50 = 0.5$$ **Answer: 0.5**

**1. Find the relative frequency of Cricket ($d$) and Basketball ($c$):** From Question 12, the total number of students is $N = 80$, and $c + d = 0.5$. - Cricket has $20$ students, so its relative frequency is: $$d = \frac{20}{80} = 0.25$$ - Therefore, the relative frequency of Basketball is: $$c = 0.5 - d = 0.5 - 0.25 = 0.25$$ (Equivalently, $a = 80 - (12 + 20 + 18 + 10) = 20$ students, so $c = 20 / 80 = 0.25$.) **Answer: Option D** A: **Incorrect:** $0.35$ would correspond to $28$ students, whereas $a = 20$. B: **Incorrect:** $0.30$ would correspond to $24$ students. C: **Incorrect:** $0.20$ would correspond to $16$ students. D: **Correct:** With $N = 80$ total students and $a = 20$ students choosing Basketball, the relative frequency is $20/80 = 0.25$.

**1. Set up the factorial equation:** The number of ways to arrange $n$ distinct objects in a line is $n!$. We are given that the number of ways to arrange $m + 2$ distinct objects is $20$ times the number of ways to arrange $m$ distinct objects: $$(m + 2)! = 20 \cdot m!$$ **2. Simplify and solve for $m$:** Divide both sides by $m!$: $$(m + 2)(m + 1) = 20$$ Since $20 = 5 \times 4$, for a non-negative integer $m$ we have: $$m + 2 = 5 \implies m = 3$$ **Answer: Option B** A: **Incorrect:** For $m = 6$, $(m+2)(m+1) = 8 \times 7 = 56 \ne 20$. B: **Correct:** For $m = 3$, $(3+2)! / 3! = 5 \times 4 = 20$. C: **Incorrect:** For $m = 2$, $(2+2)(2+1) = 4 \times 3 = 12 \ne 20$. D: **Incorrect:** For $m = 5$, $(5+2)(5+1) = 7 \times 6 = 42 \ne 20$.

**1. Find the total number of pens:** The box contains $4$ blue + $5$ black + $3$ red $= 12$ pens in total. **2. Compute the probability of drawing two blue pens without replacement:** - Probability that the first pen is blue: $\frac{4}{12} = \frac{1}{3}$ - Probability that the second pen is blue given the first was blue: $\frac{3}{11}$ - Multiply the two probabilities: $$P(\text{both blue}) = \frac{4}{12} \times \frac{3}{11} = \frac{1}{3} \times \frac{3}{11} = \frac{1}{11}$$ **Answer: Option D** A: **Incorrect:** $1/3$ is the probability that only the first pen drawn is blue. B: **Incorrect:** $1/4$ is the conditional probability on the second draw if there were 4 blue pens out of 16. C: **Incorrect:** $3/11$ is the conditional probability that the second pen is blue given the first is blue. D: **Correct:** $\frac{\binom{4}{2}}{\binom{12}{2}} = \frac{6}{66} = \frac{1}{11}$.

**1. Analyze Events A and B:** - **Event A:** The 4-member team includes **at least one engineer**. - **Event B:** The 4-member team consists **only of designers** (i.e., **zero engineers**). **2. Check each option:** - **Option A:** Since a team cannot simultaneously have *at least one engineer* and *zero engineers*, $A \cap B = \emptyset$. Thus, **Events A and B are mutually exclusive** (in fact, they are complements). (**Correct**) - **Option B:** Since $P(A) > 0$ and $P(B) > 0$, mutually exclusive events cannot be independent because $P(A \cap B) = 0 \ne P(A)P(B)$. (**Incorrect**) - **Option C:** Event B requires choosing all $4$ team members from the $4$ available designers, which can be done in $\binom{4}{4} = 1$ way. Thus, **there is only 1 favorable outcome for Event B**. (**Correct**) - **Option D:** Since Option B is false, "All of these" is false. (**Incorrect**) **Answer: Options A, C** A: **Correct:** Having at least one engineer (A) and having only designers (B) cannot occur together, so $A$ and $B$ are mutually exclusive. B: **Incorrect:** Non-empty mutually exclusive events have $P(A \cap B) = 0 \ne P(A)P(B)$, so they are dependent. C: **Correct:** Choosing 4 designers out of 4 designers can be done in $\binom{4}{4} = 1$ way. D: **Incorrect:** Option B is false.