IntStream Average Computation — OptionalDouble Handling
This 1Z0-829 practice question tests your understanding of working with streams and lambda expressions. Read the scenario carefully and evaluate each option against the stated constraints before committing to an answer. A key principle to apply: intStream.average(). 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
List<Student> students = Arrays.asList(
new Student("Alice", 85),
new Student("Bob", 92),
new Student("Charlie", 78),
new Student("David", 88)
);
OptionalDouble avg = students.stream()
.filter(s -> s.getScore() > 80)
.mapToInt(Student::getScore)
.average();
System.out.println(avg.orElse(0.0));
Refer to the exhibit. What is the output?
Exhibit
List<Student> students = Arrays.asList(
new Student("Alice", 85),
new Student("Bob", 92),
new Student("Charlie", 78),
new Student("David", 88)
);
OptionalDouble avg = students.stream()
.filter(s -> s.getScore() > 80)
.mapToInt(Student::getScore)
.average();
System.out.println(avg.orElse(0.0));
A
88.0
Why wrong: This result would occur if integer division were used (265/3 = 88), but average() always returns a double.
B
88.33333333333333
Correct. The average of 85, 92, and 88 is (85+92+88)/3 = 265/3 = 88.33333333333333.
C
85.0
Why wrong: 85.0 might be the minimum or median, but not the average. The sum is 265, so the average is 88.33...
D
85.75
Why wrong: 85.75 is not the average; it might result from a miscalculation like (85+92+88)/3 = 88.33, not 85.75.
Answer the question above first, then reveal the full breakdown to understand why each option is right or wrong.
Correct answer & explanation
✓
88.33333333333333
The code creates an IntStream with values 85, 92, and 88, then calls average() to compute the arithmetic mean. This returns an OptionalDouble, and orElse(0.0) extracts the value. The sum 85+92+88 = 265, divided by 3 gives 88.33333333333333, which matches option B exactly. No integer division occurs because average() always returns a double.
Key principle: IntStream.average()
Answer analysis
Option-by-option breakdown
For each option: why learners choose it and why it is or isn't the right answer here.
✗
88.0
Why it's wrong here
This result would occur if integer division were used (265/3 = 88), but average() always returns a double.
✓
88.33333333333333
Why this is correct
Correct. The average of 85, 92, and 88 is (85+92+88)/3 = 265/3 = 88.33333333333333.
Related concept
IntStream.average()
✗
85.0
Why it's wrong here
85.0 might be the minimum or median, but not the average. The sum is 265, so the average is 88.33...
✗
85.75
Why it's wrong here
85.75 is not the average; it might result from a miscalculation like (85+92+88)/3 = 88.33, not 85.75.
Common exam traps
Common exam trap: answer the scenario, not the keyword
The trap here is that candidates might mistakenly think integer division applies (e.g., 265/3 = 88 with truncation) or misread the output format, but the stream's `average()` method always returns a `double` with full precision.
Detailed technical explanation
How to think about this question
Under the hood, `IntStream.average()` internally uses a `DoubleSummaryStatistics` collector that accumulates the sum as a `long` and count as a `long`, then divides the sum by the count as a `double` to preserve fractional precision. This avoids integer truncation, which is a common pitfall when manually averaging integers. In real-world scenarios, such as calculating average response times from log data, using `average()` ensures accurate floating-point results without accidental integer division.
KKey Concepts to Remember
IntStream.average()
OptionalDouble.orElse()
Stream pipeline
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
IntStream.average()
Real-world example
How this comes up in practice
A practitioner preparing for the 1Z0-829 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. IntStream.average() 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.
Review intStream.average(), then practise related 1Z0-829 questions on the same topic to reinforce the concept.
Working with Streams and Lambda Expressions — This question tests Working with Streams and Lambda Expressions — IntStream.average().
What is the correct answer to this question?
The correct answer is: 88.33333333333333 — The code creates an IntStream with values 85, 92, and 88, then calls average() to compute the arithmetic mean. This returns an OptionalDouble, and orElse(0.0) extracts the value. The sum 85+92+88 = 265, divided by 3 gives 88.33333333333333, which matches option B exactly. No integer division occurs because average() always returns a double.
What should I do if I get this 1Z0-829 question wrong?
Review intStream.average(), then practise related 1Z0-829 questions on the same topic to reinforce the concept.
What is the key concept behind this question?
IntStream.average()
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