Qube Research & Technologies Interview Questions 2026

For a continuous \(U\sim\operatorname{Uniform}[0,1]\), \(P(U\ge0.5)\) equals the length of \([0.5,1]\), namely 0.5. Under the dial, the qualifying settings are exactly 0.5,0.6,0.7,0.8,0.9,1.0: six out of eleven, so \(P(\text{dial}\ge0.5)=\frac6{11}\).

With a strict >0.5 rule, the dial's 0.5 position no longer qualifies. Five settings remain, so \(P(\text{dial}>0.5)=\frac5{11}\). The continuous probability stays at \(\frac12\) because the single point \(U=0.5\) has measure zero. This distinction arises from positive probability mass at the midpoint in the discrete device, not from any change in the threshold value itself. Both dial counts use the supplied rule that all eleven labelled positions, including both endpoints, are equally likely.

Let \(G\) mean M is globally top-ranked, and \(S\) mean M is ranked first among the randomly selected shortlist of four. Since all overall rankings are equally likely, \(P(G)=1/12\). Whenever M is globally best, it must also be best within any subset containing it, so \(P(S\mid G)=1\).

By exchangeability of the strict random ordering, each of the four sampled applications is equally likely to rank first within that shortlist. Thus \(P(S)=1/4\). Bayes' rule gives
\[ P(G\mid S)=\frac{P(S\mid G)P(G)}{P(S)} =\frac{1\cdot(1/12)}{1/4}=\boxed{\frac13}. \]

For a counting cross-check, condition on M having overall rank \(r\), for \(r=1,\ldots,12\). All three sampled opponents must be worse than M, which occurs with probability
\[ P(S\mid \operatorname{rank}(M)=r)=\frac{\binom{12-r}{3}}{\binom{11}{3}}, \]
where \(\binom{k}{3}=0\) for \(k<3\). Averaging over the twelve equally likely ranks gives \(P(S)=1/4\), as above; at rank 1 the numerator is \(\binom{11}{3}\), so its contribution to the joint event \(G\cap S\) is \(1/12\).

Winning this small comparison changes the posterior from \(1/12\) to \(1/3\). It does not reveal an absolute score or imply M must be the globally best; the eight unexamined applications can include a better one.

Let \(X\) denote the compensation claim. The requested lower 80th percentile is defined as

\[q_{0.8}=\inf\{r:\Pr(X\le r)\ge0.8\}.\]

The cumulative probabilities at the possible claim amounts are \(\Pr(X\le0)=0.15\), \(\Pr(X\le100)=0.15+0.30=0.45\), \(\Pr(X\le300)=0.45+0.40=0.85\) and \(\Pr(X\le500)=1\).

Every reserve \(r<300\) can cover at most the $0 and $100 claims, or 45% of claims. Raising the reserve to exactly $300 covers 85%, meeting the 80% rule. Therefore the smallest qualifying reserve is $300. A reserve slightly below $300 would still cover only 45%, so this threshold is genuinely minimal.

Linear interpolation of a cumulative probability between $100 and $300 would implicitly treat part of the 40% probability at exactly $300 as if it were spread continuously over intermediate dollar amounts. It is not: the entire 40% atom becomes covered together when \(r\) reaches $300. Under the explicitly defined \(\inf\{r:F(r)\ge0.8\}\) quantile, interpolation is mathematically inappropriate. Other software quantile conventions can interpolate sample ranks, but they answer a different definition and do not determine the smallest reserve achieving 80% coverage here.

Write \(p_{t}\) for the fraction of visits of duration t, and \(q_{t}\) for the fraction of occupied time spent on that duration category. Under uniform sampling of busy instants,

\[q_{t}=\frac{t p_{t}}{\mathbb{E}[T]}\],

because each visit of duration t contributes t minutes to the observation pool. Inverting gives \(p_{t}=\frac{q_{t}/t}{\sum_{s}(q_{s}/s)}\). With observed \(q=(\frac{1}{4},\frac{1}{2},\frac{1}{4})\) at \(t=(10,20,40)\), the unnormalized visit weights are

\(\frac{\frac{1}{4}}{10}=\frac{1}{40}\), \(\frac{\frac{1}{2}}{20}=\frac{1}{40}\), \(\frac{\frac{1}{4}}{40}=\frac{1}{160}\).

Their sum is \(\frac{1}{40}+\frac{1}{40}+\frac{1}{160}=\frac{9}{160}\). Therefore the event-level visit probabilities are \((\frac{4}{9},\frac{4}{9},\frac{1}{9})\). The ordinary mean appointment length is

\[\mathbb{E}[T]=10\times\frac{4}{9}+20\times\frac{4}{9}+40\times\frac{1}{9}=\frac{40+80+40}{9}=\frac{160}{9}\approx17.7778\] minutes.

Averaging the cross-sectional proportions as though they were booking fractions instead gives \(10\times\frac{1}{4}+20\times\frac{1}{2}+40\times\frac{1}{4}=22.5\) minutes. That is the expected duration of the appointment seen at a random busy instant, which overweights long visits and is not the target.

