ENS6152 Steel Design Assignment Help
Delivery in day(s): 3
1. Gender – Categorical and Nominal
2. Education – Categorical and Ordinal
3. Hours of work – Continuous numerical and Interval
4. Number of children – discrete numerical and interval
5. Parents – cateorical and interval
A. Mean, Median and Standard deviation
Mean 
724.67 
Median 
720.00 
Mode 
730.00 
Standard Deviation 
114.28 
Sample Variance 
13,060.23 
Kurtosis 
0.94 
Skewness 
0.64 
The mode as the data integer value is in the small range of nearly 30 different data types
B. Yes, the measure agree to the central tendency
C. Standard deviation: 114.28
D. Sort and Standardise the data
500 
560 
570 
600 
620 
620 
650 
660 
670 
690 
690 
700 
700 
710 
720 
720 
730 
730 
730 
730 
740 
740 
760 
800 
820 
840 
850 
930 
930 
1,030 
E. Unusual data values
Yes, there are outliers which is clearly noted in the above histogram chart
F. Normal distribution
Mean 
724.67 
Standard Deviation 
114.28 
Kurtosis 
0.94 
Skewness 
0.64 
Count 
30.00 
The normal distribution possess a oveall skewness of 0 and a kurtosis of 3, From the above table the value of Skewness is 0.64 and kurtosis is 0.94. Therefore the data is normal distribution.
Part a

Mean 
Median 
Mode 
Quiz 1 
72 
72 
60 
Quiz 2 
72 
72 
65 
Quiz 3 
76 
72 
72 
Quiz 4 
76 
86.5 
90 
From the analysis the measure of central tendency agree on Quiz 1, Quiz 2 and Quiz 3. However, the Quiz 4 is not agreeing to central tendency.
Mean uses all the data and information in determining the total output, however a very large data or few data may affect the overall output
Median imparts that the high value or the small value will not affect the values like mean, however it will take more time to compute the value
Mode uses only average values and sometimes there may be more than one value of mode.

Skewness 
Kurtosis 
Quiz 1 
1.01 
0.50 
Quiz 2 
0.00 
2.21 
Quiz 3 
1.57 
1.18 
Quiz 4 
1.87 
3.40 
From the above table the skewness of Quiz 3 is 0, therefore the data is symmetric and it is normally distributed, however the values of quiz 1 and quiz 3 is 1.01 and 1.57 respectively, since the value is positive it is rightly skewed (positively skewed) and the skewness of quiz 4 is negatively skewed (skewed left).
From the analysis it is noted that mean and median of Quiz 1 and Quiz 2 possess the same mean and median but the mode is less which shows that the distribution is skewed to the right, however in case of quiz 3 the mean is greater than the median and mode so it is negatively skewed. The quiz 4 shows that the mean is lesser than the median and median is lower than the mode which shows it is skewed to the right.
The overall probability that a given alternator may fail during the 1 hour journey is 0.02, this shows that the probability not failing is 0.98
Probability that both will fail
= 0.02 x 0.02
= 0.0004
Probability that both will not fail
= 0.98 x 0.98
= 0.9604
Probability that one will fail
Probability the first will fail and the second will not fail
= 0.02 x 0.98
= 0.0196
Probability that first will not fail and second will fail
= 0.98 x 0.02
= 0.0196
Therefore that the probability that either one or other will fail
= 0.0196 + 0.0196
= 0.0392
95% confidence interval
= [(mean – 1.96 *(stdev/squareroot (18), mean + 2.11.96098 *(stdev/squareroot (18)]
=[(3278.72 – 1.96 x 97.05 / 4.24), (3278.72 – 1.96 x 97.05 / 4.24)]
= [(3278.72  44.84), (3,278.72 + 44.840]
=[3,233.89, 3,323.56]
Therefore it can be stated that the 95% confidence interval is 3,233 and 3,323 steps
Sample size
= (t x standard deviation / error) ^ 2
= (1.95 x 97.05 / 20)^ 2
= 89.5 days
This shows that the data required is nearly 90 days
Freedman, David (2010). Statistics. 4th Edition. Cengage Publishing
Sarah Boslaugh (2012). Statistics in a Nutshell. Cengage Publishing
Sincich T. Terry (2012). Statistics. 12th Edition.
Witte S. Robert, (2010). Statistics. 5th Edition. McGraw Hill
Triola (2014). Essentials of Statistics. 5th Edition. McGraw Hill