Verisk employees use SQL for querying and manipulating insurance-related databases, such as claims data, policyholder information, and risk assessment metrics, for analytical insights. It is also used for joining multiple tables to build predictive models for risk assessment, including identifying high-risk policyholders and predicting claim frequencies, the reason why Verisk asks SQL questions in interviews for Data Analyst, Data Science, and BI jobs.
To help prep you for the Verisk SQL interview, we've curated 11 Verisk Analytics SQL interview questions in this blog.
Verisk is a leading data analytics provider serving customers in insurance, energy and specialized markets, and financial services. They help customers make better decisions using data analytics and bring efficiency and integrity in tackling risk.
For the purpose of this exercise, let's assume we want to identify power users and in this context, we label them as those who have requested more than 100 reports in a month.
reports
Example Input:report_id | user_id | request_date |
---|---|---|
1001 | 1 | 01/04/2022 00:00:00 |
1002 | 2 | 01/07/2022 00:00:00 |
1003 | 1 | 01/12/2022 00:00:00 |
1004 | 3 | 01/15/2022 00:00:00 |
1005 | 2 | 01/22/2022 00:00:00 |
1006 | 1 | 01/25/2022 00:00:00 |
1007 | 3 | 01/29/2022 00:00:00 |
1008 | 1 | 01/30/2022 00:00:00 |
1009 | 1 | 01/31/2022 00:00:00 |
SELECT user_id, TO_CHAR(request_date::date, 'YYYY-MM') as monthyear, COUNT(*) as num_reports_requested FROM reports GROUP BY user_id, monthyear HAVING COUNT(*) > 100 ORDER BY monthyear, num_reports_requested DESC;
This PostgreSQL query groups the reports
by user_id
and request_date
(in 'YYYY-MM' only) and counts the number of reports requested by each user for each month-year. The HAVING clause filters out only those users who requested more than 100 reports in a particular month. The results are then sorted in ascending order by month-year and descending order of number of reports requested.
To solve a related customer analytics question on DataLemur's free interactive SQL code editor, try this Microsoft SQL Interview problem:
Given a table of Verisk employee salary information, write a SQL query to find the 2nd highest salary among all employees.
employees
Example Input:employee_id | salary |
---|---|
1 | 2500 |
2 | 800 |
3 | 1000 |
4 | 1200 |
second_highest_salary |
---|
1200 |
Solve this interview question and run your code right in the browser:
SELECT MAX(salary) AS second_highest_salary FROM employee WHERE salary < ( SELECT MAX(salary) FROM employee );
You can find a step-by-step solution with hints here: 2nd Highest Salary.
One way to find duplicatesis to use a GROUP BY
clause and then use HAVING
to find groups
SELECT column_name, COUNT(*) as dup_count FROM table_name GROUP BY column_name HAVING dup_count > 1;
You could also use the EXISTS
operator:
SELECT * FROM table_name t1 WHERE EXISTS (SELECT 1 FROM table_name t2 WHERE t1.column_name = t2.column_name AND t1.id <> t2.id);
Verisk, a leading data analytics provider, has various departments each assigned numerous risk scores for different tasks throughout the year. As a data analyst, you have been asked to write an SQL query to calculate the average risk score by month for each department. Also, capture the department's average risk score of the previous month in the same row.
risk_score
Example Input:id | department_id | date_received | risk_score |
---|---|---|---|
1 | 01 | 2022-04-10 | 80 |
2 | 01 | 2022-04-20 | 85 |
3 | 02 | 2022-04-25 | 78 |
4 | 01 | 2022-05-10 | 82 |
5 | 02 | 2022-05-15 | 78 |
6 | 01 | 2022-05-20 | 84 |
7 | 02 | 2022-05-25 | 80 |
8 | 01 | 2022-06-10 | 82 |
9 | 02 | 2022-06-15 | 84 |
month | dept_id | average_score | previous_avg_score |
---|---|---|---|
4 | 01 | 82.5 | null |
4 | 02 | 78.0 | null |
5 | 01 | 83.0 | 82.5 |
5 | 02 | 79.0 | 78.0 |
6 | 01 | 82.0 | 83.0 |
6 | 02 | 84.0 | 79.0 |
SELECT EXTRACT(MONTH FROM date_received) AS month, department_id AS dept_id, AVG(risk_score) OVER (PARTITION BY department_id ORDER BY EXTRACT(MONTH FROM date_received) RANGE BETWEEN UNBOUNDED PRECEDING AND CURRENT ROW) AS average_score, LAG(AVG(risk_score) OVER (PARTITION BY department_id ORDER BY EXTRACT(MONTH FROM date_received) RANGE BETWEEN UNBOUNDED PRECEDING AND CURRENT ROW)) OVER (PARTITION BY department_id ORDER BY EXTRACT(MONTH FROM date_received)) AS previous_avg_score FROM risk_score ORDER BY month, dept_id;
In this query, we first use the PARTITION BY clause to group the data by department_id. Then, we calculate the average risk score for each department for each month up to the current row (month). Finally, we use the LAG function to get the previous row's average risk score. The ORDER BY clause is used to sort the results by month and department id.
