Statistics for Data Science I · End Term · May 2026 · FN

Degree titles have an order (Bachelor's, Master's, PhD), so they are ordinal. State names label categories without a numerical ranking, so they are nominal. Celsius temperatures have meaningful differences but an arbitrary zero, so they are interval. Height has a meaningful zero and meaningful ratios, so it is ratio data. Thus P–2, Q–3, R–4, S–1. **Answer: B.** A: **Incorrect:** Misclassifies degree titles, states, Celsius temperatures and heights. B: **Correct:** All four classifications are correct. C: **Incorrect:** Reverses the interval and ratio classifications of Celsius and height. D: **Incorrect:** Misclassifies degree titles, states and Celsius temperatures.

Read stem as tens and leaf as units. The ordered ages are 18,19,20,21,23,24,31,32,34,35,37,40,42,43,46,51,54. There are 17 observations, so the median is the \((17+1)/2=9\)th value. The 9th age is 34. A stem-and-leaf plot already presents the data in order when the leaves are sorted. **Answer: 34.**

The 17 ordered ages are 18,19,20,21,23,24,31,32,34,35,37,40,42,43,46,51,54. Using the course's median-of-halves convention, exclude the overall median 34. The lower eight have median \(Q_1=(21+23)/2=22\); the upper eight have median \(Q_3=(40+42)/2=41\). Thus \(IQR=Q_3-Q_1=41-22=19\) years. Different software quantile conventions can differ, so the convention matters. **Answer: 19.**

Total monthly salary is \(5(4000)+2(18000)=56000\), so the mean is \(56000/7=8000\). In sorted order the first five salaries are 4000 and the last two are 18000. With seven values, the median is the fourth, 4000. The most frequent salary is also 4000, making it the mode. High supervisor salaries pull the mean above the median. **Answer: A.** A: **Correct:** Correctly gives mean 8000, median 4000 and mode 4000. B: **Incorrect:** The fourth ordered salary is 4000, not 18000. C: **Incorrect:** The mean is 8000 and the most frequent salary is 4000. D: **Incorrect:** 18000 occurs only twice; 4000 occurs five times and is the mode.

Split into disjoint cases. Exactly two women: choose them in \(\binom32=3\) ways. Choose two men from six, excluding the one forbidden pair: \(\binom62-1=14\). This gives 42 panels. Exactly three women: all three are chosen and any one of six men is allowed, giving \(\binom33\binom61=6\). Total \(42+6=48\). Four women is impossible because only three are available. **Answer: B.** A: **Incorrect:** Does not equal the valid case total 42+6. B: **Correct:** 48 counts both permitted gender compositions and excludes the forbidden pair. C: **Incorrect:** Does not result from the required case count. D: **Incorrect:** Overcounts the panels allowed by the two constraints.

First recover X's PMF from CDF jumps: \(P(X=0)=0.15\), \(P(X=1)=0.25-0.15=0.10\), \(P(X=2)=0.55-0.25=0.30\), \(P(X=3)=0.78-0.55=0.23\), \(P(X=4)=1-0.78=0.22\). The question asks for \(Y=X(X-1)\). Its values are 0,0,2,6,12. Combine both preimages of zero: \(P(Y=0)=0.15+0.10=0.25\). The remaining masses are 0.30 at 2, 0.23 at 6, and 0.22 at 12. They sum to 1. **Answer: D.** A: **Incorrect:** Uses squared values and fails to combine the two preimages of Y=0. B: **Incorrect:** Assigns zero probability to Y=0 and omits positive mass. C: **Incorrect:** Uses CDF levels as masses and the wrong support. D: **Correct:** Has the correct support 0,2,6,12 and masses 0.25,0.30,0.23,0.22.

