Descriptive Statistics #1 Count 50
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Descriptive statistics #1 count 50
Stem and Leaf plot for stem unit = 10 leaf unit = 1 Frequency 19 0 25 1 6 2 50
#1
Stem Leaf 2233334456667778899 0000000022335556666666889 001256
H −L 26 − 2 = = 3.612 1 + 3.322 log N 1 + 3.322 log 50
Frequency Distribution - Quantitative Data lower 0.5 3.5 6.5 9.5 12.5 15.5 18.5 21.5 24.5
upper midpoint < < < < < < < < <
3.5 6.5 9.5 12.5 15.5 18.5 21.5 24.5 27.5
2.0 5.0 8.0 11.0 14.0 17.0 20.0 23.0 26.0
cumulative frequency frequency 6 6 7 10 5 9 4 1 2 50
6 12 19 29 34 43 47 48 50
Histogram 50 Individuals serve in any Government Agencies 12
Frequency
10 8 6 4 2
27 .5
24 .5
21 .5
18 .5
15 .5
12 .5
9. 5
6. 5
3. 5
0. 5
0
No. of years in any Government Agencies
FrequencyPolygon 50 Individuals serve in any Government Agencies 12.0
Frequency
10.0 8.0 6.0 4.0 2.0 0.0 -3
4
10
16
22
Descriptive statistics #1 count 50 mean 11.56 1st quartile 7.00 median 10.00 3rd quartile 16.00 interquartile range 9.00 mode 10.00 low extremes 0 low outliers 0 high outliers 0 high extremes0
FIND: MEAN Dat a lower
0.5 3.5 6.5 9.5 12.5 15.5 18.5 21.5 24.5
< < < < < < < < <
uppe r
frequency
Mid. point
3.5 6.5 9.5 12.5 15.5 18.5 21.5 24.5 27.5
6 6 7 10 5 9 4 1 2
2.0 5.0 8.0 11.0 14.0 17.0 20.0 23.0 26.0
50
FORMULA: X =
∑x = 586 N
50
= 11 .72
Find: MEDIAN
cf 12 30 56 110 70 153 80 23 52 586
Data lower
upper
0.5 3.5 6.5 9.5 12.5 15.5 18.5 21.5 24.5
< < < < < < < < <
frequency
3.5 6.5 9.5 12.5 15.5 18.5 21.5 24.5 27.5
6 6 7 10 5 9 4 1 2
cf 6 12 19 29 34 43 47 48 50
50
Formula: n −CF < ~ .c = x = Lb + 2 fm
Given: Lb = 9.5
n = 50 Cf < = 19 fm =10
c =3
SOLUTION: 25 −19 ~ x = 9.5 + 3 = 11 .3 10
Find: MODE Data lower
upper
0.5 3.5 6.5 9.5 12.5 15.5 18.5
< < < < < < <
3.5 6.5 9.5 12.5 15.5 18.5 21.5
frequency 6 6 7 10 5 9 4
21.5 24.5
< <
24.5 27.5
1 2 50
FORMULA: d1 Xˆ = Lmo + d1 + d 2
.c
GIVEN:
d1 = f mo − f 1 = 10 − 7 = 3 d 2 = f mo − f 2 = 10 − 5 = 5 Lmo = 9.5
c =3
SOLutION: 3 Xˆ = 9.5 + .3 = 10 .625 3 +5
Data lower
0.5 3.5 6.5 9.5 12.5 15.5 18.5 21.5 24.5
upper
< < < < < < < < <
3.5 6.5 9.5 12.5 15.5 18.5 21.5 24.5 27.5
frequency 6 6 7 10 5 9 4 1 2 50
FIND: Q1 FORMULA:
cf 6 12 19 29 34 43 47 48 50
Q1 = LQ1
n − Cf
GIVEN:
.c
n 50 = = 12 .5 4 4
LQ1 = 3.5 CF
FQ1 = 6 C =3
SOLUTION: 12 .5 − 6 Q1 = 3.5 + .3 = 6.75 6 This mean that 25% or 13 among 50 individual who’s serve in the Government Agencies have a service year of 7 years or less.