To solve a related window function SQL problem on DataLemur's free online SQL code editor, try this Google SQL Interview Question:
A primary key is a column or group of columns that uniquely identifies a row in a table. For example, say you had a database of Verisk marketing campaigns data:
CREATE TABLE MarketingCampaigns ( CampaignID INTEGER PRIMARY KEY, CampaignName VARCHAR(255), StartDate DATE, EndDate DATE, Budget DECIMAL(8,2) );
In this Verisk example, the CampaignID column is the primary key of the MarketingCampaigns table. The PRIMARY KEY
constraint ensures that no two rows have the same CampaignID. This helps to maintain the integrity of the data in the table by preventing duplicate rows.
Verisk is a leading data analytics company that serves customers in insurance, energy markets, financial services, and others. Suppose you are part of the insurance department and you have been asked to analyze the insurance policy details for different customers.
You are required to find out the total insurance premiums for each customer, their number of policies and the average premium per policy for each customer.
The following are your primary tables:
customers
Example Input:customer_id | customer_name |
---|---|
101 | Alice |
102 | Bob |
103 | Charlie |
104 | David |
insurance_policy
Example Input:policy_id | customer_id | premium |
---|---|---|
201 | 101 | 1200 |
202 | 101 | 1500 |
203 | 102 | 1000 |
204 | 103 | 500 |
205 | 103 | 750 |
SELECT c.customer_name, COUNT(p.policy_id) AS number_of_policies, SUM(p.premium) AS total_premium, ROUND(AVG(p.premium), 2) AS average_premium_per_policy FROM customers c JOIN insurance_policy p ON c.customer_id = p.customer_id GROUP BY c.customer_name;
customer_name | number_of_policies | total_premium | average_premium_per_policy |
---|---|---|---|
Alice | 2 | 2700 | 1350 |
Bob | 1 | 1000 | 1000 |
Charlie | 2 | 1250 | 625 |
This query would give you the total premiums for each customer, the number of insurance policies each of them has, and the average premium per policy they pay.
COALESCE()
do, and when would you use it?The COALESCE()
function accepts an unlimited number of input arguments, and returns the first argument from that list which isn't null.If all input arguments are null, COALESCE
will return null too.
For example, suppose you had data on Verisk salespeople, and the amount of deals they closed. This data was exported from a 3rd-party system CRM, which exports a NULL
value if the salesperson didn't close any deals.
sales_person | closed_deals |
---|---|
Jason Wright | 4 |
Drew Jackson | NULL |
Chris Ho | 2 |
Adam Cohen | NULL |
Samantha Perez | 3 |
To get rid of these NULLs, and replace them with zero's (so you can do some further analytics like find the average number of closed deals), you would use the COALESCE()
function as follows:
SELECT name, COALESCE(closed_deals, 0) as closed_deals FROM verisk_salespeople;
You'd get the following output:
sales_person | closed_deals |
---|---|
Jason Wright | 4 |
Drew Jackson | 0 |
Chris Ho | 2 |
Adam Cohen | 0 |
Samantha Perez | 3 |
Verisk is a leading data analytics provider serving customers in Insurance, Energy and other Specialized Markets. They are interested in analyzing their customer behavior based on multiple factors.
For this question, Verisk wants to filter their customers based on multiple conditions:
Use the SQL language to answer this question.
customer_policy
Example Input:cust_id | policy_id | policy_type | policy_start_date | policy_end_date |
---|---|---|---|---|
1001 | 40001 | Home | 01/01/2022 | 01/01/2023 |
1001 | 40002 | Auto | 01/05/2022 | 01/05/2023 |
1001 | 40003 | Life | 01/09/2022 | 01/09/2023 |
1002 | 40004 | Auto | 01/02/2022 | 01/02/2023 |
1003 | 40005 | Auto | 01/03/2022 | 01/03/2023 |
customer_details
Example Input:cust_id | first_name | last_name | country |
---|---|---|---|
1001 | John | Doe | USA |
1002 | Jane | Doe | CANADA |
1003 | Jim | Beam | AUSTRALIA |
SELECT cd.cust_id, cd.first_name, cd.last_name, cd.country, COUNT(cp.policy_id) AS total_policies FROM customer_details cd INNER JOIN customer_policy cp ON cd.cust_id = cp.cust_id WHERE cp.policy_end_date > CURRENT_DATE + INTERVAL '30 DAY' AND (cd.country = 'USA' OR cd.country = 'CANADA') GROUP BY cd.cust_id HAVING COUNT(cp.policy_id) > 2;
This query will fetch the customer_id, first_name, last_name, country and total number of policies taken by customers who have taken more than 2 insurance policies, live in either USA or Canada and do not have any policy that's expiring within next 30 days.
