Data Scientist Interview Questions

Lucas G. - "I would conduct a sample z-test because we have enough samples and the population variance is known. H1: average monthly spending per user is $50 H0: average monthly spending per user is greater $50 One-sample z-test x_bar = $85 mu = $50 s = $20 n = 100 x_bar - mu / (s / sqrt(n) = 17.5 17.5 is the z-score that we will need to associate with its corresponding p-value. However, the z-score is very high, so the p-value will be very close to zero, which is much less than the standa"See full answer

Ruth A. - "Spoiled food In a process I improved, I streamlined how tasks were assigned to reduce delays and confusion."See full answer

Rajeev K. - "I’ve spent over 6 years building and scaling e-commerce products across EMEA and APAC. At Jumia, I led product initiatives on the checkout and payments side. For example, I launched gamified promotions on PDP and checkout that improved engagement and delivered a 2.3x uplift in conversion. I also introduced automated installment payments and order cancellation flows, which not only improved user trust but also reduced complaints by 30% and lowered operational costs. Before that, at Lazada, I work"See full answer

Anonymous Eagle - "I generate insights through stakeholder requirements and the data I have in hand"See full answer

Nick S. - "[I'm not sure whether the answer below is the best, as I have not gotten result and feedback from my interview] Ans: I would solve by first using a VAE-style model, to create a latent space embedding that translates user description to generate images. Training would be done on the 1000 avatar images and 100000 descriptions, following this scheme: VAE: description -> encoder -> latent space -> decoder -> image Q: "OK, but that means you're limiting the generated images to be only the 1000 imag"See full answer

Amparo L. - "The only time I felt unhappy at work was when I didn’t get to pick the job I want to do at the jobsite."See full answer

Himani E. - "The cases where data is under heavy outlier influence. Since mean fluctuates due to the presence of an outlier, median might be a better measure"See full answer

Chetak C. - "Probability that one of the coupons is used = 1 - Probability that no coupon is used = 1 - nC0 p^0 * (1-p)^n = 1 -(1-p)^n"See full answer

Yuexiang Y. - "A Random Forest works by building an ensemble of decision trees, each trained on a slightly different version of the data. The key mechanism is bagging: for each tree, we sample the training data with replacement (bootstrapping), so every tree sees a different subset of examples. On top of that, at each split the algorithm randomly selects a subset of features, so trees explore different predictors. These two sources of randomness decorrelate the trees. When we aggregate them — by averag"See full answer

Vaibhav D. - "Make current as root. 2 while current is not null, if p and q are less than current, go left. If p and q are greater than current, go right. else return current. return null"See full answer

Anmol R. - "The example given is wrong. The 2nd test case should have answer 3, as we can get to 6 by using 3 coins of denomination 2."See full answer

Saurabh J. - "The algorithm calculates certain metrics like entropy & Gini Impurity. The goal of the decision tree algorithm is to find the most optimal value for these metrics, lowest values for Gini Impurity & Entropy. Once it converges on the minima, it creates a split & grows the branches."See full answer

Chen J. - "Leetcode 347: Heap + Hashtable Follow up question: create heap with the length of K instead of N (more time complexity but less space )"See full answer

Sai R. - "def countuniqueoutfits(totalpants: int, uniquepants: int, totalshirts: int, uniqueshirts: int, totalhats: int, uniquehats: int) -> int: """ Number of unique outfits can simply be defined by (uniquepantschoose1uniqueshirtschoose1uniquehatschoose_1) (uniquepantschoose1*uniqueshirtschoose1) # Not wearing a hat nchoosek is n """ res = (uniquepants*uniqueshirtsuniquehats) + (uniquepantsunique_shirts) return res print(countuniqueoutfits(2, 1, 1, 1, 3, 2))"See full answer