Statistics for Data Science II · Quiz 2 · Jan 2026

**1. Identify the item:** This zero-mark item supplies a reference formula sheet rather than a separate calculation. The tables list probability functions, means and variances for standard discrete and continuous distributions. **2. Use it correctly:** In a Normal distribution written Normal(mu,sigma squared), the second parameter is variance; the standard deviation is its square root. For independent variables, the variance of a weighted sum is the sum of squared weights times variances. These distinctions matter in the later questions. **3. Interpret the choices:** The acknowledgement that useful data has been supplied is the keyed response to this reference item; it is not an extra numerical problem. **Answer: A.** A: **Correct:** This acknowledges the supplied formula and useful-data reference. B: **Incorrect:** The source keys the acknowledgement in A for this zero-mark reference item.

**1. Use the rectangular support:** Both x and y independently range from 0 to 17; there is no restriction coupling them. **2. Compute the marginals:** \[f_X(x)=\int_0^{17}\frac{4xy}{17^4}\,dy=\frac{2x}{17^2},\qquad f_Y(y)=\frac{2y}{17^2}.\] **3. Compare product and joint density:** Their product is 4xy/17 to the fourth power, exactly the joint density throughout the rectangular support, and both are zero outside the corresponding supports. Therefore X and Y are independent. **Answer: B.** A: **Incorrect:** The computed marginals do factor the joint density on its rectangular support. B: **Correct:** The joint density equals the product of the two marginal densities.

**1. Define the weighted sum:** Let S=3X1+X2+2X3+2X4. Each Xi has mean 1 and variance 18, and the variables are independent. A linear combination of independent normal variables is exactly normal. **2. Find mean and variance:** \[E[S]=3+1+2+2=8,\qquad\operatorname{Var}(S)=18(3^2+1^2+2^2+2^2)=324.\] The standard deviation is 18, not 324. **3. Standardize the event:** \[P(S<12)=P\left(Z<\frac{12-8}{18}\right)=F_Z(2/9).\] Since normal variables are continuous, using less than or less than or equal to makes no difference. **Answer: B.** A: **Incorrect:** This is the complementary upper-tail probability rather than the requested lower tail. B: **Correct:** The standardized threshold is (12-8)/18=2/9. C: **Incorrect:** The threshold is 2/9, not 1/9. D: **Incorrect:** This has both the wrong threshold and the wrong tail.

**1. Invert the transformation:** Y=3X+1 implies X=(Y-1)/3. At y=4, x=1. **2. Include the Jacobian:** For a monotone change of variable, \[f_Y(y)=f_X((y-1)/3)\left|\frac{dx}{dy}\right|=\frac23e^{-2(y-1)/3},\quad y>1.\] The factor 1/3 is essential because a density changes when the horizontal scale changes. **3. Evaluate:** \(f_Y(4)=(2/3)e^{-2}\approx0.0902235\), giving 0.09 to two decimal places. This is a density value, not the point probability P(Y=4), which is zero. **Answer: 0.09.**

**1. List the support pairs:** With X,Y in {1,2,3}, the event X+Y=4 consists of (1,3), (2,2) and (3,1). **2. Add their joint masses:** \[P(X+Y=4)=\frac{1\cdot3}{36}+\frac{2\cdot2}{36}+\frac{3\cdot1}{36}=\frac{10}{36}=\frac5{18}.\] **3. Round:** 5/18 is approximately 0.2777778, so the required two-decimal answer is 0.28. No independence assumption is needed because the joint mass function is already given. **Answer: 0.28.**

**1. Expand the MGF around zero:** \[M_X(t)=\frac{e^t-1}{t}=1+\frac t2+\frac{t^2}{6}+\frac{t^3}{24}+\cdots.\] The apparent singularity at zero is removable; the limiting value is 1. **2. Read the first two moments:** In an MGF, the coefficient of t is E[X] and the coefficient of t squared is E[X squared]/2. Hence E[X]=1/2 and E[X squared]=1/3. **3. Compute variance:** \(\operatorname{Var}(X)=1/3-(1/2)^2=1/12=0.083333\ldots\), giving 0.083 to three decimal places. This MGF also identifies a Uniform(0,1) variable. **Answer: 0.083.**