Find: q3 Formula: Q3 = LQ3
3N − CF
.C
Given: 3n 3(50 ) = = 37 .5 4 4 LQ 3 =15 .5
CF
FQ3 = 9 C =3
Solution: 37 .5 − 34 Q3 = 15 .5 + .3 = 16 .67 9
This mean that 75% or 38 among 50 individual who’s serve in the Government Agencies have a service year of 17 years or less.
Find: p80 Formula: L P 80
N + ( N ) −CF
.C
Given: N 80 (50 ) = (50 ) = 40 100 100
LP80 = 15 .5 CF
FP80 = 9 C =3
Solution: 40 P80 = 15 .5 + − 34 .3 = 17 .5 9 This mean that 80% or 40 among 50 individual who’s serve in the Government Agencies have a service year of 18 years or less.
Find: p30 Formula: LP30
N + 100 ( N ) −CF
.C
Given: N 30 (N ) = (50 ) = 15 100 100
LP30 = 6.5
CF
Solution: 15 −12 p30 = 6.5 + .3 = 7.78 7 This mean that 30% or 15 among 50 individual who’s serve in the Government Agencies have a service year of 8 years or less.
FIND: D2 FORMULA: Dn = LDn
n ( N ) − CF
GIVEN: n 2 (N ) = (50 ) = 10 10 10 L PN = 3.5 n = 10 10 CF
SOLUTION: 10 − 6 Dn = 3.5 + .3 = 5.5 6
.c
This mean that 20% or 10 among 50 individual who’s serve in the Government Agencies have a service year of 6 years or less.
Find: d6 Formula: Dn = LDn
n ( N ) − CF
.c
Given: n 6 (N ) = (50 ) = 30 10 10 LPN = 12 .5 n = 30 10 CF
Solution: 30 − 29 Dn = 12 .5 + .3 = 13 .1 5 This mean that 60% or 30 among 50 individual who’s serve in the Government Agencies have a service year of 13 years or less.
GROUP 1
No. of years in service in any Government Agencies
Angelito Malayao Marva Joppa V. Consegra Alex Jessafe L. Divinasflores Jowelle Krys C. Brioso Mylen Orobia Marivic Miranda
BSBM 2-1
Stem and Leaf plot for stem unit = 10 leaf unit = 1 Frequency 19 0 25 1 6 2 50
#1
Stem Leaf 2233334456667778899 0000000022335556666666889 001256
H −L 26 − 2 = = 3.612 1 + 3.322 log N 1 + 3.322 log 50
Frequency Distribution - Quantitative Data lower 0.5 3.5 6.5 9.5 12.5 15.5 18.5 21.5 24.5
upper midpoint < < < < < < < < <
3.5 6.5 9.5 12.5 15.5 18.5 21.5 24.5 27.5
2.0 5.0 8.0 11.0 14.0 17.0 20.0 23.0 26.0
cumulative frequency frequency 6 6 7 10 5 9 4 1 2 50
6 12 19 29 34 43 47 48 50
Histogram 50 Individuals serve in any Government Agencies 12
Frequency
10 8 6 4 2
27 .5
24 .5
21 .5
18 .5
15 .5
12 .5
9. 5
6. 5
3. 5
0. 5
0
No. of years in any Government Agencies
FrequencyPolygon 50 Individuals serve in any Government Agencies 12.0
Frequency
10.0 8.0 6.0 4.0 2.0 0.0 -3
4
10
16
22
Descriptive statistics #1 count 50 mean 11.56 1st quartile 7.00 median 10.00 3rd quartile 16.00 interquartile range 9.00 mode 10.00 low extremes 0 low outliers 0 high outliers 0 high extremes0
FIND: MEAN Dat a lower
0.5 3.5 6.5 9.5 12.5 15.5 18.5 21.5 24.5
< < < < < < < < <
uppe r
frequency
Mid. point
3.5 6.5 9.5 12.5 15.5 18.5 21.5 24.5 27.5
6 6 7 10 5 9 4 1 2
2.0 5.0 8.0 11.0 14.0 17.0 20.0 23.0 26.0
50
