Data Researcher Interview Questions
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Data Researcher interview questions shared by candidates
Are there any areas listed in the job description that you are less skilled at or inexperienced in?
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What are your future goals?
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Want to grow like Tree .
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1 weakness My weakness is that I see a person in pain and I cannot help him, and my strength is when I feel a better person on my way, whatever of his color, religion or gender. I only care that he is a human being. it’s to see some one Less
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My future ambition (goals) is to be a heart surgeon for children and to gain many experiences that qualify me for human service Less
Ten people are bidding on a stock at 90, while 100 people are offering to sell it at 91. What price is the next trade?
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If someone wants to sell at $91 and the only bid offers are at $90..it looks like there is no next trade until someone either moves their bid price or asking price. Less
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10 people want to buy at 90 and 100 people want to sell at 91... that's not enough information for a precise answer. Are these "people" all bidding equal amount of shares? 10 people bidding for blocks of 100 shares and 100 people asking to sell at 91 but each selling 1 share. What's the tick increment? If the bid-ask spread was 90-91 then the trade is whoever crosses the spread first. It doesn't matter if one side's limit orders outnumbers the other. There is no right answer. I'm pretty sure this question is gauging how well you understand how the order book works. Trade on Level 2 (Depth of Market) and you'll see how things work. Less
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I think Ryan may be correct on this, but here's a simpler way to look at it. There are fewer buyers than sellers, so no trades are happening. Because there are more sellers, I would expect a seller to drop his price to that of the closest buyer. So, the seller would drop his price to 90 where both parties would agree on price. The next trade would be at 90 dollars. Less
Why you want to work here?
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Because this is big platform to start my career and I am very interested to work in this company please give me a chance to show my talent Less
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A Dynamic environment along with the experts that is what every aspirant wants to have and sunpharma is the place to experience it. Less
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Wants to famous sun pharma in terms of reputation
You play rock, paper, scissors with an opponent, but your opponent cannot play rock. What should you play to maximize your expected profit if every time you win you win $1, lose you lose $1, draw you win $0?
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You made a mistake in calculating the expected profit. It would be Profit=P=a+b-3ab. We want to maximize this in a and minimize in b. Now for a fixed a, the minimum expected profit you can get is min(a,1-2a). Moreover, for a fixed b, the maximum profit would be max(b,1-2b). So player two should choose b=1/3 to minimize the max of the profit. Player one should also choose a=1/3 to maximize the min of the profit. Less
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If you play rock with probability 'a' and scissors with probability (1-a), expect your opponent to play paper with probability b and scissors with probability (1-b), then our expected profit is: p = a*((1-b) - b) + (1-a)*(b) = a - 3ab + 1. To maximize profit, take derivative w.r.t. a, set to 0 dp/da = 1 - 3b = 0 ==> b = 1/3 So you expect your opponent to play paper with 1/3 probability, you should throw rock with only 1/3 probability. Thus, play scissors 2/3 of the time, rock 1/3 of the time. This makes sense, you draw most of the time, but then other times you maximize your chance of winning. Less
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For player two, expected payout would be E = (1-a)b+a(1-b)-(1-a)(1-b) = a(2-3b)+2b-1, note that player two will lose money if he plays rock while player one plays paper. So clearly, we need to eliminate the impact from a, which gives b = 2/3. Then no matter what a is, we can have E = 1/3. Less
Suppose you want to gamble in Vegas. In a game, you win $x if the number is prime and lose $x/2 if composite. The number is uniformly randomly generated by a machine between 1 and 10 inclusive. Will you play this game? Follow up: What if you can play n number of times and then stop. Will you play it?
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I think everybody is wrong here. 1 is neither prime or composite, so when 1 turns out I assume nothing happens and effectively you has a space of size 9: 4 primes, 5 composites Less
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Question is a bit ambiguous, but I believe x is the roll you get, not the amount you bet. Therefore there are four primes (2,3,5,7) and six composite(1,4,6,8,9,10). Probability of rolling a prime is 0.4, composite is 0.6. The expected winnings when rolling a prime is (2+3+5+7)/4 = 17/4. The expected loss when rolling composite is (1+4+6+8+9+10)/6 * (1/2) = 38/12 = 19/6. Therefore the total expected winning per roll is 0.4*(17/4) - 0.6*(19/6) = 1.7 - 1.9 = -0.2. You are expected to lose 0.2$ each bet. Less
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Above answers are inaccurate. 1 is not considered a prime number. Primes - 2, 3, 5, 7 Not primes - 1, 4, 6, 8, 9, 10 (Question says inclusive) So 4 primes, 6 composites. Expected return for 1 attempt = 0.4*x - 0.6*x/2 = 0.1x > 0 So, I will play 1 round. Follow up - I would think yes. But I have to ask the interviewer more questions to be sure. Less
Minimize the number of comparisons for finding minimum and maximum of a given set of numbers.
