Statistics for Data Science II · Quiz 1 · Sep 2025

**1. Identify the entry:** This is a zero-mark acknowledgement attached to the exam’s useful-data sheet, which lists probability distributions, moments and inequalities. It is not a numerical statistics problem. **2. Select the intended acknowledgement:** Choose the statement confirming that useful data has been provided above. In later questions, identify the appropriate formula and substitute the problem’s own parameters. **Answer: A — Useful Data has been mentioned above..** **Convention note:** This is an administrative response choice, not a claim that the second sentence is mathematically false. A: **Correct:** This is the acknowledgement marked as the accepted response in the paper. B: **Incorrect:** This describes the reference role of the sheet, but it is not the acknowledgement selected by the paper’s scoring key.

**1. Differentiate each nonconstant CDF branch:** For \(-1<x<0\), \(F(x)=(x+1)^2/2\), so \(f(x)=x+1\). For \(0<x<1\), \(F(x)=1-(1-x)^2/2\), so \(f(x)=1-x\), using the chain rule. **2. Check the entire density:** Outside \([-1,1]\), the CDF is constant and the density is zero. The two pieces are nonnegative and each has area \(1/2\), so the total area is one. The CDF has no jumps at \(-1,0,1\), hence no atoms. **Answer: D — D — f(x)=x+1 for -1<x<0; f(x)=1-x for 0<x<1; 0 otherwise.** A: **Incorrect:** It reverses the two derivative branches and has total area 3, rather than one. B: **Incorrect:** \(x+1\) is correct only on the negative half; on \((0,1)\) the derivative is \(1-x\). Its total integral is 2. C: **Incorrect:** It is negative on \((-1,0)\), so it cannot be a probability density. D: **Correct:** It gives \(x+1\) on the negative half and \(1-x\) on the positive half, matching both derivatives.

**1. Find the mean and variance of the sum:** Independence gives \(E[X+Y]=10+15=25\) and \(\operatorname{Var}(X+Y)=4+9=13\). **2. Convert variance into a second moment:** For any finite-variance variable \(S\), \(E[S^2]=\operatorname{Var}(S)+(E[S])^2\). With \(S=X+Y\), \[E[(X+Y)^2]=13+25^2=638.\] **3. Cross-check by expansion:** \(E[X^2]=104\), \(E[Y^2]=234\), and independence yields \(E[XY]=10(15)=150\). Thus \(104+2(150)+234=638\). **Answer: 638.**

**1. Use the complement of a minimum event:** Since both variables take values 1 and 2, \(\min(X,Y)>1\) happens only if \(X=2\) and \(Y=2\). **2. Apply the stated independence:** \(P(X=2,Y=2)=0.6(0.5)=0.3\). Consequently \(P(Z\le1)=1-0.3=0.7\), already correct to one decimal place. **Answer: 0.7.**

**1. Normalize the PMF:** The probabilities at \(1,2,3,4\) are \(0.1,a,0.4,b\). Their sum gives \(a+b=0.5\). **2. Use the specified expectation:** \(1(0.1)+2a+3(0.4)+4b=2.5\), hence \(2a+4b=1.2\). Subtract \(2(a+b)=1.0\) to obtain \(2b=0.2\), so \(b=0.1\). **3. Verify validity:** Then \(a=0.4\); all four probabilities are nonnegative, sum to one, and give mean \(0.1+0.8+1.2+0.4=2.5\). **Answer: 0.1.**

**1. Read the joint table:** In the row \(Y=1\), the probabilities at \(X=0,1,2\) are 0.12, 0.30 and 0.18. Thus \(P(Y=1)=0.60\), while \(P(X\ge1,Y=1)=0.30+0.18=0.48\). **2. Identify the conditional denominator:** \[P(X\ge1\mid Y=1)=\frac{P(X\ge1,Y=1)}{P(Y=1)}.\] Blank A is \(P(Y=1)\), which is option (5). The denominator is the probability of the condition, not the target event or the intersection. **Answer: 5.**

**1. Read the joint table:** In the row \(Y=1\), the probabilities at \(X=0,1,2\) are 0.12, 0.30 and 0.18. Thus \(P(Y=1)=0.60\), while \(P(X\ge1,Y=1)=0.30+0.18=0.48\). **2. Split the numerator into disjoint cells:** \(P(X\ge1,Y=1)=P(X=1,Y=1)+P(X=2,Y=1)\). Blank B is the first cell, \(P(X=1,Y=1)\), which is option (2). Options (1) and (3) use \(Y=0\), outside the condition. **Answer: 2.**