FORMULA: X =
∑x = 586 N
50
= 11 .72
Find: MEDIAN
cf 12 30 56 110 70 153 80 23 52 586
Data lower
upper
0.5 3.5 6.5 9.5 12.5 15.5 18.5 21.5 24.5
< < < < < < < < <
frequency
3.5 6.5 9.5 12.5 15.5 18.5 21.5 24.5 27.5
6 6 7 10 5 9 4 1 2
cf 6 12 19 29 34 43 47 48 50
50
Formula: n −CF < ~ .c = x = Lb + 2 fm
Given: Lb = 9.5
n = 50 Cf < = 19 fm =10
c =3
SOLUTION: 25 −19 ~ x = 9.5 + 3 = 11 .3 10
Find: MODE Data lower
upper
0.5 3.5 6.5 9.5 12.5 15.5 18.5
< < < < < < <
3.5 6.5 9.5 12.5 15.5 18.5 21.5
frequency 6 6 7 10 5 9 4
21.5 24.5
< <
24.5 27.5
1 2 50
FORMULA: d1 Xˆ = Lmo + d1 + d 2
.c
GIVEN:
d1 = f mo − f 1 = 10 − 7 = 3 d 2 = f mo − f 2 = 10 − 5 = 5 Lmo = 9.5
c =3
SOLutION: 3 Xˆ = 9.5 + .3 = 10 .625 3 +5
Data lower
0.5 3.5 6.5 9.5 12.5 15.5 18.5 21.5 24.5
upper
< < < < < < < < <
3.5 6.5 9.5 12.5 15.5 18.5 21.5 24.5 27.5
frequency 6 6 7 10 5 9 4 1 2 50
FIND: Q1 FORMULA:
cf 6 12 19 29 34 43 47 48 50
Q1 = LQ1
n − Cf
GIVEN:
.c
n 50 = = 12 .5 4 4
LQ1 = 3.5 CF
FQ1 = 6 C =3
SOLUTION: 12 .5 − 6 Q1 = 3.5 + .3 = 6.75 6 This mean that 25% or 13 among 50 individual who’s serve in the Government Agencies have a service year of 7 years or less.
Find: q3 Formula: Q3 = LQ3
3N − CF
.C
Given: 3n 3(50 ) = = 37 .5 4 4 LQ 3 =15 .5
CF
FQ3 = 9 C =3
Solution: 37 .5 − 34 Q3 = 15 .5 + .3 = 16 .67 9
This mean that 75% or 38 among 50 individual who’s serve in the Government Agencies have a service year of 17 years or less.
Find: p80 Formula: L P 80
N + ( N ) −CF
.C
Given: N 80 (50 ) = (50 ) = 40 100 100
LP80 = 15 .5 CF
FP80 = 9 C =3
Solution: 40 P80 = 15 .5 + − 34 .3 = 17 .5 9 This mean that 80% or 40 among 50 individual who’s serve in the Government Agencies have a service year of 18 years or less.
Find: p30 Formula: LP30
N + 100 ( N ) −CF
.C
Given: N 30 (N ) = (50 ) = 15 100 100
LP30 = 6.5
CF
Solution: 15 −12 p30 = 6.5 + .3 = 7.78 7 This mean that 30% or 15 among 50 individual who’s serve in the Government Agencies have a service year of 8 years or less.
FIND: D2 FORMULA: Dn = LDn
n ( N ) − CF
GIVEN: n 2 (N ) = (50 ) = 10 10 10 L PN = 3.5 n = 10 10 CF
SOLUTION: 10 − 6 Dn = 3.5 + .3 = 5.5 6
.c
This mean that 20% or 10 among 50 individual who’s serve in the Government Agencies have a service year of 6 years or less.
Find: d6 Formula: Dn = LDn
n ( N ) − CF
.c
Given: n 6 (N ) = (50 ) = 30 10 10 LPN = 12 .5 n = 30 10 CF
Solution: 30 − 29 Dn = 12 .5 + .3 = 13 .1 5 This mean that 60% or 30 among 50 individual who’s serve in the Government Agencies have a service year of 13 years or less.
GROUP 1
No. of years in service in any Government Agencies
Angelito Malayao Marva Joppa V. Consegra Alex Jessafe L. Divinasflores Jowelle Krys C. Brioso Mylen Orobia Marivic Miranda
BSBM 2-1
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