ch5_processing_of_large_data_sets.studentview

# The median is a better indicator of the centre than

This preview shows page 1. Sign up to view the full content.

This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: 6.2 7.5 8.3 9.1 total f 15 25 30 10 N=80 N=80 This is UNGROUPED data. This UNGROUPED Mean = 7.75625 Mean s.d. = 0.9006 s.d. Variance = 0.8112 Mode = 8.3 Range = 2.9 MEDIAN = 7.9 It’s MULTIPLE data. MULTIPLE Now compare Q4 and Q5. Now What do we do for the class boundaries? What Q4. Q4. Q5. Class Class 4.6 - 9.5 Frequency 5 Time Cumulative Cumulative frequency frequency 9.6 - 14.5 14.6 - 19.5 7 10 Below 5 Below 10 0 5 19.6 - 24.5 24.6 - 29.5 12 6 Below 15 Below 20 14 30 29.6 -34.5 total 1 41 Below 25 Below 30 37 40 Q4. ( Exercise Ch5 ) Q4. (a) Find the mean, s.d, variance, C.V. (b) It was later discovered that in the raw data list, 21.2 has been mistakenly entered into the frequency table as 12.2. Find the correct s.d. from the frequency table. Find Class 4.6 - 9.5 Frequency 5 9.6 - 14.5 14.6 - 19.5 7 10 19.6 - 24.5 24.6 - 29.5 12 6 29.6 -34.5 total 1 41 Solution for Q4(a) Solution Find the mean, s.d, variance, C.V Class Class boundaries boundaries Class Class mark mark x Frequency f 4.55-9.55 7.05 5 9.55-14.55 12.05 7 14.55-19.55 17.05 10 19.55-24.55 22.05 12 24.55-29.55 27.05 6 29.55-34.55 32.05 1 total XXX 41 This is grouped data This grouped Mean = 18.2695 s.d. = 6.5127 Variance = 42.4152 C.V.= 35.6479% For interest: Mode = 20.8 Median = 18.8 Skewness=-24.44% Q4(b) Delete 12.2 and replace with correct value 21.2 21.2 Class Class boundaries boundaries Class mark x Frequency f Class Class boundaries boundaries Class mark x Frequency f 4.55-9.55 7.05 5 4.55-9.55 7.05 5 9.55-14.55 12.05 7 wrong 9.55-14.55 12.05 6 correct 14.55-19.55 17.05 10 14.55-19.55 17.05 10 19.55-24.55 22.05 12 wrong 12 19.55-24.55 22.05 13 correct 24.55-29.55 24.55-29.55 27.05 6 24.55-29.55 27.05 6 29.55-34.55 32.05 1 29.55-34.55 32.05 1 total XXX 41 total XXX 41 Use the Use Q4 (b) solution Q4 and ∆∇ Now you can press: To go to Line 2 and To change the freq = 7 which is change wrong to 6 by pressing 6 EXE to Then go to Line 4 changing the Then freq = 12 to 13 by pressing 13 EXE. 13 SHIFT S-VAR SHIFT in the usual way to give the corrected values for the mean , s.d. and variance. mean Mean = 18.48902439 s.d. = 6.45585 Variance = 41.678 Q5. What do we do for the class boundaries? (This was a past class-test question) Time Time Cumulative Cumulative frequency frequency Below 5 Below 10 0 5 Below 15 Below 20 14 30 Below 25 Below 30 37 40 Solution for Q5 Solution Class Class boundary Class Class mark mark x Freq Cum Cum freq freq f Below 5 0-5 2.5 0 0 Below 10 5-10 7.5 5 5 Below 15 10-15 12.5 9 14 Below 20 15-20 17.5 16 30 Below 25 20-25 22.5 7 37 Below 30 25-30 27.5 3 40 total XXX XX 40 XX This is grouped data This grouped Mean = 16.75 s.d. = 5.4256 Variance = 29.4375 C.V.= 32.39% For interest: Mode = 17.1875 Median = 16.875 Special notes: Special To clear M symbol from calculator screen press 0 SHIFT RCL (STO) M+ (M)...
View Full Document

## This document was uploaded on 09/28/2013.

Ask a homework question - tutors are online