To verify the inversion, weight the reconstructed visit probabilities forward by duration: \(10(\frac{4}{9})=\frac{40}{9}\), \(20(\frac{4}{9})=\frac{80}{9}\), \(40(\frac{1}{9})=\frac{40}{9}\). Divide these occupied-time contributions by their sum \(\frac{160}{9}\) to recover \((\frac{1}{4},\frac{1}{2},\frac{1}{4})\) exactly. Positive duration categories and the stable-occupancy assumption are essential; empty-time observations were expressly excluded.

Set \(H_0:\mu_R\ge\mu_S\) against \(H_1:\mu_R<\mu_S\), and calculate using the boundary \(\mu_R-\mu_S=0\) under the null. Because samples and individual jobs are independent and the population variances are known, the difference in sample means has variance

\(\operatorname{Var}(\bar X_R-\bar X_S)=16/100+144/9=0.16+16=16.16\) hours squared.

The observed difference is \(9-13=-4\) hours, so the correct standard error is \(\sqrt{16.16}\approx4.01995\) hours and \(z=-4/4.01995\approx\boxed{-0.995}\). This is above the one-sided 5% cutoff \(-1.645\), so we do not reject the null. The specialist group's contribution of \(16\) to the variance dwarfs the routine group's \(0.16\) despite its smaller sample.

The colleague's equal-variance pooling would instead produce \(s_p^2=(99\cdot16+8\cdot144)/107=2736/107\approx25.5701\) and \(SE_{\rm pooled}=\sqrt{(2736/107)(1/100+1/9)}\approx1.75978\). The corresponding statistic is \(z_{\rm pooled}=-4/1.75978\approx\boxed{-2.273}\), incorrectly crossing \(-1.645\). Pooling weights the low-variance team's data heavily, then wrongly assigns that averaged variance to the small, noisy specialist team. The known heterogeneous variances explicitly rule out this shortcut. A non-rejection is not proof the population means are equal.

The population regression slope on reported mileage is

\[ \beta_{W}=\frac{\operatorname{Cov}(W,Y)}{\operatorname{Var}(W)}. \]

Substitute \(W=X+U\), \(Y=200+20X+\varepsilon\). By the stated independence/zero-covariance assumptions,

\[ \operatorname{Cov}(W,Y)=\operatorname{Cov}(X+U,20X+\varepsilon)=20\operatorname{Var}(X)=20(16)=320, \]

while

\[ \operatorname{Var}(W)=\operatorname{Var}(X)+\operatorname{Var}(U)=16+9=25. \]

Thus the slope fitted on the erroneous covariate is \(320/25=\boxed{12.8}\) dollars per 1,000 kilometres. In general, classical additive predictor noise multiplies a true slope \(\beta\) by \(\lambda=\operatorname{Var}(X)/[\operatorname{Var}(X)+\operatorname{Var}(U)]\). Here \(\lambda=16/25=0.64\), so the corrected structural slope is \(12.8/(16/25)=\boxed{20}\) dollars per 1,000 kilometres.

This is a population statement; a sample estimate still has sampling error. If instead mileage \(X\) were observed exactly and only the response contained an additional mean-zero noise term independent of \(X\), the slope numerator \(\operatorname{Cov}(X,Y)\) would retain \(20\operatorname{Var}(X)\) and the denominator would remain \(\operatorname{Var}(X)\), so the population slope would stay 20. Response noise increases uncertainty about individual costs, but unlike predictor error it does not inflate the regression's explanatory-variable variance.

The admissible predictor at each historical decision date may depend on \(X\), but not on the future random variable \(Z\). Since \(Y=X+Z\) and \(X,Z\) are independent and centred,
\[ E[Y\mid X]=X,\qquad E[(Y-X)^2]=E[Z^2]=\boxed{1}. \]
This is the minimum mean-square error among all predictors measurable from \(X\), not merely linear predictors, because conditional expectation minimises squared error.

The revised indicator is \(X^*=X+Z/2\). Its variance is \(1+1/4=5/4\), the outcome has variance \(\mathrm{Var}(Y)=2\), and their covariance is
\[ \mathrm{Cov}(Y,X^*)=\mathrm{Cov}(X+Z,X+Z/2)=1+1/2=3/2. \]
Because these variables are jointly normal, the conditional mean on revised data is linear:
\[ E[Y\mid X^*]=\frac{3/2}{5/4}X^*=\boxed{\frac65X^*}. \]
Its minimum MSE is the conditional variance,
\[ \mathrm{Var}(Y)-\frac{\mathrm{Cov}(Y,X^*)^2}{\mathrm{Var}(X^*)} =2-\frac{(3/2)^2}{5/4}=2-\frac95=\boxed{\frac15}. \]
The revised history appears to reduce MSE by \(1-1/5=\boxed{4/5}\), or 80% relative to the live attainable risk. But \(X^*=X+Z/2\) was only constructed after part of the future outcome noise \(Z\) became known. It is not in the original information set and cannot be used as a live forecast input. Even applying the revised coefficient to unrevised \(X\) would give MSE \(\mathrm{Var}(Z-X/5)=1+1/25=26/25\), worse than the legitimate risk of 1. The observed retrospective gain is an information-leakage artefact, not executable forecasting skill.