**1. Read the joint table:** In the row \(Y=1\), the probabilities at \(X=0,1,2\) are 0.12, 0.30 and 0.18. Thus \(P(Y=1)=0.60\), while \(P(X\ge1,Y=1)=0.30+0.18=0.48\). **2. Match the missing numerical cell:** The expression is \(0.30+C\), so \(C=P(X=2,Y=1)=0.18\). In the supplied list, 0.18 is option (8). Enter the option number 8, rather than the probability 0.18. **Answer: 8.**

**1. Read the joint table:** In the row \(Y=1\), the probabilities at \(X=0,1,2\) are 0.12, 0.30 and 0.18. Thus \(P(Y=1)=0.60\), while \(P(X\ge1,Y=1)=0.30+0.18=0.48\). **2. Match the numerator total:** Blank D is \(0.30+0.18=0.48\), the probability of the intersection \(X\ge1,Y=1\). This value is option (12). It is not yet the conditional probability because division by 0.60 still remains. **Answer: 12.**

**1. Read the joint table:** In the row \(Y=1\), the probabilities at \(X=0,1,2\) are 0.12, 0.30 and 0.18. Thus \(P(Y=1)=0.60\), while \(P(X\ge1,Y=1)=0.30+0.18=0.48\). **2. Match the full conditioning row:** Blank E is \(P(Y=1)=0.12+0.30+0.18=0.60\). This is option (9). Include \(X=0\) in this denominator even though it is excluded from the numerator. **Answer: 9.**

**1. Read the joint table:** In the row \(Y=1\), the probabilities at \(X=0,1,2\) are 0.12, 0.30 and 0.18. Thus \(P(Y=1)=0.60\), while \(P(X\ge1,Y=1)=0.30+0.18=0.48\). **2. Divide numerator by denominator:** \[P(X\ge1\mid Y=1)=\frac{0.48}{0.60}=0.80.\] Blank F is 0.80, listed as option (10). Enter 10, as the question requests the option number. **Answer: 10.**

**1. Integrate both density pieces:** The density is \(k(1+x)\) on \([0,1]\), \(k(3-x)\) on \([1,2]\), and zero elsewhere. Normalization requires \[1=k\int_0^1(1+x)\,dx+k\int_1^2(3-x)\,dx.\] **2. Solve for the scale:** Each integral is \(3/2\). Therefore \(1=3k\), giving \(k=1/3\). The two formulas agree at \(x=1\); integrating the adjacent intervals does not double-count any probability at that point. **Answer: B — 1/3.** A: **Incorrect:** The total area would be \(3(1/2)=1.5\), exceeding one. B: **Correct:** The total area is \(3(1/3)=1\), and the density is nonnegative. C: **Incorrect:** The total area would be 9, not one. D: **Incorrect:** The total area would be 6, not one.

**1. Normalize and identify the intersection:** The density has scale \(k=1/3\). Since \(1<X<2\) implies \(X>1/2\), the conditional numerator is simply \(P(1<X<2)\). **2. Evaluate the two required probabilities:** \[P(1<X<2)=\frac13\int_1^2(3-x)\,dx=\frac12.\] Also, \[P(X>1/2)=1-\frac13\int_0^{1/2}(1+x)\,dx=1-\frac5{24}=\frac{19}{24}.\] **3. Divide and round:** \(P(1<X<2\mid X>1/2)=(1/2)/(19/24)=12/19=0.631578\ldots\), so enter 0.63. **Answer: 0.63.**

**1. Convert conditional masses into joint masses:** Use \(f_{XY}(x,0)=P(X=x)P(Y=0\mid X=x)=f_X(x)(1-x/2)\). **2. Evaluate the two requested cells:** At \(x=1\), the mass is \(0.5(1-1/2)=0.25\). At \(x=2\), it is \(0.3(1-2/2)=0\). Thus \(f_{XY}(1,0)+f_{XY}(2,0)=0.25\). **Answer: D — 0.25.** A: **Incorrect:** 0.20 is the separate cell \(f_{XY}(0,0)\), which is not requested. B: **Incorrect:** 0.30 is the marginal \(P(X=2)\); its joint mass with \(Y=0\) is zero. C: **Incorrect:** 0.45 also includes the unrequested \(X=0,Y=0\) cell. D: **Correct:** The requested cells contribute 0.25 and zero.

