Please answer question 1 A. Normal No Spac.. Heading1 Hea Parageaph Exam 2-Chapt
ID: 331387 • Letter: P
Question
Please answer question 1 A. Normal No Spac.. Heading1 Hea Parageaph Exam 2-Chapters 4,5.6 and 7 (100 points) You should prepare your responses in the manner they would be presented to management. Explain all statements made. Prepare your answers off line. When complete click on the above tab and attach your file. There are 105 points available. Complete all parts of each question. You will have one attempt to post your results If you have a problem in posting please let me know and I will clear that attempt. Do not copy and paste your results in the text box- attach your file. 1. Determine the: Explain all terms and their values. (10 Points) a. Mean, mode, and median of the following numbers b. Calculate 1, 2, and 3 standard deviations (plus and minus). The standard deviation for the data set is equal to 2.42 Data: 225, 226, 227, 226, 227, 228, 228, 229, 222, 223, 224, 226, 227, 228,225, 221, 227, 229, 230 2. Calculate the Z scores from the data in question 1. (10 points) a. Are any of these values outliers or those that would be outside of the control limits? Explain. 3. Using the data in Question 1 prepare a frequency table and frequency chart (20 points) a. Explain your values for the number of classes and class size. b. Interpret the Histogram. Shape, Skewness etc
Explanation / Answer
1. Mean - It is the simple average of the data, which can be obtained by dividing the sum of all values by the number of events
Xbar = sum ( x values) / n = 4298 /19 = 226.21
2. Median - It appers on the middle of the ordered sequence of values woth half of data are lower than this and half are higher than this.
It is the size of n+1/2 th value when the values are arranged in ascending order
Here it will be 19+1/2 = 10th value, which is 227.
3. Mode = It is the value which has the highest frequency, or which has occured the highest times. In this case it is 227 which has occured four times.
b. Standard deviation = 2.42 ( given)
one sigma values will correspond to
Xbar +- 2.42
= 226.21 +2.42 = 228.63 ( upper limit)
and 226.21 -2.42 = 223.79
Two sigma limits
= Xbar +- 2x 2.42
= 226.21+ 4.84 = 231.05 ( UCL ) and 226.21-4.84 = 221.37 ( LCL)
Three sigma limits
= Xbar +- 3x2.42
= 226.21 +7.26 = 233.47 ( UCL) and 226.21 - 7.26 = 218.95 ( LCL)