FS - Fundamentals of Surveying Applied Mathematics and Statistics Questions and Answers

Question 1

A distance is measured in three separate segments. The lengths and their respective standard errors are: Segment 1 = 150.25 ft (±0.03 ft), Segment 2 = 310.50 ft (±0.04 ft), and Segment 3 = 225.10 ft (±0.02 ft). What is the total length of the line and its propagated standard error?

Accepted Answer

685.85 ft (±0.054 ft)

Answer

The total length is the sum of the individual segments: 150.25 + 310.50 + 225.10 = 685.85 ft. The standard error of a sum is found by the square root of the sum of the squares of the individual errors (Law of Propagation of Errors). Error = sqrt((±0.03)^2 + (±0.04)^2 + (±0.02)^2) = sqrt(0.0009 + 0.0016 + 0.0004) = sqrt(0.0029) ≈ ±0.054 ft. [13, 18]

Question 2

In the statistical analysis of survey measurements, what is the relationship between the weight of an observation and its standard error (σ)?

Accepted Answer

Weight is inversely proportional to the variance (σ²).

Answer

The weight of an observation is a measure of its relative worth or precision. A more precise measurement (i.e., one with a smaller standard error) should have a greater influence on the result. Therefore, weight is inversely proportional to the variance (the square of the standard error). A smaller variance leads to a larger weight. [7, 30]

Question 3

A surveyor makes five repeated measurements of an angle with the following results: 92° 15' 33", 92° 15' 28", 92° 15' 31", 92° 15' 35", and 92° 15' 28". What is the sample standard deviation of these measurements?

Accepted Answer

3.1"

Answer

1. Calculate the mean: (33+28+31+35+28)/5 = 155/5 = 31". So the mean is 92° 15' 31".
2. Calculate the residuals (measurement - mean): +2", -3", 0", +4", -3".
3. Square the residuals: 4, 9, 0, 16, 9.
4. Sum the squared residuals: 4+9+0+16+9 = 38.
5. Calculate the variance (sum of squares / (n-1)): 38 / (5-1) = 38 / 4 = 9.5 sec².
6. Calculate the standard deviation (sqrt(variance)): sqrt(9.5) ≈ 3.1". [2, 12]

Question 4

A simple circular horizontal curve has a radius of 1000.00 ft and a central angle (delta) of 30°00'00". What is the length of the tangent (T) from the Point of Intersection (PI) to the Point of Curvature (PC)?

Accepted Answer

267.95 ft

Answer

The formula for the tangent length (T) of a simple circular curve is T = R * tan(Δ/2). Given R = 1000.00 ft and Δ = 30°00'00". First, divide the central angle by two: Δ/2 = 15°00'00". Then, calculate the tangent: T = 1000.00 * tan(15°) = 1000.00 * 0.267949 ≈ 267.95 ft. [6, 11, 25]

Question 5

A symmetric vertical curve is 600 feet long (6 stations). The initial grade (G1) is +2.0% and the final grade (G2) is -4.0%. What is the rate of change of grade (r) per station?

Accepted Answer

-1.00% per station

Answer

The rate of change of grade (r) for a vertical curve is calculated by the formula: r = (G2 - G1) / L, where L is the length of the curve in stations. In this case, G1 = +2.0%, G2 = -4.0%, and L = 6 stations. So, r = (-4.0 - 2.0) / 6 = -6.0 / 6 = -1.00% per station. [3, 4, 22]

FS - Fundamentals of Surveying Exam Practice Test

FS - Fundamentals of Surveying Exam Practice Test

FS - Fundamentals of Surveying Applied Mathematics and Statistics Questions and Answers