An exponential lifetime with mean 8 has rate \(\lambda=1/8\). The survival probability is \(P(T>t)=e^{-t/8}\). Therefore \[P(5<T<12)=P(T>5)-P(T\ge12)=e^{-5/8}-e^{-12/8}\approx0.3121.\] Since the distribution is continuous, including or excluding either endpoint does not change the probability. **Answer: B.** A: **Incorrect:** Equals one minus the desired interval probability. B: **Correct:** Subtracts the two survival probabilities in the correct order. C: **Incorrect:** Uses the wrong signs and can produce a negative value. D: **Incorrect:** Subtracts both tails without applying the correct CDF difference.

Perfect negative correlation means every point lies on one decreasing straight line. From (2,14) and (4,10), the slope is \((10-14)/(4-2)=-2\). The line is \(y=18-2x\), which also gives 6 at x=6 and 2 at x=8. At x=10 it gives \(k=18-20=-2\). **Answer: -2.**

Let A be clearing quiz 1 and B clearing quiz 2. Inclusion-exclusion says \(P(A\cup B)=P(A)+P(B)-P(A\cap B)\). Rearranging gives \(P(B)=0.53-0.38+0.25=0.40\). The overlap must be added back because it was subtracted in the union formula. To one decimal place, enter 0.4. **Answer: 0.4.**

Let P mean part-time, F full-time and L attending all live sessions. Then \(P(P)=0.4\), \(P(F)=0.6\), \(P(L\mid P)=0.2\), \(P(L\mid F)=0.5\). Total attendance probability is \(P(L)=0.4(0.2)+0.6(0.5)=0.38\). Bayes' rule gives \[P(P\mid L)=\frac{P(L\mid P)P(P)}{P(L)}=\frac{0.08}{0.38}\approx0.210526.\] Round to 0.21. The denominator must include attendees from both groups. **Answer: 0.21.**

The CDF jumps from 0.20 just before 1 to 0.50 at 1. A discrete point mass equals its jump: \(P(X=1)=F(1)-F(1^-)=0.50-0.20=0.30\). Read the filled point as the value at the endpoint. To one decimal place, enter 0.3. **Answer: 0.3.**

At most 2 means \(X\le2\), exactly what the CDF measures: \(F(2)=P(X\le2)\). The filled point at x=2 is at 0.80, while the open point at 0.50 belongs to the previous interval. Thus the probability is 0.8. **Answer: 0.8.**

For \(X\sim\mathrm{Binomial}(n,p)\), the mean is np and variance is np(1-p). Dividing variance by mean gives \(1-p=1.26/1.8=0.7\), so p=0.3. Then \(n=1.8/0.3=6\). Thus the model has six trials, corresponding here to six balls bowled. **Answer: 6.**

From mean np=1.8 and variance np(1-p)=1.26, obtain p=0.3 and n=6. At least one wicket is the complement of zero wickets: \[P(X\ge1)=1-P(X=0)=1-(1-p)^n=1-0.7^6=0.882351.\] Correct to two decimal places, this is 0.88. Using the complement avoids summing six binomial terms. **Answer: 0.88.**

The number of white balls has a hypergeometric distribution. Its expectation is sample size times the white proportion: \(E[X]=nK/N=2(9/15)=1.2\). Equivalently, write X as the sum of indicators for a white ball on each draw; each draw has marginal probability 9/15 of white, so linearity gives 9/15+9/15. The draws need not be independent for this expectation calculation. **Answer: 1.20.**

Variance is \(\sigma^2=E[X^2]-\mu^2\), so \(E[X^2]=\sigma^2+\mu^2\). Standard deviation is the square root of variance and therefore has the same units as X and its mean. For independent random variables, cross-covariances vanish, giving \(\operatorname{Var}(\sum_iX_i)=\sum_i\operatorname{Var}(X_i)\), assuming the variances exist. **Answer: B, D.** A: **Incorrect:** The displayed second-moment identity is wrong; it should be sigma squared plus mu squared. B: **Correct:** Taking the square root of variance restores the original units. C: **Incorrect:** The mean and standard deviation both have X's units. D: **Correct:** Independence makes the covariance terms zero, so the variance sum identity holds.