Data Analysts & Data Scientists at Penn Mutual uses SQL to analyze policyholder data patterns, such as identifying high-risk segments and tracking policyholder behavior, for risk assessment. It is also used to query databases for insightful reports, including claims analysis and policyholder profiling, to aid the decision-making process in premium determination. For this reason, Penn Mutual asks prospective hires SQL coding interview questions.
As such, to help you practice for the Penn Mutual SQL interview, we've collected 10 Penn Mutual Life Insurance SQL interview questions – scroll down to start solving them!
As a life insurance company, Penn Mutual considers its power users to be individuals who buy their policies frequently. Your task is to write a SQL query to identify customers who have bought policies more than three times in the last six months.
Provide:
Let's create a hypothetical table and table for the problem.
policy_id | customer_id | purchase_date | policy_type |
---|---|---|---|
101 | 4 | 02/11/2022 | Term Life |
102 | 3 | 03/12/2022 | Whole Life |
103 | 4 | 04/17/2022 | Variable Universal Life |
104 | 5 | 01/29/2022 | Universal Life |
105 | 4 | 06/06/2022 | Term Life |
customer_id | full_name |
---|---|
1 | John Doe |
2 | Jane Smith |
3 | Peter Griffin |
4 | James Patrick |
5 | Mickey Mouse |
This SQL query joins customers and policies tables based on customer_id, counts the number of policies each customer has purchased in the last six months, and checks if it is greater than 3. If it is greater than 3, then it returns the id, name of the customers, the total count of their policies, and their purchase dates.
To solve a related customer analytics question on DataLemur's free online SQL coding environment, try this Microsoft SQL Interview problem:
Imagine there was a table of Penn Mutual employee salaries, along with which department they belonged to. Write a query to compare the average salary of employees in each department to the company's average salary. Return the comparison result as 'higher', 'lower', or 'same' for each department. Display the department ID, payment month (in MM-YYYY format), and the comparison.
Code your solution to this interview question and run your code right in DataLemur's online SQL environment:
The answer is LONG – 30+ lines of SQL. You can find a detailed solution with hints here: Department Salaries.
A non-correlated sub-query is not linked to the outer query and can be run on its own. It does not reference any columns from the outer query and is used to retrieve additional data for the outer query. On the other hand, a correlated sub-query is one that relies on the outer query and cannot be run independently. It uses the outer query to filter or transform data by referencing a column from it, while the outer query uses the results of the inner query.
Here is an example of a non-correlated sub-query:
The sub-query in this case is non-correlated, as it does not reference any columns from the outer query.
Here is an example of a correlated sub-query:
This query selects the and total sales of all Penn Mutual customers in the sales table whose are greater than the average of their own customer group. The sub-query in this case is correlated with the outer query, as it references the column from the outer query ().
Non-correlated sub-queries are typically faster, as they only have to be executed once, wheareas correlated sub-queries are slower since they have to be re-evaluated for each row of the outer query.
Using Penn Mutual's sales database, calculate the total premium amount per month per insurance policy for each insurance policy type using SQL. For simplification, assume that the policy_type and amount columns are not null.
sale_id | policy_id | sale_date | policy_type | amount |
---|---|---|---|---|
1001 | 2001 | 01/17/2022 | Term Life | 500 |
1002 | 2002 | 01/18/2022 | Whole Life | 1000 |
1003 | 2003 | 02/04/2022 | Term Life | 800 |
1004 | 2004 | 02/08/2022 | Whole Life | 1500 |
1005 | 2005 | 02/15/2022 | Term Life | 700 |
month | policy_type | total_amount |
---|---|---|
1 | Term Life | 500 |
1 | Whole Life | 1000 |
2 | Term Life | 1500 |
2 | Whole Life | 1500 |
In this query, we first extract the month from the sale date using the DATE_TRUNC function. Next, we use a window function, specifying 'policy_type' and 'month' as the partitioning criteria in order to generate a sum of 'amount' for each distinct pair of 'policy_type' and 'month'. The GROUP BY clause is then used to ensure the results are arranged by 'month' and 'policy_type'. Finally, the results are ordered by 'month' and 'policy_type' to get the desired output.
To practice a similar window function interview problem which uses RANK() on DataLemur's free interactive coding environment, try this Amazon SQL question asked in a BI Engineer interview:
Denormalization is the process of modifying a database schema in a way that deviates from the typical rules of normalization (1NF, 2NF, 3NF, etc.).
Denormalization is often used to improve the performance of a database, particularly when it is being used for reporting and analytical purposes (rather than in an Online Transaction Processing (OLTP) manager).
By duplicating data, denormalization can reduce the number of expensive joins required to retrieve data, which can improve query performance. However, denormalization can also cause problems such as increased data redundancy and the need for more complex update and delete operations.
Penn Mutual is a life insurance company. They offer different types of life insurances. They have three types of clients: Individual, Business, and Trust. A client can have multiple policies, and each policy can belong to one of the three categories. The categories are: Term Life, Whole Life, and Universal Life.
