Quantitative Researcher Interview Questions in Dublin, Ireland | Glassdoor

# Quantitative Researcher Interview Questions in Dublin, Ireland

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Quantitative researcher interview questions shared by candidates

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### Quantitative Research Associate at Susquehanna International Group (SIG) was asked...

Jun 2, 2013
 toss 100 coins, what is the probability of getting more than 60 heads?7 Answersuse normal dist to estimate binary dist, but notice the tailbinomial distribution (100 60) (1/2)^60 (1/2)^40100 C 60 (0.5)^100 will get one prob of 60 heads, not 60 and above. Interview candidate is correct. You will most likely have to use normal dist. to estimate the binom dist given time constraints. A google search can inform one how to do thisShow More Responsessimply use CLTdon't have latex installed, so sorry in advance. answer is: sum[i=60 to i=100] of (100 choose i)/(2^100)As pointed out, use a normal approximation for the sum. Define an RV called X. X = 0 for tails. X = 1 for heads. We have: E[X] = 1/2, Var[X] = 1/4. Now denote S = sum(X), where the sum runs over 100 realizations. So: E[S] = 50, Var[S] = 25. (The event are uncorrelated, so their covariance is zero, hence we can just sum variances here.) Then P(S >= 60) corresponds to P ( (S - 50)/sqrt(25) >= (60 - 50) / sqrt(25)). Note that we just subtracted the mean and divided by the standard deviation. This renders the term at the LHS a standard normal variable. So we need: P(N >= 2), where N is standard normal. If no computer is allowed, one can remember that the 97.5% percentile lies at 1.96. (If you often compute two-sided 95% confidence intervals, then you might know this number.) Or remember that within 2 sigma of the mean, 95 percent of the realizations lie. Since the distribution is symmetric, you will come to the same conclusion. So we estimate: P(S > 60) = 2.5%. In summary: make two approximations: one for going from Binomial to Normal, and a second for estimating the probability.Using the normal approximation to a binomial distribution... Find the z value (i.e. z=x-u/st) u=mean u=np=100x0.5=50 st=standard deviation st=sq(npq)=sq(100x0.5x0.5)=5 x= 60-0.5=59.5 (i.e. continuity correction) z=59.5-50/5=1.9 Therefore P(x≥ 60)=P(z≥ 1.9)= 1-P(z≤1.9)= 1- 0.9713=0.0287 Answer: 2.87 % chance of getting more than 60 heads

### Quantitative Research Associate at Susquehanna International Group (SIG) was asked...

Jul 26, 2013
 None really.1 AnswerI think if they get back to you only if you're selected for the next stage.

### Quantitative Researcher at Susquehanna International Group (SIG) was asked...

Jun 14, 2017
 How many ways are there to place number 1 to 6 on a dice ?4 Answers30180 C(6,2)* C(4,2)* C(2,2) *26!=720Show More Responses150

### Quantitative Researcher at Susquehanna International Group (SIG) was asked...

Feb 8, 2016
 Expected pay out of two dice games.Be the first to answer this question

### Quantitative Researcher at Susquehanna International Group (SIG) was asked...

Feb 19, 2017
 What is the expected payoff of a game in which you win whatever a die roll shows? What if you can re-roll the die?2 Answers3.5, 4.25http://math.stackexchange.com/questions/179534/the-expected-payoff-of-a-dice-game
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