As a Verisk analyst, you have been tasked with analysing an insurance company's data. The data contains various policy types and each policy is associated with a risk score. The risk score is calculated based on several factors like age of the insured, the area where they live, the type of policy and the coverage amount. Can you write a SQL query to find the average risk score for each policy type?
insurance_policies
Example Input:policy_id | insured_person_id | policy_type | coverage_amount | risk_score |
---|---|---|---|---|
171 | 123 | car | 10000 | 5.2 |
280 | 256 | home | 50000 | 3.4 |
293 | 362 | boat | 20000 | 7.1 |
352 | 192 | car | 15000 | 6.2 |
517 | 981 | home | 40000 | 4.1 |
policy_type | avg_risk_score |
---|---|
car | 5.70 |
home | 3.75 |
boat | 7.10 |
SELECT policy_type, AVG(risk_score) AS avg_risk_score FROM insurance_policies GROUP BY policy_type;
In this query, we first specify the policy_type
column, which will be included in the output table. Next, we specify the AVG
function to calculate the average of the risk_score
column and alias this as avg_risk_score
. Because we are dealing with averages that depend on groupings by policy_type
, we use the GROUP BY
clause with the policy_type
column. The FROM
clause specifies the insurance_policies
table as our data source.
This would give us the average risk score per policy type.
To practice a very similar question try this interactive Amazon Average Review Ratings Question which is similar for calculating averages based on categories or this Alibaba Compressed Mean Question which is similar for finding mean values from data.
FOREIGN KEY
constraint?A FOREIGN KEY
is a field in a table that references the PRIMARY KEY
of another table. It creates a link between the two tables and ensures that the data in the FOREIGN KEY
field is valid.
Say for example you had sales analytics data from Verisk's CRM (customer-relationship management) tool.
CREATE TABLE verisk_accounts ( account_id INTEGER PRIMARY KEY, account_name VARCHAR(255) NOT NULL, industry VARCHAR(255) NOT NULL ); CREATE TABLE opportunities ( opportunity_id INTEGER PRIMARY KEY, opportunity_name VARCHAR(255) NOT NULL, account_id INTEGER NOT NULL, FOREIGN KEY (account_id) REFERENCES verisk_accounts(account_id) );
In this example, the opportunities
table has a foreign key field called account_id
that references the "account_id" field in the verisk_accounts
table (the primary key). This helps to link the data about each opportunity to the corresponding account information in the accounts table.
This makes sure the insertion of rows in the opportunities
table that do not have corresponding entries in the verisk_accounts
table. It also helps to enforce the relationship between the two tables and ensures that data is not deleted from the verisk_accounts
table if there are still references to it in the opportunities
table.
As a data analyst at Verisk, a leading data analytics provider specializing in insurance, your task is to analyze the amount claimed by every policy holder as part of insurance claims. Given a table of insurance claims, write an SQL query to find the total sum of claims for each policy holder in the last month.
claims
Example Input:claim_id | policy_holder_id | claim_date | claim_amount |
---|---|---|---|
101 | 23 | 2022-06-08 | 15000 |
105 | 54 | 2022-06-10 | 20000 |
203 | 23 | 2022-06-18 | 5000 |
305 | 77 | 2022-07-26 | 10000 |
306 | 23 | 2022-07-05 | 12000 |
policy_holder_id | total_claimed_last_month |
---|---|
23 | 5000 |
54 | 20000 |
SELECT policy_holder_id, SUM(claim_amount) AS total_claimed_last_month FROM claims WHERE DATE_PART('month', claim_date) = DATE_PART('month', CURRENT_DATE - interval '1' month) GROUP BY policy_holder_id;
In the above SQL query, we are grouping by policy_holder_id
and summing up claim_amount
to find out the total amount claimed by each policy holder in the last month. DATE_PART
is used to get the month part of the claim date and to compare it with the last month of the current date. CURRENT_DATE - interval '1' month
gives us the date corresponding to last month.
The best way to prepare for a Verisk SQL interview is to practice, practice, practice. Beyond just solving the earlier Verisk SQL interview questions, you should also solve the 200+ SQL coding questions which come from companies like Netflix, Airbnb, and Amazon.
Each problem on DataLemur has hints to guide you, step-by-step solutions and most importantly, there's an online SQL coding environment so you can instantly run your query and have it checked.
To prep for the Verisk SQL interview it is also helpful to practice interview questions from other fintech companies like:
Discover how Verisk is leveraging Artificial Intelligence to stay ahead of emerging issues and shape the future of risk management!
However, if your SQL coding skills are weak, forget about diving straight into solving questions – improve your SQL foundations with this SQL tutorial for Data Scientists & Analysts.
This tutorial covers SQL concepts such as window functions and creating pairs via SELF-JOINs – both of which come up routinely in Verisk SQL interviews.
In addition to SQL interview questions, the other topics to practice for the Verisk Data Science Interview are:
To prepare for Verisk Data Science interviews read the book Ace the Data Science Interview because it's got:
Also focus on the behavioral interview – prepare for it with this guide on acing behavioral interviews.