Question 348 of 500
Advanced Searching and StatisticshardMultiple ChoiceObjective-mapped

Quick Answer

The answer is hours where any user's logon count is more than double the average for that hour. This is correct because the `eventstats` command computes a per-hour average logon count across all users, appending that statistic to each event without reducing the dataset. The subsequent `where` clause then filters for events where an individual user’s logon count for that hour exceeds twice that hourly average, isolating the outliers. On the SPLK-1003 exam, this question tests your ability to distinguish `eventstats` from `stats` and to interpret how appended aggregations interact with filtering commands—a common trap is confusing the per-hour average with a global average or thinking the result shows users with the highest counts. Remember the memory tip: “Double the average, double-check the `eventstats`” to recall that the command keeps row-level detail while adding a calculated field, enabling precise comparisons like this per-hour threshold filter.

SPLK-1003 Advanced Searching and Statistics Practice Question

This SPLK-1003 practice question tests your understanding of advanced searching and statistics. Read the scenario carefully and evaluate each option against the stated constraints before committing to an answer. After answering, compare your reasoning against the explanation and wrong-answer breakdown below. Once you have made your selection, read the full explanation to reinforce the concept and understand why each distractor is designed to mislead on exam day.

Exhibit

index=security sourcetype=windows_logon | eval logon_hour=strftime(_time, "%H") | stats count by logon_hour, user | eventstats avg(count) as avg_count by logon_hour | where count > avg_count * 2

Refer to the exhibit. What does the final result represent?

Question 1hardmultiple choice
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Exhibit

index=security sourcetype=windows_logon | eval logon_hour=strftime(_time, "%H") | stats count by logon_hour, user | eventstats avg(count) as avg_count by logon_hour | where count > avg_count * 2

Answer choices

Why each option matters

Answer the question above first, then reveal the full breakdown to understand why each option is right or wrong.

Correct answer & explanation

Hours where any user's logon count is more than double the average for that hour.

The `eventstats` command calculates a per-hour average logon count across all users. The `where` clause then filters for events where a specific user's logon count for that hour is more than double that hourly average. This directly matches option C: hours where any user's logon count exceeds twice the average for that hour.

Key principle: Answer the scenario, not the keyword: identify the specific constraint before choosing the most familiar-sounding option.

Answer analysis

Option-by-option breakdown

For each option: why learners choose it and why it is or isn't the right answer here.

  • Users who log on more than twice on average.

    Why it's wrong here

    No averaging across hours

  • Hours where the total logon count is more than double the average.

    Why it's wrong here

    Eventstats averages per hour, not total

  • Hours where any user's logon count is more than double the average for that hour.

    Why this is correct

    Correct: per hour, per user comparison to hour average

    Related concept

    Read the scenario before looking for a memorised answer.

  • Users who have a logon count greater than twice their personal average.

    Why it's wrong here

    Average is per hour, not per user

Common exam traps

Common exam trap: answer the scenario, not the keyword

The trap here is that candidates confuse `eventstats ... by hour` (which computes a global average per hour) with a per-user average, leading them to incorrectly select option D or A.

Detailed technical explanation

How to think about this question

Under the hood, `eventstats` computes the average logon count for each hour across all users and appends that value (`avg_logons`) to every event in that hour. The `where` clause then filters events where the individual user's `logon_count` exceeds twice that appended average. This is a common pattern for detecting outliers or anomalies in time-series data, such as identifying hours with unusually high activity from a single user compared to the norm.

KKey Concepts to Remember

  • Read the scenario before looking for a memorised answer.
  • Find the constraint that changes the correct option.
  • Eliminate answers that are true in general but not in this case.

TExam Day Tips

  • Watch for words such as best, first, most likely and least administrative effort.
  • Review why wrong options are wrong, not only why the correct option is correct.

Key takeaway

Answer the scenario, not the keyword: identify the specific constraint before choosing the most familiar-sounding option.

Real-world example

How this comes up in practice

A practitioner preparing for the SPLK-1003 exam encounters this exact type of scenario on the job. The correct answer here is not the most general option — it is the best answer for the specific constraint described. Answer the scenario, not the keyword: identify the specific constraint before choosing the most familiar-sounding option. Real exam questions reward reading the full scenario before eliminating options, because the constraint defines which answer fits.

What to study next

Got this wrong? Here's your next step.

Identify which exam domain this question belongs to, review the core concept, then practise similar questions from the same domain.

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FAQ

Questions learners often ask

What does this SPLK-1003 question test?

Advanced Searching and Statistics — This question tests Advanced Searching and Statistics — Read the scenario before looking for a memorised answer..

What is the correct answer to this question?

The correct answer is: Hours where any user's logon count is more than double the average for that hour. — The `eventstats` command calculates a per-hour average logon count across all users. The `where` clause then filters for events where a specific user's logon count for that hour is more than double that hourly average. This directly matches option C: hours where any user's logon count exceeds twice the average for that hour.

What should I do if I get this SPLK-1003 question wrong?

Identify which exam domain this question belongs to, review the core concept, then practise similar questions from the same domain.

What is the key concept behind this question?

Read the scenario before looking for a memorised answer.

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Last reviewed: Jun 25, 2026

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This SPLK-1003 practice question is part of Courseiva's free Splunk certification practice question bank. Courseiva provides original exam-style practice questions with explanations, topic-based practice, mock exams, readiness tracking, and study analytics to help learners prepare for the SPLK-1003 exam.