Statistics for Data Science II · Quiz 2 · May 2026

This is a reference formula card rather than a scored calculation. It summarises common probability distributions, moment formulas, Markov and Chebyshev bounds, and the CLT. When using it, distinguish a distribution's rate from its scale and a sample mean's standard error from an individual observation's standard deviation. The supplied normal approximation includes P(Z>1.5)=0.07. **Answer: A.** A: **Correct:** Acknowledges the provided formulas and numerical reference values. B: **Incorrect:** The reference description is also true, but the source selects the first administrative acknowledgement rather than this alternative.

Linearity gives E(A)=3mu, E(B)=6mu and E(C)=(2+0.7+0.3)mu=3mu. Since mu>0, E(B)>E(C). Independence gives variance as sigma squared times the sum of squared coefficients: Var(A)=3sigma², Var(B)=(4+9+1)sigma²=14sigma², and Var(C)=(4+0.49+0.09)sigma²=4.58sigma². Therefore Var(A)<Var(C)<Var(B) for the intended nondegenerate case sigma²>0. If zero variance were allowed, these become equal rather than strictly ordered; the source's strict comparison assumes positive variance. **Answer: A, B, D.** A: **Correct:** Both coefficient sums are 3, giving equal expectations. B: **Correct:** 6mu exceeds 3mu because mu>0. C: **Incorrect:** The variance ordering of B and C is reversed. D: **Correct:** The squared-coefficient totals are 3,4.58,14, in increasing order when variance is positive.

Y=min(X,1) for X uniform on [-3,2]. Below -3 the CDF is zero. For -3<=y<1, Y<=y exactly when X<=y, so \(F_Y(y)=(y+3)/5\). At y=1 every possible Y is included, so the CDF jumps to 1 and stays there. The jump has size 1-4/5=1/5 because the interval 1<X<=2 is all mapped to the point Y=1. **Answer: A.** A: **Correct:** Has the correct support, uniform denominator 5 and jump at 1. B: **Incorrect:** Uses the wrong lower-end shift and interval length. C: **Incorrect:** The shift y+2 would not give zero at y=-3. D: **Incorrect:** Uses denominator 4 instead of the original interval length 5.

The joint distribution is uniform on the square 0<x<2,0<y<2, of area 4. The complement X-Y>=1 is the triangle below y=x-1 for 1<x<2. Its legs have length 1, so its area is 1/2. Hence P(X-Y<1)=1-(1/2)/4=1-1/8=7/8=0.875. Boundary inclusion does not matter for this continuous distribution. **Answer: 0.875.**

Condition on the row Y=2. Its probability is 0.15+0.25+0.05=0.45. The event X<2 includes X=0 and X=1, whose joint probabilities in that row sum to 0.40. Therefore \(P(X<2\mid Y=2)=0.40/0.45=8/9\approx0.888889\), rounding to 0.89. The denominator is the conditioning row total, not the whole table's total 1. **Answer: 0.89.**

The mean of a sample average equals the population mean: \(E[\bar X]=(1/n)\sum_iE[X_i]=(1/n)n(1200)=1200\). Blank A is therefore the expression 1200, labelled option 4. Enter the option number 4, not the value 1200 itself, because this subquestion requests a label. **Answer: 4.**

For independent bulb lifetimes, \(\operatorname{Var}(\bar X)=180^2/n\), so \(SD(\bar X)=180/\sqrt n\). The standard error decreases with square root of sample size, not linearly with n. Blank B is this expression, which is option 10. Enter 10 as the requested option number. **Answer: 10.**

Standardise the threshold using the sample-mean standard error: \[P(\bar X>1230)\approx P\left(Z>\frac{1230-1200}{180/\sqrt n}\right).\] Thus C is (1230-1200)/(180/sqrt(n)), listed as option 3. A positive deviation from the mean must give a positive z-score. **Answer: 3.**

The supplied normal table says P(Z<1.5)=0.93, hence P(Z>1.5)=0.07. Therefore D is the positive threshold 1.5, labelled option 6. A negative threshold -1.5 would give an upper-tail probability near 0.93 rather than 0.07. Enter the option label 6. **Answer: 6.**

Using the paper's supplied rounded z-table, require \((1230-1200)/(180/\sqrt n)\ge1.5\). This simplifies to sqrt(n)/6>=1.5, hence sqrt(n)>=9 and n>=81. Thus blank E is 81, and the table-based minimum is 81. This answer uses the stated approximation P(Z>1.5)=0.07; using the exact 0.93 quantile would yield a different integer threshold, so do not mix exact and supplied rounded tables midway. **Answer: 81.**

The joint uniform rectangle is [1,4]×[3,7], with area 12 and density 1/12. Integrating over y gives fX(x)=4/12=1/3 for 1<x<4. Integrating over x gives fY(y)=3/12=1/4 for 3<y<7. Thus X is uniform [1,4] and Y uniform [3,7]. The density is zero outside each stated support. **Answer: A, C, D.** A: **Correct:** X's marginal is uniform over the rectangle's x interval [1,4]. B: **Incorrect:** Y's interval begins at 3, not 4. C: **Correct:** The x-marginal density is 1/3 on its interval. D: **Correct:** The y-marginal density is 1/4 on its interval.