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By the same logic it is n-1 when only maximum or minimum has to be found out from n nos., the correct answer will be n/2 + (n/2-1)+(n/2-1) = 3n/2-2 Less
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Given n numbers, the optimal number of comparison should be 3n/2 Pairwise compare all numbers = n/2 comparisons It is easy to see that the max, min lies among the n/2 elements which are greater, smaller respectively. Pairwise compare the n/2 greater elements = n/4 comparisons. Again pairwise compare the n/4 greater elements obtained in previous step = n/8 comparions. ... You get max(min) in n/4+n/8+...1 = n/2 comparisons. initial comparisons = n/2 comparisons to get max number = n/2 comparisons to get min number = n/2 Total = 3n/2 Less
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Here is matlab code according to the suggested algorithm : clear; clc; vec=[3,8,7,2,1,6,5,4]; [max12,min12]=max_min(vec(1),vec(2)); [max34,min34]=max_min(vec(3),vec(4)); [max56,min56]=max_min(vec(5),vec(6)); [max78,min78]=max_min(vec(7),vec(8)); [max1234,~]=max_min(max12,max34); [max5678,~]=max_min(max56,max78); [max12345678,~]=max_min(max1234,max5678); fprintf('max=%d\n',max12345678); % Find Min [~,min1234]=max_min(min12,min34); [~,min5678]=max_min(min56,min78); [~,min12345678]=max_min(min1234,min5678); fprintf('min=%d\n',min12345678); Less
Recent Economic Activity - Throw some light on the facts Stock Exchange based questions Accounting principles and terms Terms related to financial statements
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The process takes time, can be from 7 - 30 days. You can always write to them in case it's been longer than this. Wish you luck! Less
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HI...have they directly informed you on the spot that you have been selected or the HR has caleed you back after some days?? Less
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They update later not at the spot, generally they drop you an e-mail to inform you - When did you have your last round ? Less
Flip the coins, the expected time to get HHT
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Let x_i be the probability a random infinite binary sequence has its first HHT ending at position i. Clearly x_i =0 for i=1,2 and x_i=1/8 for i=3,4,5 and we can deduce the recurrence x_i=1/8(1-x_{i-3}-...-x_1) for i>=3. Since the sum of all x_i is 1 we can write x_i = 1/8(x_{i-2}+x_{i-1}+x_i+...) so x_3+x_4+x_5+... = 1/8 (x_1+2x_2+3x_3+...) so multiplying both sides by 8 we get that the expected number of flips is 8. Less
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This is incorrect. You cannot bracket like this. Condition on the first 2 steps and you will see that the expected number of steps to get HHT is 14. Less
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Here is another approach to this problem using conditional expectation. Let X be the minimum number of flips needed to get the first run of HHT. Let HH denote the event that the first two flips land on heads and similarly define TH, HT, TT. Then using a well known theorem, we can write E[X] as E[X] = E[X | HT] P(HT) + E[X | TT] P(TT) + E[X | HH] P(HH) + E[X | TH] P(TH) where E[X | Z] is the conditional expectation of random variable X with respect to random variable Z and P is the probability measure. It is easy to see that E[X | HT] = E[X | TT] = E[X] + 2, and P(HH) = P(HT) = P(TT) = P(HT) = 0.25. Using the above theorem for E[X | HH], E[X | TH] and a little observation, we obtain E[X | HH] = 4 and E[X | TH] = 0.5 * E[X] + 4 The first equation then implies E[X] = 8. Less
1) Tow coins, P(head)=1/3, P(tail)=2/3, design a way to get the effect of fair coin
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I guess Play 2 games , TH or HT = outcome 1, TT = outcome 2 . Both of probability 4/9 disregard HH Less
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We need unbiased decision out of a biased coin. Throw the coin twice. Classify it as "heads" if we get HT and "tails" if we get TH. Disregard the other two occurrences i.e. HH and TT. Less
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Swift and anon are both correct, but Swift's solution is twice as efficient because 8/9 of the time, Swift only requires 2 flips, while 4/9 of the time, anon requires only two flips. Indeed, for Swift, we can show that the expected number of flips is 2.25, while for anon, the expected number of flips is double that, 4.5. Let X be the expected number of flips. Then, for Swift, EX = 2 + 1/9*EX ==> EX = 18/8 = 2.25, while for anon, EX = 2 + 5/9*EX ==> EX = 18/4=4.5. Less