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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 4.559.55 7.05 5 9.5514.55 12.05 7 14.5519.55 17.05 10 19.5524.55 22.05 12 24.5529.55 27.05 6 29.5534.55 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 4.559.55 7.05 5 4.559.55 7.05 5 9.5514.55 12.05 7 wrong 9.5514.55 12.05 6 correct 14.5519.55 17.05 10 14.5519.55 17.05 10 19.5524.55 22.05 12 wrong
12 19.5524.55 22.05 13 correct 24.5529.55
24.5529.55 27.05 6 24.5529.55 27.05 6 29.5534.55 32.05 1 29.5534.55 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 SVAR
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 classtest 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 05 2.5 0 0 Below 10 510 7.5 5 5 Below 15 1015 12.5 9 14 Below 20 1520 17.5 16 30 Below 25 2025 22.5 7 37 Below 30 2530 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)...
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 Fall '13
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