Estadistica

2 | 5 | 6 | 8 | 8 |
2 | 5 | 6 | 8 | 8 |
2 | 5 | 6 | 8 | 8 |
2 | 6 | 6 | 8 | 9 |
3 | 6 | 7 | 8 | 9 |
4 | 6 | 7 | 8 | 9 |
DATOS A ANALIZAR NO AGRUPADOS
Media Aritmética:
X = n=1∞X! = 2+2+2+2+3+4+5+5+5+6+6+6+6+6+6+6+7+7+8+8+8+8+8+8+8+8+8+9+9+9
30
X=185
30

X= 6.16

Media Geométrica
G=X*X2*X3…X37
G= 2* 2*2* 2* 3*4* 5*5* 5* 6*6*6* 6*6* 6* 6* 7*7*8*8*8*8*8*8*8* 8* 8* 9* 9*9ݒ

G= 3.221094644x1022

MEDIA CODIFICADA
Xc=1=1nx!-c
_____________
N

2 – 2 = 0 185 / 30 = 6.16
2 – 2 = 0
2 – 2 = 0 125 / 30 = 4.16 + 2 = 6.16
2 – 2 = 0
3 – 2 = 1
4 – 2 = 2
5 – 2 = 3
5 – 2 = 3
5 – 2 = 3
6 – 2= 4
6 – 2= 4
6 – 2= 4
6 – 2= 4
6 – 2= 4
6 – 2= 4
6 – 2= 4
7 – 2 = 5
7 – 2 = 5
8 – 2 = 6
8 – 2 = 6
8 – 2 =6
8 – 2 = 6
8 – 2 = 6
8 – 2 = 6
8 – 2 = 6
8 – 2 = 6
8 – 2 = 6
9 – 2 = 7
9 – 2 = 7
9 – 2 = 7
185 125

Análisis Estadísticos
GRADO DEL COMPORTAMIENTO DEL ALUMNO ANTES DE PROBAR EL CIGARRO
DATOS A ANALIZAR
2 , 2 , 2 , 2 , 3 , 4 , 5 , 5 , 5 , 6 , 6 , 6 , 6 , 6 , 6 , 6 , 7 , 7 , 8 , 8 , 8 , 8 , 8 , 8 , 8 , 8 , 8 ,9 , 9 , 9

A= 9 – 2
A= 7

K= 1 + 3.333 + log (n)RANGO DE CLASE= AMPLITUD
K= 1 + 3.333 + log (30) = (1.4771) K
K= 1 + 4.8010 RC = 7 = 1.4
K= 5.801 5
K= 5 RC= 2

CLASE | MARCA | FRECUENCIA | MK | F.RELATIVA | F. ACUM. | F. COMPLEM. |
0 – 0 | 0 | 0 | 0 | 0/30=0X100=0% | 0 | 30 |
1 – 2 | IIII | 4 | 1.5 | 4/30=0.1333X100=13.33% | 4 | 26 |
3 – 4 | II | 2 | 3.5 | 2/30=0.0666X100=6.66% | 6 | 24 |
5 – 6 | IIIIIIIIII | 10 | 5.5 | 10/30=0.3333X100=33.33% | 16 | 14 |
7 – 8 | IIIIIIIIIII | 11 | 7.5 | 11/30=0.3666X100=36.66% | 27 | 3 |
9 – 10 | III | 3 | 9.5 | 3/30=0.1X100=10% | 30 | 0 |
11 –12 | 0 | 0 | 11.5 | 0/30=0X100=0% | 30 | 0 |
30 |

OBTENCIÓN DE TENDENCIAS CENTRALES

a. MEDIA.
M=k=0nXi*MK
__________
N

X= (4 X 1.5) + (2 X 3.5) + (10 + 5.5) + (11 X 7.5) + (3 X 9.5) + (0 X 11.5)=

X= 6 + 7 + 55 + 82.5 + 28.5 + 0 = 179

X=179
30

X= 5.966

a. MEDIANA

X=Li+ n2-faacm*C
_____________
Fcm

Cm= n = 30 =
2 2
X = 5 + 302- 6 * 2
__________
10


X= 5 + (15 – 6)
10

X= 5 + (9) x 2
10

X= 5 + 0.9 (2) =

X= 1.8 + 5=

X= 6.8

MODA

X=Li+D1D1-D2*C

D1= 11 – 10 = 1
D2= 11 – 3 = 8

X= X = 7 + 11+ 8*2

X = 7 + 19*2

X = 7 + (0.11) * 2 = 0.22

X= 7 + 0.22

X= 7.22










TABLAS DE FRECUENCIAS
GRADO DEL COMPORTAMIENTO DEL ALUMNO ANTES DE PROBAR EL CIGARRO
GRAFICA DE BARRAS

GRAFICA DE PASTEL


A) 0 X 360° = 0
B) 0.1333 X 360° = 47.98
C) 0.6666 X 360° = 23.976 = 71.956D) 0.33 X 360° = 119.988 = 191.944
E) 0.3666 X 360° = 131.976 = 323.92
F) 0.1 X 360° = 36 = 359.92
G) 0 X 360° = 0
COMPROBACION = 360 – 359.92 = 0.08

POLIGONOS DE FRECUENCIA

OJIVAS ASCENDENTES Y DESCENDENTES

GRAFICA DE HISTOGRAMA



MODA DE DISPERCION


1.- MEDIDA DE DISPERCION 185/30 = 6.16
AMPLITUD: VM – VMA = 9 – 2 = 8 DESVIACION RESPECTO A LA MEDIA
A = 8 ((DR)) = i=InX-X

2.-M= i=InXi
_________________
n
m=i=InXi
M= 185 / 30 = µ 6.16
OBTENER DMR = nX-n
1. 2 – 2.16 = - 4.16
2. 2 – 2.16 = - 4.16
3. 2 – 2.16 = - 4.16
4. 2 – 2.16 = -...