Question

Class Pulses Part 4 -Class.Pulses and Group Pulses e mean, median, mode, range, and standard deviation for the data. Us e the correct symbols to describe the mean and standard deviation. Find the mean, median, mode, range, and standard deviation for your group pulse data, deviation. Use the correct symbols to describe the mean and standard Was the sample mean and sample standard deviation a good predictor of the related population parameters? Do you expect the graph to be skewed left, skewed right, or symmetric? Discuss. C Pyld Part 5- Survey Pulses Create an ordered stem-and-leaf plot. . Discuss the graph. Is it skewed left, skewed right, or approximately symmetric? Part 6-Class Pulses Create classes. Split your data into 5 classes of equal width. Show the formula used to create the classes. Create a relative frequency distribution and a relative frequency histogram. Using the formulas for grouped data, find the approximate population mean and approximate population standard deviation. . Is the distribution skewed left, skewed right, or approximately symmetric? Explain how you know. Part 7 - Class/Survey Combined Pulses Create a larger sample by appending your survey pulses to the class pulses Create a box plot. Give the five-number summary, upper and lower fences, and outliers. . .Discuss the graph. Is it skewed left, skewed right, or approximately symmetric? Compare the mean and the median. Which seems to provide a better measure of central tendency in this case? Why? Overall: Use sentences, proper grammar and proper punctuation. All discussion items must be typewritten. Graphs may be drawn by hand or using StatCrunch. Symbols may be typed or written in by hand. Remember to provide appropriate titles, labels, and keys (where required.) Include the evaluation rubric with your project
Use the data on the white sheet on the left to complete PART 6. Show formula used to create the classes and complete all of part 6

Answer 1

Class pulses : 75,77,69,72,74,56,59,69,71,72,77,68,54,89,84,80,74,65,68,73,75,93,79,85,71,65,78,80,56

Maximum = 93

Minimum = 54

Class width = Max - Min/ No. of classes = 93 -54/ 5 = 7.8

So first class= 54 - 61.8 and then so on

Class Interval	Frequency	Relative Frequency
54-61.8	4	0.14
61.8-69.6	5	0.17
69.6-77.4	11	0.38
77.4-85.2	7	0.24
85.2-93	2	0.07
Total	29	1

Relative Frequency = Frequency / Total Frequency

3) . M is the midpoint of class interval and f is frequency

Classes	M	f	M · f	M2 · f
54-61.8	57.9	4	231.6	13409.64
61.8-69.6	65.7	5	328.5	21582.45
69.6-77.4	73.5	11	808.5	59424.75
77.4-85.2	81.3	7	569.1	46267.83
85.2-93	89.1	2	178.2	15877.62
	Sum =	29	2115.9	156562.29

Now Frome the table

Mean = Sum of M*f / n = 2115.9/29 = 72.962

Variance = s² = 1/N sum(m²f) - Mean² = 1/29*156562.29 - 72.962²  = 75.24

Standard Deviation = s = (75.24)^1/2 = 8.674

To find skewness where y is midpoint so m and n is no. of observation

skewness (n - 1)3

Classes	M	frequency	(M-x)3*f
54-61.8	57.9	4	-13668.09287
61.8-69.6	65.7	5	-1914.867544
69.6-77.4	73.5	11	1.712929592
77.4-85.2	81.3	7	4057.735293
85.2-93	89.1	2	8405.80148
	Sum =	29	-3117.710714

Since our sum is negative our data is left-skewed.