Investigate applications of statistical techniques

Course: HND in Construction and the Built Environment

Unit 8: Mathematics for Construction in the Higher National Certificate/Diploma in Construction and the Built Environment equips students with the mathematical tools to tackle real-world construction problems. Through applying concepts like dimensional analysis, statistics, and calculus, students develop the ability to analyze data, solve complex scenarios, and model engineering situations. The unit covers areas like calculating material quantities, analyzing project delays with financial penalties, and even using trigonometry for design elements. By building a strong foundation in these mathematical applications, graduates gain a significant edge in the construction industry.

Unit 8 Mathematics for Construction

Assignment Title - Building Mathematics

Task 1

In this section, we briefly describe about distribution of the variables using group data and also find the probability of values using normal distribution. Classical probability technique deals with relative frequency of each and every event in the sample space. Let us consider the experiment of rolling a die twice and the possible outcomes are 2, 3, 4, 5, 6, 7, 8, 9, 10, 11 and 12.

For normally distributed random variables, the z test was used to calculate the probability values. Here, we see that z is a standard normal distribution with mean 0 and variance 1. That is,
Z=(X-μ)/σ~N(0,1)

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Task 2

Scenario 1

Produce a histogram for each of the distributions scaled such that the area of each rectangle represents frequency density and find the mode.

Histogram

Mode=L+(f_m-f_(m-1))/((f_m-f_(m-1) )+(f_m-f_(m+1) ) )*w
Mode=10+(40-38)/((40-38)-(40-22) )*5
= 10 + 2/20*5
= 10.5
Thus, the required mode for the Number of customers in January is 10.5
Mode=L+(f_m-f_(m-1))/((f_m-f_(m-1) )+(f_m-f_(m+1) ) )*w
Mode=10+(69-39)/((69-39)-(69-41) )*5
= 10 + 2.586
= 12.586
Thus, the required mode for the Number of customers in July is 12.586

For each distribution find the:
• the mean
• the range
• the standard deviation

c)

The descriptive statistics is given below
Table - January

Table - July

The mean and standard deviation for grouped data is calculated by using the formula given below
(x_Jan ) ¯=(∑fx)/(∑f)=(2.5*27+7.5*38+12.5*40+17.5*22+25*13+35*4)/(24+38+40+22+13+4)=11.82
(x_Jul ) ¯=(∑fx)/(∑f)=(2.5*22+7.5*39+12.5*69+17.5*41+25*20+35*5)/(22+39+69+41+20+5)=13.28
s_Jan=√((∑?fx^2 ?)/(∑f)-((∑fx)/(∑f))^2 )=√(28318.75/144-(1702.5/144)^2 )=7.54
s_Jul=√((∑?fx^2 ?)/(∑f)-((∑fx)/(∑f))^2 )=√(44293.75/144-(2602.5/144)^2 )=7.05

Scenario 2

Information given
Population mean = μ = 360 days
Standard Deviation = σ = 60 days
If the population standard deviation, sigma is known, then the mean has a normal (Z) distribution

How could the assumption that the bulb life is a normal distribution betested?

a) Error analysis can be used to validate the normality assumption for the bulb life

If it is decided to replace all bulbs at one specified time, what interval must be allowed between replacements if not more than 10% of bulbs should fail beforereplacement?

b)
P(X>A)=P((X-μ)/σ>(A-360)/60)=0.1
1-P((X-μ)/σ<(A-360)/60)=0.1
P(Z<(A-360)/60)=0.9---(1)
where Z=(X-μ)/σ~N(0,1)
Using normal distribution table, we have
P (Z < 1.282) = 0.9 ---------------- (2)
On comparing (1) and (2). We have
(A-360)/60=1.282
A = 360 + 60 * 1.282 = 436.89
The required interval is (0, 436.89)
Thus, it requires 437 days to replace the bulbs

What practical considerations might dictate such a replacement policy?

c)
It seems to be difficult to replace all the 5000 bulbs at a time, as customer do not prefer to wait till that time
Also, if a bulb is not working, it should be replaced immediately and it would be in great problem if we wait till the given interval

The supplier offers a new type of bulb, Type B, that has a mean life of 450 days and the same standard deviation (60 days) as the present type. If these bulbs were to be used how would the replacement time be affected?

d)
The new Type B bulb has a mean time of 450 hours which seems to be high mean hours of lifetime when compared with Type A bulb. Thus, customer would prefer to buy this new Type B bulb which has less chance of being replaced

Determine whether the new type of bulb is preferable given that is costs 25% more thanthe existing Type A. Present and explain your conclusions.