**1. Recall uniform density over a region:** If a joint distribution is uniform over a finite region R, its constant density must integrate to 1. Thus fXY=1/Area(R). **2. Match the blank:** In the supplied solution, fXY=1/A, so A must stand for Area of R. The question asks for the listed option number, not the area value. Area of R is option 3. **Answer: 3.**

**1. Identify the region:** x squared+y squared<9 with x>0,y>0 describes the first-quadrant quarter of a disk of radius 3. **2. Compute its area:** \(\operatorname{Area}(R)=\pi(3)^2/4=9\pi/4\). **3. Match the requested option:** Blank B is the area value. In the provided list, 9pi/4 is option 1, so enter 1 rather than 9pi/4. **Answer: 1.**

**1. Find the upper boundary:** The circular boundary is x squared+y squared=9. For a fixed x between 0 and 3, its positive y-coordinate is sqrt(9-x squared). **2. Match C:** The displayed upper limit is sqrt(C-x squared), so C=9. The choice labelled 9 appears as option number 9 in the supplied list; the required entry is therefore 9. **Answer: 9.**

**1. Form the X marginal:** To retain x and remove y, integrate the joint density over the permitted y-values: \[f_X(x)=\int_0^{\sqrt{9-x^2}} f_{XY}(x,y)\,dy,\qquad0<x<3.\] **2. Match the integrand and differential:** Blank D must be fXY(x,y) dy. Integrating with dx would remove the wrong variable. This expression is option 7. **Answer: 7.**

**1. Use the constant joint density:** Since the quarter-disk area is 9pi/4, its density is 4/(9pi). **2. Integrate over y:** \[f_X(x)=\int_0^{\sqrt{9-x^2}}\frac4{9\pi}\,dy=\frac{4\sqrt{9-x^2}}{9\pi},\quad0<x<3.\] **3. Match the coefficient:** In the supplied form E sqrt(C-x squared)/(9pi), E=4. The number 4 is listed as option 10, so the requested answer is 10. **Answer: 10.**

**1. Locate the event:** X<1,Y<2 within the first quadrant gives the rectangle 0<x<1,0<y<2. Its furthest corner from the origin has squared distance 1 squared+2 squared=5<9, so the entire rectangle lies inside the quarter-disk support. **2. Divide areas:** A uniform planar distribution assigns probabilities in proportion to area. The rectangle has area 2 and the support has area 9pi/4. Hence \[P(X<1,Y<2)=\frac{2}{9\pi/4}=\frac8{9\pi}\approx0.282942.\] **3. Round:** The two-decimal answer is 0.28. **Answer: 0.28.**

**1. Normalize the density:** A valid PDF must integrate to 1 over its support. Therefore \[1=k\int_0^1(x^2-x+1/2)\,dx=k\left[\frac{x^3}{3}-\frac{x^2}{2}+\frac x2\right]_0^1.\] **2. Evaluate:** The bracket is 1/3-1/2+1/2=1/3. Thus k/3=1 and k=3. **3. Check nonnegativity:** x squared-x+1/2=(x-1/2) squared+1/4 is positive, so this normalization produces a nonnegative density. **Answer: 3.**

**1. Use the normalized density:** The previous normalization gives fX(x)=3(x squared-x+1/2) on [0,1]. **2. Integrate to one-half:** \[P(X\leq1/2)=3\left[\frac{x^3}{3}-\frac{x^2}{2}+\frac x2\right]_0^{1/2}=3\left(\frac1{24}-\frac18+\frac14\right)=\frac12.\] **3. Explain the symmetry:** The polynomial is unchanged by replacing x with 1-x, so exactly half the probability is on either side of 1/2. The answer to one decimal place is 0.5. **Answer: 0.5.**

**1. Define the statistic:** The sample mean is the random variable \(\overline X_n=(X_1+\cdots+X_n)/n\). Different random samples generally produce different means. **2. Compute its moments:** For independent identically distributed observations, \[E[\overline X_n]=\mu,\qquad\operatorname{Var}(\overline X_n)=\frac{n\sigma^2}{n^2}=\frac{\sigma^2}{n}.\] **3. Distinguish parameter and statistic:** In this ordinary sampling model, mu is the fixed population mean, whereas the sample mean is random. Increasing n reduces its variance and makes it concentrate around mu. **Answer: B, C, D.** A: **Incorrect:** The variance is sigma squared divided by n, not sigma squared independently of n. B: **Correct:** The realized mean depends on the random sample selected. C: **Correct:** The variance decreases as 1/n, reducing sampling dispersion. D: **Correct:** The population expectation is a fixed parameter of the given distribution.

