Question

Please answer the following

3. Piles are commonly used in foundations of civil worlk structures. Test piles were driven and used to measure pile strength at a selected site. The following strength data (in kips) were collected as shown in Table 2: Table 2. Pile number and strength in kips ile N Strength 8829 10236 5101 9144 7790 9327 8470 10570 10186 9305 10746 9069 11436 10044 10281 12311 0639 0215 8723 8877 20 a. Plot the histogram and frequency diagrams for pile strength b. For the data, determine the central tendency measures (average value, median, and mode). You may use the functions in Excel for the calculations, but you must show the equations. c. For the data, determine the dispersion measures: the variance, standard deviation, and coefficient of variation. You may use the functions in Excel for the calculations, but you must show the equations. d. Discuss the data with respect to the statistical tests (skewness, normally distributed, etc.)

Answer 1

All my answers are based on the image of the question attached and all the formulas used are in the above images.

1: The best way to represent data graphically is to used tools. I have used Excel for the first part of this question.

Following is the table I used to generate my histogram.

Pile no	Pile Strength(in kips)
1	8829
2	10236
3	5101
4	9144
5	7790
6	9327
7	8470
8	10570
9	10186
10	9305
11	10746
12	9069
13	11436
14	10044
15	10281
16	12311
17	639
18	215
19	8723
20	8877
21	20

We have n=21 observations.

1. First input and select the data in excel.

2. Go to insert ---> Charts ---> Columns(2-D) , and you will obtain a histogram as shown below.

b: Mean=$\sum x_{i}/n$= 171319/21 =8158.047619 kips

For Median you need to arrange your observations either in ascending or descending order of sequence.

I have arranged them in ascending order. Data in ascending order is:

20,215,639,5101,7790,8470,8723,8829,8877,9069,9144,9305,9327,10044,10186,10236,10281,10570,10746,11436,12311

now, as n=21. Median =(n+1)/2 th observation = (21+1)/2= 11th observation

Therefore, Median= 9144 kips

As, none of the observation occurs more than once mode of the sample does not exist.

(c) Variance=$\sigma^{2}$= $[(\sum x_{i}-\bar{x})^{2}/(n-1)]$ = [(1655834119-21*8158.047619^2)}/(21-1)=12910277.95 kips square

Standard deviation=$\sigma$=$\sqrt{[(\sum x_{i}-\bar{x})^{2}/(n-1)]}$= $\sqrt{Var(x)}$=$\sqrt{12910277.95}$=3593.087533 kips

Coefficient of variation = ( $\sigma$*100)/$\bar{x}$=(3593.087533*100)/8158.047619=44.04347328

(d) Skewness= b₁= $(\sum(x_{i}-\bar{x})^{3}/n)/\sqrt{\sum (x_{i}-\bar{x})^{2}/(n-1)}$ =(-1316319781873.42)/(3593.087533^3) = -1.35125938935568

Kurtosis=g2=$(\sum(x_{i}-\bar{x})^{4}/n)/\sum (x_{i}-\bar{x})^{2}/(n)^{2}$-3=(12216357878800700/21)/(158738358747408)-3 =3.66471804390782-3=0.66471804390782

This distributions is negatively skewed or left skewed that means the left tail of the distribution is longer and is leptokurtic.