For Data Science positions, during the interview process, is common to ask questions about Statistics and Probabilities. We will provide some potential interview questions and their indicative solutions.

**Question 1: **Assuming that X follows the Normal Distribution with **Mean=0.545** and **Standard Deviation=0.155** find the probability that X exceeds 0.395.

**Answer 1:**

\(Pr(X>0.395) = 1-Pr(X \leq 0.395) = 1- Pr(\frac{X-0.545}{0.155} \leq \frac{0.395-0.545}{0.155} )=\)

\(=1-Z( \frac{0.395-0.545}{0.155})=1-Z(-0.96744) = 1-0.166587 = 0.833\)

We can provide also the solution in R.

1-pnorm(0.395, mean=0.545, sd=0.155) [1] 0.8334134

**Question 2:** The probability that a patient recovers from a rare blood disease is 0.4. If 15 patients are known to have contracted the disease, what is the probability that exactly 5 fail to recover?

**Answer 2:**

The probability to recover is 0.4 and the probability to fail is 0.6. We want to calculate the probability that exactly 5 out of 15 failed to recover. All the possible combinations to get 5 out of 15 people is: \( {15\choose 5} = 3003\).

So the probability we would like to calculate is \(Pr(X=5) = 3003 \times 0.4^{10} \times 0.6^5 = 0.024486\)

We can provide also the solution in R.

dbinom(5,15,0.6) [1] 0.02448564

**Question 3: ** A secretary makes 2 errors per page on average. What is the probability that on the next 2 pages she makes not more than 3 errors?

**Answer 3:** We can argue that the mistakes per page follow the Poisson distribution with parameter λ=2. Now we can argue that the number of mistakes of every 2 pages follow a Poisson distribution with parameter λ=4. The probability to make not more than 3 errors is:

Paragraph

\(Pr(X \leq 3) = \sum_{k=1}^{3} \frac{4^k e^{-4}}{k!}=0.43347\)

We can provide also the solution in R.

ppois(3,4) {1] 0.4334701

**Question 4: **A homeowner plants 6 flower bulbs selected at a random from a box containing 5 tulips and 4 roses. What is the probability that he planted 2 roses and 4 tulips?

**Answer 4:** The probability is given by: \(\frac{ {4 \choose 2} \times {5 \choose 4} }{9 \choose 6 }=0.3571429\)

We can provide also the solution in R.

dhyper(x=4, m=5, n=4, k=6) [1] 0.3571429

## 1 thought on “Interview questions about Stats and Probabilities”