The company wants to know the total face amount per client type and per policy type in a specific year.
client_id | client_type |
---|---|
1 | Individual |
2 | Business |
3 | Trust |
policy_id | policy_type |
---|---|
1 | Term Life |
2 | Whole Life |
3 | Universal Life |
client_id | policy_id |
---|---|
1 | 1 |
2 | 3 |
3 | 2 |
1 | 2 |
2 | 1 |
3 | 3 |
client_id | policy_id | face_amount | issue_date |
---|---|---|---|
1 | 1 | 50000 | 2020-01-01 |
2 | 3 | 100000 | 2020-05-01 |
3 | 2 | 150000 | 2020-07-01 |
1 | 2 | 200000 | 2020-11-01 |
2 | 1 | 250000 | 2020-12-01 |
3 | 3 | 300000 | 2021-02-01 |
Given these tables, write a PostgreSQL query to find the total face amount per client type and per policy type in the year of 2020.
The above SQL query joins the four tables on their respective keys and filters the rows where the policy issue date is in the year of 2020. It then groups the data by client type and policy type to calculate the total face amount for each combination of client type and policy type.
Both window functions are used to find rows at a given offset from the current row. However, will give you the rows AFTER the current row you. On the other hand, will give you the rows BEFORE the current row.
Penn Mutual is interested to analyze active life insurance policies. They want to filter down customer records of those who have life insurance policies and are affiliated with the branch located in Pennsylvania. They also want to differentiate the records based on whether the customer has a single policy or more than one.
The table has the following structure:
customer_id | name | branch | policy_count | status |
---|---|---|---|---|
101 | John Doe | Pennsylvania | 3 | Active |
202 | Jane Smith | New Jersey | 1 | Inactive |
303 | Doe Alen | California | 2 | Active |
404 | Mike Jr | Pennsylvania | 1 | Active |
505 | Alice Test | Florida | 2 | Inactive |
Please write a PostgreSQL query that returns all customer names, branches, and policy count for those customers who have "Active" status, are located at a Pennsylvania branch, and have more than one policy.
The above query first filters 'Active' status customers from the 'status' column using WHERE 'status' clause. It then filters customers from the Pennsylvania branch using the AND clause with condition 'branch = Pennsylvania'. Finally, it filters customers having more than one policy using the AND clause with condition 'policy_count > 1'. The result will return names, branches, and policy counts of the filtered customers.
Penn Mutual, a major life insurance company, leverages a database system to record and manage its insurance policy and client information. As a data analyst working at Penn Mutual, find out the average premium amount for life insurance policies bought by their clients.
For this question, assume Penn Mutual's database has a table named with the following structure and data:
policy_id | client_id | issue_date | insurance_type | premium |
---|---|---|---|---|
2198 | 521 | 05/04/2022 00:00:00 | Life Insurance | 400 |
3297 | 361 | 04/16/2022 00:00:00 | Life Insurance | 550 |
2085 | 492 | 06/12/2022 00:00:00 | Life Insurance | 380 |
4571 | 256 | 07/01/2022 00:00:00 | Life Insurance | 450 |
3915 | 688 | 08/08/2022 00:00:00 | Life Insurance | 500 |
Your task is to write a SQL query that would retrieve average premium amount of life insurance policies.
The following PostgreSQL query accomplishes the task:
This query calculates the average premium of all life insurance policies. It filters the policies table to only consider rows where the insurance_type is 'Life Insurance' using the WHERE clause, and then uses the AVG function to find the average of the premium column from this filtered data.
To practice a very similar question try this interactive Amazon Average Review Ratings Question which is similar for averaging and grouping data or this Alibaba Compressed Mean Question which is similar for < calculating a mean.
Here's an example of a clustered index on the column of a table of Penn Mutual payments table:
This creates a clustered index on the column, which will determine the physical order of the data rows in the table.
Here is an example of a non-clustered index on the column of the same table:
This will create a non-clustered index on the column, which will not affect the physical order of the data rows in the table.
In terms of query performance, a clustered index is usually faster for searches that return a large number of records, while a non-clustered index is faster for searches that return a small number of records. However, updates to a clustered index are slower, as they require the data rows to be physically rearranged, while updates to a non-clustered index are faster, as they only require the index data structure to be updated.
The best way to prepare for a SQL interview, besides making sure you have strong SQL fundamentals, is to practice a ton of real SQL questions that were asked in recent job interviews. Besides solving the earlier Penn Mutual SQL interview questions, you should also solve the 200+ SQL coding questions which come from companies like Microsoft, Google, and Meta.
Each DataLemur SQL question has hints to guide you, detailed solutions and best of all, there is an online SQL coding environment so you can instantly run your query and have it checked.
To prep for the Penn Mutual SQL interview it is also helpful to practice SQL problems from other insurance companies like:
Dive into the latest news and updates from Penn Mutual and discover how they're shaping the future of the insurance industry!
However, if your SQL foundations are weak, forget about going right into solving questions – refresh your SQL knowledge with this SQL interview tutorial.
This tutorial covers things like CASE/WHEN/ELSE statements and manipulating string/text data – both of these show up routinely during SQL job interviews at Penn Mutual.
Besides SQL interview questions, the other types of problems covered in the Penn Mutual Data Science Interview are:
To prepare for Penn Mutual Data Science interviews read the book Ace the Data Science Interview because it's got:
Don't ignore the behavioral interview – prep for it using this list of behavioral interview questions for Data Scientists.