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Intelligence analyst Interview Questions in United States

Intelligence analyst Interview Questions in United States

Intelligence analyst Interview Questions in United States

An intelligence analyst evaluates data and information to identify security risks and mitigate them for various organizations. Some intelligence analysts work for government agencies, but the field is broad and spans across various industries. When interviewing for a position as an intelligence analyst, you may face questions about the tools you use to identify risks.

4,676 Intelligence Analyst interview questions shared by candidates

Top Intelligence Analyst Interview Questions & How to Answer

Here are three top intelligence analyst interview questions and how to answer them.

Question #1: Describe your method for identifying and mitigating security risks.

How to answer: An interviewer uses this question to understand your methods and processes for managing important tasks. Use the STAR method to describe a specific situation that demonstrates how you analyze data to identify potential security risks for an organization and take action.

Question #2: How do you track the data you use to make decisions and share information with others on the team?

How to answer: Data analysis is part of the role of an intelligence analyst, but this individual also needs to be able to track the data they use and disseminate it across the intelligence team. When answering this question, talk about the software or tools you have experience with that allow for accurate data analysis and management.

Question #3: What do you think is the greatest responsibility of an intelligence analyst?

How to answer: This question allows you to share your perspective on the role of an intelligence analyst and their responsibilities to an organization. Your perspective might outline the importance of mitigating security risks or describe the responsibility of protecting a group of people.

Top Interview Questions

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Wayfair
Business Intelligence Analyst was asked...November 21, 2014

50,000 shoppers with a 0.5% conversion rate for a chair that costs $250. Wayfair makes a 27% profit. Next, 50,000 shoppers will get a 10% discount. What is the conversion rate they must achieve to achieve the same profits as before?

13 Answers

This is incorrect Old revenue = 50000*250*.005=250*250 New revenue with conversion rate r% = 50000*r*250*.9 = old revenue = 250*250 r = .55% Less

July 9 is wrong because the profit margin changes(As sale price changes, but the cost doesn't change) Less

The new conversion rate is 0.944, Profit from case 2 = $16,875 and the third part, if this scenario actually occurred I would give a 10% discount (with the new conversion rate .944%) because the overall profit margin would remain same i.e. $16,875 Less

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Accenture

Describe the different parts of an SQL statement

2 Answers

SELECT for which columns to include in the result, (tables must be present in the FROM clause), can also include aggregates like COUNT(), SUM(), AVG() FROM for which tables to select rows/attributes from WHERE which is used to filter rows based on a given criteria with comparisons (=, !=, , and keywords LIKE and BETWEEN), INNER JOIN and LEFT/RIGHT OUTER JOIN to combine data across multiple tables ORDER BY DESC/ASC to specify ordering of a specific column name GROUP BY to group data based on the tables passed HAVING to filter rows based on an aggregate used in the SELECT statement that cannot be used with the WHERE clause Less

I described a simple statement: Select (Name all columns from tables or use * for all) From (table name(s)) Where (any conditions, join statements ect) Group By / Order By to sort or aggregate the data Less

Universal Health Services, Inc.

What is a clustered/non clustered index.

2 Answers

I explained how a clustered index was in sequential physical order on disk and a non clustered index used pointers or a hash table for index lookup. I started to explain the advantages/disadvantages of each index type but the two guys interviewing me just seemed so confused and lost. I think I would have been better of simply saying "it makes lookup faster". Less

I explained how a clustered index was in sequential physical order on disk and a non clustered index used pointers or a hash table for index lookup. I started to explain the advantages/disadvantages of each index type and how I would use them. Less

World Wide Technology

Whats the difference between inner join and outer join?

2 Answers

Inner Join returns values where the key between the two tables are the same, and values are present in both tables. Outer Join returns the Values from both tables, based on the key, even if there is not any data the joining table. If not value is available, then NULL is returned for that specific Row Data based on the Key Less

Specified in the WHERE clause, joins simply combine data from multiple tables in the result. INNER JOIN, the most common, returns the rows for which the given ON condition is satisfied for both tables. LEFT/RIGHT OUTER JOIN statements return all the rows from the specified table regardless if there is a match in the unspecified table, with the matching rows specified in the ON condition in the unspecified table. Less

IBM

What reporting tools have you used? Actuate, Brio, Crystal Reports? How much experience have you had in scheduling jobs on Unix? Can you write XML and implement it?

