CTFL_Syll_4.0 Exam Dumps : ISTQB Certified Tester Foundation Level (CTFL) v4.0

The three-point test estimation technique is a method of estimating the test effort based on three initial estimates: the most optimistic, the most likely, and the most pessimistic. The technique uses a weighted average of these three estimates to calculate the final estimate, which is also known as the expected value. The formula for the expected value is:

Expected value = (most optimistic + 4 * most likely + most pessimistic) / 6

Using the given values, the expected value is:

Expected value = (6 + 4 * 30 + 54) / 6 Expected value = 30 person hours

However, the expected value is not the only factor to consider when estimating the test effort. The technique also calculates the standard deviation, which is a measure of the variability or uncertainty of the estimates. The formula for the standard deviation is:

Standard deviation = (most pessimistic - most optimistic) / 6

Using the given values, the standard deviation is:

Standard deviation = (54 - 6) / 6 Standard deviation = 8 person hours

The standard deviation can be used to determine a range of possible values for the test effort, based on a certain level of confidence. For example, using a 68% confidence level, the range is:

Expected value ± standard deviation

Using the calculated values, the range is:

30 ± 8 person hours

Therefore, the final estimate is between 22 person hours and 38 person hours, which is option A.

The correct outcome of a two-value boundary value analysis applied to the password length is the set of test cases represented by option D. Boundary value analysis is a test design technique that focuses on the values at the boundaries of an equivalence partition, such as the minimum and maximum values, or the values just above and below the boundaries. A two-value boundary value analysis uses two values for each boundary, one representing the valid value and one representing the invalid value. For example, if the valid range of values is from 4 to 7, then the two values for the lower boundary are 3 and 4, and the two values for the upper boundary are 7 and 8. The test cases in option D use these values for the password length, while also satisfying the other requirements of the password, such as containing at least one numeric character, one capital letter, and one lowercase letter. The test cases in option D are:

1RhT: a 4-character password that is valid

rSp53: a 5-character password that is valid

3N3e10: a 6-character password that is valid

8sBdby: an 8-character password that is invalid The test cases in the other options are incorrect, because they either use values that are not at the boundaries of the password length, or they do not meet the other requirements of the password. For example, the test cases in option A are:

1xA: a 3-character password that is invalid, but it does not contain a capital letter

aB11: a 4-character password that is valid

Pq1ZZab: a 7-character password that is valid

7iDD0a1x: an 8-character password that is invalid References: ISTQB Certified Tester Foundation Level (CTFL) v4.0 sources and documents:

ISTQB® Certified Tester Foundation Level Syllabus v4.0, Chapter 2.2.1, Black-box Test Design Techniques1

ISTQB® Glossary of Testing Terms v4.0, Boundary Value Analysis, Equivalence Partition2

To determine the correct execution order that minimizes the number of configuration switches and respects the dependencies, we need to consider the following:

Initial Configuration:CONF1.

Dependencies:

TC1 depends on nothing.

TC2 depends on TC4.

TC3 depends on TC4.

TC4 depends on nothing.

TC5 depends on TC1.

Configuration Requirements:

TC1 requires CONF2.

TC2 requires CONF2.

TC3 requires CONF1.

TC4 requires CONF1.

TC5 requires CONF2.

Given the initial configuration is CONF1, start with test cases that can run on CONF1 and respect the dependencies. Then switch to CONF2 only when necessary. The optimal order to minimize configuration switches is:

Start withTC4(no dependencies, CONF1).

Continue withTC3(depends on TC4, CONF1).

Switch to CONF2.

ExecuteTC2(depends on TC4, CONF2).

Continue withTC1(no dependencies, CONF2).

Finally, executeTC5(depends on TC1, CONF2).

Therefore, the correct order is:

TC4(CONF1)

TC3(CONF1)

TC2(CONF2)

TC1(CONF2)

TC5(CONF2)

Thus, the answer isA. TC4, TC3, TC2, TC1, TC5.