The event 2<X<4 and 3<Y<6 is a subrectangle of width 2 and height 3, hence area 6. Divide by the full rectangle's area (4-1)(7-3)=12. Probability=6/12=0.5. Uniformity makes area ratios valid, and open endpoints have no effect on a continuous probability. **Answer: 0.5.**

Bernoulli(0.4) gives P(X=1)=0.4 and P(X=0)=0.6. Exponential(1) has conditional density e^-y, while Exponential(2) has density 2e^-2y for y>0. The law of total probability gives \(f_Y(y)=0.6e^{-y}+0.4(2e^{-2y})=0.6e^{-y}+0.8e^{-2y}\) for y>0, and zero otherwise. **Answer: A.** A: **Correct:** Uses both correct mixture weights and the rate-2 normalisation. B: **Incorrect:** Omits the factor 2 in the rate-2 conditional density. C: **Incorrect:** Omits the mixing probabilities and does not integrate to 1. D: **Incorrect:** Reverses the probabilities assigned to the two rates.

For a continuous observation Y=1, use Bayes' density formula rather than ordinary point probabilities (both P(Y=1) terms are zero). \[P(X=1\mid Y=1)=\frac{0.4\cdot2e^{-2}}{0.6e^{-1}+0.4\cdot2e^{-2}}=\frac{0.8e^{-2}}{0.6e^{-1}+0.8e^{-2}}\approx0.329.\] Rounded to two decimal places, enter 0.33. The denominator combines both possible mixture components. **Answer: 0.33.**

Each Xi is Bernoulli(1/3), so Yi=Xi-1/3 takes -1/3 with probability 2/3 and 2/3 with probability 1/3. Its MGF is \(M_Y(t)=(2e^{-t/3}+e^{2t/3})/3\). For the average of three independent Yi, multiply the three MGFs at t/3: \[M_{\bar Y}(t)=\left(\frac{2e^{-t/9}+e^{2t/9}}3\right)^3.\] Both averaging and independence are essential: averaging rescales the argument and independence produces the cube. **Answer: B.** A: **Incorrect:** Uses the rescaled one-variable MGF but omits the product over three variables. B: **Correct:** Correctly rescales by 3 and cubes the individual MGF. C: **Incorrect:** Uses the argument for a sum rather than a mean. D: **Incorrect:** Interchanges the probabilities/centralised values in the exponential terms.

For Xi Bernoulli(1/3), \(M_X(t)=2/3+e^t/3\). The average of three independent variables has \(M_{\bar X}(t)=[M_X(t/3)]^3\). At t=9, \[M_{\bar X}(9)=\left(\frac{2+e^3}{3}\right)^3\approx398.99.\] This asks for the MGF of the uncentred X average, not the centred Y average from the preceding question. The two are related by the appropriate exponential shift factor. **Answer: 398.99.**

For each Xi, \(E[X_i^2]=\operatorname{Var}(X_i)+(E[X_i])^2=25+10^2=125\). By linearity, \[E\left[\frac{X_1^2+\cdots+X_8^2}{8}\right]=\frac{8\cdot125}{8}=125.\] This is an average of squares, not the square of the sample average; these have different expectations. **Answer: 125.**

For eight independent variables each with variance 25, the sum has variance 8×25. Scaling that sum by 1/8 scales variance by 1/64. Thus \(\operatorname{Var}(\bar X)=8(25)/64=25/8=3.125\). The population mean 10 does not enter this variance calculation. **Answer: 3.125.**

Integrate the density f(x)=x/8 over the requested interval: \[P(2<X<4)=\int_2^4\frac{x}{8}\,dx=\left[\frac{x^2}{16}\right]_2^4=\frac{16-4}{16}=0.75.\] Using the density at one point instead of its area would not give an interval probability. **Answer: 0.75.**

The expectation of a continuous variable is \(E[X]=\int xf_X(x)dx\). Here \[E[X]=\int_0^4x\frac{x}{8}\,dx=\left[\frac{x^3}{24}\right]_0^4=\frac{64}{24}=\frac83\approx2.666667.\] Rounded to two decimals, daily expected rainfall is 2.67 cm. Weight the density by x before integrating; integrating the density alone gives total probability 1. **Answer: 2.67.**