1 Answers

The interviewer was interested in breath of experience rather than specific technical skill. Unix work with scheduling and XML skills are necessary, but not heavy (the same procedures are done over and over again). Emphasis was on selling yourself through experience. Less

Nestlé USA

Explain/define "Weighted Average".

1 Answers

An average in which each quantity to be averaged is assigned a weight. These weightings determine the relative importance of each quantity on the average. Weightings are the equivalent of having that many like items with the same value involved in the average. Less

Baylor Scott & White Health

The typical "difficult" question that comes up in Business Intelligence is (and I'm paraphrasing), "What's the best data warehouse solution?".

1 Answers

It's a great question but analogous to asking, "What's the best house or the best car?". The answer depends on several factors. I'd start with requirements and budget and discuss what is possible within those parameters. If you watch "Property Virgins" on HGTV, the educational process is similar, as are the "buyers" perceptions. Less

Amazon

Basic SQL questions. Describe a join to a non-technical person. How do you handle a query that does not perform quickly? They want to know that you can use 'explain plans', which I currently do not use (I'm still entry level). Select all customers who purchased at least two items on two separate days. Given a table with a combination of flight paths, how would you identify unique flights if you don't care which city is the destination or arrival location.

11 Answers

Customer problem: select customerId from orders group by customerId having count(distinct date(orderDate)) > 1; -- Assuming the orderDate has time associated with it. Flights problem: select arrival, departure from flights union select departure, arrival from flights; Less

select distinct(b.id) from ( select a.id, a.d, a.#items, row_number() over (partition by id order by d) as rn from ( select id, d, sum(q) as #items from cust group by id, d having sum(q) >=2 ) a)b where b.rn>2 Less

CREATE TABLE test_flights ( origin VARCHAR(255), destination VARCHAR(255) ); INSERT INTO test_flights (origin, destination) VALUES ('Boston', 'Los Angeles'), ('Los Angeles', 'Boston'), ('New York', 'Pittsburgh'), ('Pittsburgh', 'New York') SELECT * FROM test_flights WHERE origin < destination Less

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Amazon

probability of the product coming from location A is 0.8 and from location B is 0.6. What is the probability the customers will receive the product from location A or location B SQL - tested on different join , lead , lag, pivoting in sql , sub query, group by having, where, aggregate and think about how you would find outliers)

10 Answers

probability of the product coming from location A is 0.8 and from location, B is 0.6. What is the probability the customers will receive the product from location A or location B P(A)=0.8 P(B)=0.6 Assuming the events are independent: P(A OR B) = 1 - P(not A AND not B) = 1-(0.2*0.4) = 1-0.08 = 0.92 The other ways: P(A or B) = P(A) + P(B) - P(A AND B) = 0.8 + 0.6 - (0.8*0.6) = 1.4 - 0.48 = 0.92 OR P(A or B) = P(A) + P(B )*P(not A) = 0.8 + (0.6*0.2) = 0.8 + 0.12 = 0.92 OR P(A OR B) = P(B) + P(A)*P(not B) = 0.6 + (0.8*0.4) = 0.6 + 0.32= 0.92. Less

helpful

P(A)=0.8 P(A')=1-0.8= 0.2 P(B)= 0.6 P(B')= 0.40 P(A or B)= P(A)P(B')+P(B)P(A') It means from location A and not from B, OR, from location B and not from A Ans: 0.8*0.4+0.6*0.2 => 0.44 Less

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Amazon

Derive customer's account status as of month end for all the months in 2019. If for given month, there are more than one rows, pick the data from the latest date within the month. If for given month, there is no data, pick the data from latest date prior to the month You can use last_day function to get month ending date(Eg: last_day(01/01/2015) = 01/31/2015) customer_id event_date status credit_limit 1 1/1/2019 C 1000 1 1/5/2019 F 1000 1 3/10/2019 1000 1 3/10/2019 1000 1 8/27/2019 L 1000 2 1/1/2019 L 2000 2 1/5/2019 2500 2 3/10/2019 2500 3 1/1/2019 S 5000 3 1/5/2019 6000 3 3/10/2019 B 5000 4 3/10/2019 B 10000

