Analytical Methods: Modelling Engineering Situations Using Statistics and Probability
TASK 1
- 8 classes
Weight (Kg) | 4.2 | 4.3 | 4.4 | 4.5 | 4.6 | 4.7 | 4.8 | 4.9 | Total |
Frequency | 1 | 3 | 7 | 10 | 12 | 10 | 5 | 2 | 50 |
- Frequency distribution
Weight
Kgs |
Frequency
|
F (X) |
4.2 | 1 | 4.2 |
4.3 | 3 | 12.9 |
4.4 | 7 | 30.8 |
4.5 | 10 | 45.0 |
4.6 | 12 | 55.2 |
4.7 | 10 | 47.0 |
4.8 | 5 | 24.0 |
4.9 | 2 | 9.8 |
Totals | 50 | 228.9 |
- Frequency histogram and frequency polygon.
- Frequency histogram
- Frequency polygon
- Mean, median and inter quartile range
- Mean (M)
Mean = = 228.9/50 = 4.578 Kgs
- The median is the weight of the middle casting arranged from the least weight to the highest weight. The middle cast is between 25 and 26. This is the average of the two.
25th and the 26th casts all weighs 4.6 Kg in the modal class of 12.
Therefore, the median weight = 4.6 Kg
- The inter quartile range is the difference between the 75 % and the 25% of the highest cast weight all arranged in ascending order. In this case, it is the range between 12.5 and 37.5 of the total 50 arranged in ascending order.
At 12.5, the weight is 4.5 Kg and at 37.5 the weight is 4.7 Kg.
Therefore, the range is 4.7-4.5 = 0.2 Kg
- Standard Deviation
Weight
(Kg) X |
Frequency
(F)
|
F (X) | X-M
M=4.578 |
(X-M)2 | f(X-M)2 |
4.2 | 1 | 4.2 | -0.378 | 0.1429 | 0.1429 |
4.3 | 3 | 12.9 | -0.278 | 0.0773 | 0.2319 |
4.4 | 7 | 30.8 | -0.178 | 0.0317 | 0.2219 |
4.5 | 10 | 45.0 | -0.078 | 0.0061 | 0.0610 |
4.6 | 12 | 55.2 | 0.022 | 0.0005 | 0.0060 |
4.7 | 10 | 47.0 | 0.122 | 0.0149 | 0.1490 |
4.8 | 5 | 24.0 | 0.222 | 0.0493 | 0.2465 |
4.9 | 2 | 9.8 | 0.322 | 0.1037 | 0.2074 |
Totals | 50 | 228.9 | 0.4264 | 1.2666 |
σ 2= where n=50 and σ 2= variance
σ 2= 1.2666/50 = 0.0253
S.D = = 0.1592
- Limits of 95% and 99% of data
- Limits of 95%
95% of 50 = 47.5 items
From the table, only one item is 4.2 Kg hence it does not form part of the majority 95%.
2 items are also 4.9 Kg and also does not form part of the majority 95%.
Therefore, the limits for the majority 95% will be between 4.3 Kg and 4.8 Kg.
- Limits of 99%
99% of 50 = 49.5 items
From the table, only one item falls on its own class = 4.2 Kg cast. The rest around 98% percent falls on the limits 4.3 Kg and 4.9 Kg.
Therefore, the limits for 99% are between 4.3 Kg and 4.9 Kg.
TASK 2
- Scatter diagram
- Correlation coefficient
Temperature and harness are negatively correlated. The hardness decreases with increase in temperature.
Temperature
X |
Hardness
Y |
X2 | Y2 | XY | |
240 | 550 | 57600 | 302500 | 132000 | |
275 | 532 | 75625 | 283024 | 146300 | |
300 | 525 | 90000 | 275625 | 157500 | |
335 | 514 | 112225 | 264196 | 172190 | |
360 | 474 | 129600 | 224676 | 170640 | |
395 | 470 | 156025 | 220900 | 185650 | |
420 | 458 | 176400 | 209764 | 192360 | |
455 | 446 | 207025 | 198916 | 202930 | |
∑x=2780 | ∑y=3969 | ∑x2 =1004500 | ∑y2 =1979601 | ∑xy=1359570 |
n = 8
Correlation coefficient = r
8*1359570 – (2780 *3969)
10876560– (11033820) = -157260
= -157260/9871015
r = – 0.1593
- Least-squares linear regression analysis
The equation is given by
.a= {3969 – b (2780)} * 1/8
b= [8*1359570 – 2780*3969]
[8*1004500 – 27802 ]
b= [10876560 – 11033820]
[307600 ]
b= – 0.511
.a = 673
y= 673 – 0.511 x
- 260°C and 370°C.
- 260°C
Y = 673 – 0.511 (260) = 543 V
- 370°C.
Y = 673 – 0.511 (370) = 482 V
TASK 3
- Normal distribution and confidence intervals
Given data;
Normal distribution mean = 0.15 cm
Standard deviation = 0.01 cm
Sample size (n) = 150
Computed mean = 0.147 cm
5% confidence level = 0.95, therefore σ = 0.05
Using the POM software to find the distribution given the mean and standard deviation, the distribution is as shown in the figure below.
Confidence interval (c.i) =
Mean = 0.147
The true standard error of the mean = = = 0.00082
σ = 0.05 = σ/2 = 0.025
Z σ/2 = 1.96 since F (1.96) = 1 – α/2 = .975).
c.i = 0.147 1.96 *0.00082
= 0.1486 OR 0.1454
Therefore, the mean of the components have changed since the mean of 0.15 does not lie in the interval computed
TASK 4
- Actual size 120mm * 500mm
- Let probability that length will begreater be PL= 0.025
Also let the Probability that width will be greater be PW= 0.03
Therefore, the probability that both sides will be oversize = PL and PW = PL*PW
=0.025*0.03 =0.00075
- Probability that blank is unacceptable = PL and (not PW) or PW and (not PL) or PL*PW
= PL*0.97 +PW*0.975 + 0.025*0.03
= 0.025*0.97 +0.03*0.975 +0.025*0.03 = 0.02425 + 0.02925+ 0.00075= 0.05425
- Binomial distribution formula
P(X = r) = nCr p r (1-p) n-r
Where nCr = (n! / (n-r)! ) / r!
r = Number of successful events =4
n = Number of events = 6
p = Probability of success on a single trial. = 0.94575
1-p = Probability of failure. = 0.05425.
nCr =[(6*5*4*3*2)/2]/4*3*2 = 360/24 =15
P(X = r) = 15* (0.94575)4 * (0.05425)2 = 0.03532
- Poisson distribution formula
f(x) = e-λλx / x! Where,
λ= an average rate of value = 20
e =the base of logarithm (e=2.718)
x = Poisson random variable =2
e-λ = e-20 = 2.061E-9
λx = 202 = 400
f(x) = e-λλx / x! = 0.0000412
References
Devore, J. (2012). Probability & Statistics for Engineering and the Sciences. CengageBrain. com.
Johnson, R. A., Miller, I., & Freund, J. E. (2011). Probability and statistics for engineers. Prentice-Hall.
Milton, J. S., & Arnold, J. C. (2002). Introduction to probability and statistics: principles and applications for engineering and the computing sciences. McGraw-Hill, Inc..
Montgomery, D. C., & Runger, G. C. (2007). APPLIED STATISTICS AND PROBABILITY FOR ENGINEERS, (With CD). John Wiley & Sons.
Use the order calculator below and get started! Contact our live support team for any assistance or inquiry.
[order_calculator]