**1. Sum over all possible defect counts:** The marginal probability is \[f_Y(0)=\sum_{x=0}^2 f_X(x)P(Y=0\mid X=x).\] The conditional mass at zero is \(1-x/2\). **2. Substitute the three weights:** \(f_Y(0)=0.2(1)+0.5(1/2)+0.3(0)=0.20+0.25=0.45\). Unlike the preceding two-cell sum, this marginal includes \(X=0\). **Answer: 0.45.**

**1. Set the threshold:** Processing time \(X\) is nonnegative, so Markov’s inequality applies. The threshold is \(1.3E[X]=1.3(500)=650\) milliseconds. **2. Apply the requested inequality:** \[P(X\ge650)\le\frac{E[X]}{650}=\frac{500}{650}=\frac{10}{13}.\] The variance is not required for this Markov bound; it is an upper bound rather than the exact probability. **Answer: B — 10/13.** **Convention note:** The question specifically requests Markov’s bound. Other inequalities using the supplied variance can produce tighter bounds; the smaller alternatives are not the result of this Markov calculation. A: **Incorrect:** \(3/13\) is the complement of the Markov upper bound, not the upper bound itself. B: **Correct:** It is the Markov ratio \(500/650=10/13\). C: **Incorrect:** \(8/13\) is not the ratio of the mean to the stated threshold. D: **Incorrect:** \(5/13\) is half the required mean-to-threshold ratio.

**1. Center the interval at the mean:** With \(\mu=500\) and \(\sigma^2=100\), the interval 480 to 520 corresponds to deviations at most 20 milliseconds. **2. Apply Chebyshev and complement:** \(P(|X-500|\ge20)\le100/20^2=1/4\), hence \(P(|X-500|<20)\ge3/4=0.75\). The same lower bound holds if endpoints are included. **3. State the requested bound:** Enter 0.75 to two decimal places. No normal-distribution assumption is needed. **Answer: 0.75.**

**1. Count the operating ovens:** Exactly one of the three ovens is off, so the daily total is the sum of two operating counts. Each count has support \(\{0,1,2,3\}\). **2. Form the support of the sum:** The minimum is \(0+0=0\) and maximum is \(3+3=6\). Every intermediate integer is attainable, and independence with positive individual masses makes each such sum have positive probability. Thus the range is \(\{0,1,2,3,4,5,6\}\). **Answer: A — {0, 1, 2, 3, 4, 5, 6}.** A: **Correct:** It contains exactly all attainable totals from two operating ovens. B: **Incorrect:** Totals 7, 8 and 9 require more than the two operating ovens can produce. C: **Incorrect:** It omits zero, which occurs when both operating ovens have zero defects. D: **Incorrect:** It omits attainable totals 0 and 1 and includes impossible totals above 6.

**1. Condition on which ovens operate:** With \(O_3\) shut down, \(Y=X_1+X_2\). The operating counts are independent with probabilities \(p(0)=0.2,p(1)=0.3,p(2)=0.3,p(3)=0.2\). **2. Enumerate sums equal to two:** The disjoint pairs are \((0,2),(1,1),(2,0)\). Thus \[P(Y=2)=0.2(0.3)+0.3(0.3)+0.3(0.2)=0.06+0.09+0.06=0.21.\] **3. Interpret the conditioning:** Do not multiply by the probability of choosing \(O_3\) for maintenance; the question already specifies that it is off. Enter 0.21. **Answer: 0.21.**

**1. Identify the geometric tail:** The PMF is \(P(X_1=k)=(3/4)(1/4)^{k-1}\), \(k\ge1\). Therefore \(X_1>3\) means \(k=4,5,\ldots\). **2. Sum the geometric series:** \[P(X_1>3)=\frac34\sum_{k=4}^{\infty}\left(\frac14\right)^{k-1}=\frac34\frac{(1/4)^3}{1-1/4}=\frac1{64}.\] **3. Read the numerator:** Comparing \(1/64=t/64\) gives \(t=1\). **Answer: 1.**

**1. Use the minimum possible values:** Both geometric variables have support \(1,2,3,\ldots\). Thus \(Z=X_1+X_2=2\) can occur only through \((X_1,X_2)=(1,1)\). **2. Multiply by independence:** \(P(X_i=1)=3/4\), so \[P(Z=2)=P(X_1=1)P(X_2=1)=\left(\frac34\right)^2=\frac9{16}.\] No other pair contributes. **Answer: B — 9/16.** A: **Incorrect:** \(3/16\) is a geometric mass at value 2 for one variable, not the required joint event. B: **Correct:** It is the product \((3/4)(3/4)\) for the unique favorable pair. C: **Incorrect:** \(3/4\) is the probability that just one variable equals 1. D: **Incorrect:** \(1/2\) does not equal the probability \(9/16\) of the unique favorable pair.