**1. Apply Chebyshev to the sample mean:** Its variance is sigma squared/n=1/n. With deviation threshold 0.2, \[P(|\overline X_n-\mu|\geq0.2)\leq\frac{1/n}{(0.2)^2}=\frac{25}{n}.\] **2. Enforce the requested guarantee:** Require 25/n<=0.05. Multiplying by positive n and dividing by 0.05 gives n>=500. **3. Choose the smallest integer:** n=500 makes the bound exactly 0.05. This is the smallest sample size guaranteed by Chebyshev, not necessarily the smallest possible for a specifically known distribution. **Answer: 500.**

**1. Use a mixture density:** Let H mean healthy and M mean mild metabolic disorder. The population proportions are P(H)=0.6 and P(M)=0.4. The marginal density is 0.6 fX|H+0.4 fX|M. **2. Evaluate the two normal densities:** For healthy participants, mu=50 and variance=36, so at x=50 the density is 1/(6 sqrt(2pi)). For the disorder group, mu=60 and variance=25, so the density is e to the power -2 divided by 5 sqrt(2pi). **3. Apply the weights:** \[f_X(50)=\frac1{\sqrt{2\pi}}\left(\frac{0.6}{6}+\frac{0.4}{5}e^{-2}\right)=\frac1{\sqrt{2\pi}}\left(\frac1{10}+\frac2{25}e^{-2}\right).\] A density value is not a point probability for a continuous variable. **Answer: D.** A: **Incorrect:** This omits the population mixing weights 0.6 and 0.4. B: **Incorrect:** The exponential factor is attached to the wrong component. C: **Incorrect:** This also attaches the exponential factor to the healthy component and omits weights. D: **Correct:** The normal-density values are correctly weighted by the two population proportions.

**1. Use density-based Bayes' rule:** For an observed continuous measurement x=50, \[P(H\mid X=50)=\frac{P(H)f_{X|H}(50)}{f_X(50)}.\] This is interpreted through the conditional density model, not by dividing zero point probabilities. **2. Substitute the component values:** The common 1/sqrt(2pi) factor cancels, giving \[\frac{1/10}{1/10+(2/25)e^{-2}}=\frac1{10}\left(\frac1{10}+\frac2{25}e^{-2}\right)^{-1}\approx0.9023.\] **3. Interpret:** A reading at the healthy group's mean is much more likely under that group, so the posterior probability is higher than its prior 0.6. **Answer: A.** A: **Correct:** This correctly normalizes the weighted healthy density by the full mixture density. B: **Incorrect:** The numerator/normalizing structure does not match the correct Bayes expression. C: **Incorrect:** This gives the mixture coefficient itself rather than the conditional healthy probability. D: **Incorrect:** This does not divide the healthy contribution by the marginal density.

**1. Separate observations and their average:** Individual bulb lifetimes have mean 800 and standard deviation 40. The sample contains n=100 observations. **2. Compute the sampling scale:** \[E[\overline X]=800,\qquad\operatorname{SD}(\overline X)=\frac{40}{\sqrt{100}}=4.\] **3. Apply the Central Limit Theorem:** For a large independent sample with finite variance, the sample mean is approximately normal even if individual lifetimes are not normally distributed. Hence the intended approximation is Normal(800,16), with standard deviation 4. **Answer: A.** A: **Correct:** The mean stays 800 and the standard deviation is reduced to 4. B: **Incorrect:** This uses the individual standard deviation instead of the sample mean's standard error. C: **Incorrect:** The CLT does not give a uniform sampling distribution. D: **Incorrect:** The original observations need not themselves be normal for the CLT approximation.

**1. Use the sample-mean approximation:** The CLT gives mean 800 and standard error 40/sqrt(100)=4. **2. Standardize 808:** \[P(\overline X>808)\approx P\left(Z>\frac{808-800}{4}\right)=P(Z>2).\] **3. Take the upper tail:** The supplied table gives FZ(2)=0.9772. Therefore the probability is approximately 1-0.9772=0.0228, or 0.02 to two decimal places. The question asks for exceeds, so the upper tail rather than the CDF is required. **Answer: 0.02.**