9 Answers

select customer_id, last_day(a.event_date) month, coalesce( a.status,b.latest_status_ever) latest_status_per_month from table a join( -- latest if no record select customer_id, last_day(event_date) month, status as latest_status_ever from table QUALIFY ROW_NUMBER() OVER (PARTITION BY customer_id ORDER BY event_date desc) = 1 group by 1,2 )b on a.customer_id = b.customer_id and last_day(a.event_date) > b.month QUALIFY ROW_NUMBER() OVER (PARTITION BY customer_id, month(event_date) ORDER BY event_date desc) = 1 group by 1,2 Less

Try this - Assumptions - The first record for a customer begins in the month it first had a status. Data is restricted to 2019. End month for all customers will be Dec 2019. All the status provided are valid status even including NULL status. with temp_data as ( SELECT 1 as customer_id, '1/1/2019'::date as event_date, 'C' as status, 1000 as credit_limit union all SELECT 1, '1/5/2019', 'F', 1000 union all SELECT 1, '3/10/2019','', 1000 union all SELECT 1, '3/10/2019','', 1000 union all SELECT 1, '8/27/2019', 'L', 1000 union all SELECT 2, '1/1/2019', 'L', 2000 union all SELECT 2, '1/5/2019', '', 2500 union all SELECT 2, '3/10/2019', '', 2500 union all SELECT 3, '1/1/2019', 'S', 5000 union all SELECT 3, '1/5/2019', '', 6000 union all SELECT 3, '3/10/2019', 'B', 5000 union all SELECT 4, '3/10/2019', 'B', 10000 ) ,tab2 as ( select customer_id ,last_day(event_date) as event_month ,status ,credit_limit ,dense_rank() over (partition by customer_id, event_month order by event_date desc ) as status_rank from temp_data group by customer_id,event_date,status,credit_limit ) ,cal as ( select cal_month_end_date from calendar_table where cal_year_id = 2019 group by 1 ) ,tab3 as ( select t.* ,event_month as start_month ,coalesce(lead(add_months(event_month,-1)) over(partition by customer_id order by event_month),year_end) as end_month from tab2 t ,(select max(cal_month_end_date) as year_end from cal) cal_end_month where status_rank = 1 ) select customer_id, cal_month_end_date, status, credit_limit from tab3, cal where cal_month_end_date between start_month and end_month group by 1,2,3,4 order by 1,2 Less

Assumptions - The first record for a customer begins in the month it first had a status. Data is restricted to 2019 End month for all customers will be Dec 2019 All Values in Status column is a valid status including NULLs with temp_data as ( SELECT 1 as customer_id, '1/1/2019'::date as event_date, 'C' as status, 1000 as credit_limit union all SELECT 1, '1/5/2019', 'F', 1000 union all SELECT 1, '3/10/2019','', 1000 union all SELECT 1, '3/10/2019','', 1000 union all SELECT 1, '8/27/2019', 'L', 1000 union all SELECT 2, '1/1/2019', 'L', 2000 union all SELECT 2, '1/5/2019', '', 2500 union all SELECT 2, '3/10/2019', '', 2500 union all SELECT 3, '1/1/2019', 'S', 5000 union all SELECT 3, '1/5/2019', '', 6000 union all SELECT 3, '3/10/2019', 'B', 5000 union all SELECT 4, '3/10/2019', 'B', 10000 ) ,tab2 as ( select customer_id ,last_day(event_date) as event_month ,status ,credit_limit ,dense_rank() over (partition by customer_id, event_month order by event_date desc ) as status_rank from temp_data group by customer_id,event_date,status,credit_limit ) ,cal as ( select cal_month_end_date from calendar_table where cal_year_id = 2019 group by 1 ) ,tab3 as ( select t.* ,event_month as start_month ,coalesce(lead(add_months(event_month,-1)) over(partition by customer_id order by event_month),year_end) as end_month from tab2 t ,(select max(cal_month_end_date) as year_end from cal) cal_end_month where status_rank = 1 ) select customer_id, cal_month_end_date, status, credit_limit from tab3, cal where cal_month_end_date between start_month and end_month group by 1,2,3,4 order by 1,2 Less

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