e)
The new Type B bulb has a mean time of 450 hours which seems to be high mean hours of lifetime when compared with Type A bulb. Thus, customer would prefer to buy this new Type B bulb which has less chance of being replaced. Even though the cost is high, the lifetime is almost 1.5 times higher when compared to that of Type A bulb and therefore, it can be preferred when compared to that of Type A

A rival supplier now offers a third type of bulb, Type C, that has a mean life of 432 days and a standard deviation of 45 days. If these bulbs were to be used how would the replacement timebe affected?

f)
Type C is better than Type A and slightly lower than Type B. Therefore, if the company looks for minimal cost with good mean lifetime, then, he can choose Type C ahead of Type B

A simple random sample of 10 people from a certain population has a mean age of 27 years. Can we conclude that the mean age of the population is not 30 years? The variance of the populate ages is known to be 20. Test your chosen hypothesis at a 5% level of significance using both a two tailed test and a one tailed test and explain your conclusions.

B
Null Hypothesis: H0: μ = 30
That is, the mean age do not differ significantly from 30 years
Alternate Hypothesis: Ha: μ ≠ 30
That is, the mean age differ significantly from 30 years
Level of Significance:
Let the level of significance be α = 0.05
Test Statistic
The z test statistic is
z=(x ¯-μ)/(σ⁄√n)=(27-30)/√(20/10)=2.12
Here, the critical value is 1.96
The p - value of z test statistic is 0.0169
Since the p - value falls well below 0.05, we conclude that the mean age differ significantly from 30 years

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Task 3

Scenario 1

State the amplitude, phase, frequency and periodic time of each of these waves.

The Amplitude is defined as the distance between the centre line and peak (also termed as height)
The Phase Shift is defined as how far the function stated is moved or shifted horizontally from its original position

Here,
??1 = 3.75 sin (100???? + 2??/9)
??2 = 4.42 sin (100???? - 2??/ 5)
Amplitude A = 2
Phase shift = -0.5
Vertical shift D = 3
Frequency = 4
Period = ¼
(ii) When both machines are switched on, how many seconds does it take for each machine to produce its maximum displacement?

When both machines are switched on, it takes 0.05 seconds to get the maximum displacement

(iii) 2 seconds
(iv)
Here,
??1 = 3.75 sin (100???? + 2??/9)
??2 = 4.42 sin (100???? - 2??/ 5)

The table showing the values of x1 and x2 is given below

(vii)
Yes, it is clearly seen that the path does not seems to be changed

Scenario 2
Calculate the distance AB.
Distance=√((40-0)^2+(0-0)^2+(-20-0)^2 )=60
Find the angle between the sections AB and BC.
The required angle, ? is 1480
Cosθ=(((-2@-2@1))*((3@4@1)))/(√(2^2+2^2+1^2 ) √(3^2+4^2+1^2 ))=cos(13/(3*√26))=148
Write down a vector equation of the line BC. Hence find a and b.
Vector Equation
Here, we have
Bc=((40@0@-20))+λ((3@4@1))
When z = 0, we have λ = 20
On substituting these values, we have
a = 40 + 3 * 20 = 100 and b = 20 * 4 + 0 = 80

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Task 4

Scenario 1
(ai)

(a ii) Investigate and state the range of values where the above Bending Moment Function ls maximum or Minimum, decreasing or increasing or neither.

The points are given below

(b) Determine the range of the temperature for positive t

The range of the temperature is given below

(c) Note that in the thermodynamic system provided herein, the expression given is equated to 0 to solve the problem given to be solved.

Here
dv/dx=q
Also, we see that q = - ω0
Integrating, V = - ω0x + C1
The function is
Log (P) + n log (V) - log (C)
Thus, we have
PV = C
The rate of change of V is 10
Scenario 2
The given function is
C=16t-2+2t

(b) Explain how calculus may be used to find an analytical solution to this problem of optimisation.

The integral function was used to integrate the function which is evaluated along a curve. In real life situations, it was mainly used to improve the architecture, particularly do not depends on building only but also relates to other infrastructures like bridge, etc

(c) Use calculus to find the production time at which the cost is at a turningpoint.

Using integration, we see that the product time is 2 minutes at which the cost is at a turning point

(d) Show that the turning point is a mathematical minimum.

In order to reach the minimum cost point, the second differentiation should be less than 0. Here, we see that the second differentiation is less than zero, indicating that turning point is a mathematical minimum.

FAQ: Statistical Techniques for Data Analysis and Presentation with Software

• 1. What are statistical techniques used for?
• 2. What are some common statistical techniques?
• 3. What are some popular computer software packages for statistical analysis?
• 4. How do I choose the right software package?
• 5. How can I learn more about statistical techniques and software?
• 6. Why are sinusoidal waves and vector functions important in construction?
• 7. What construction applications use sinusoidal waves?
• 8. How are vector functions used in construction?

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