Ejercicio De Estadistica

DESARROLLO DE MEDIDAS DE TENDENCIA CENTRAL

DATO MAYOR: 65
DATO MENOR:30
RECORRIDO DE RANGO:35

DETERMINAR EL NUMERO DE INTERVALOS:M=1+3.3*40=6.88= aproxima a 7

AMPLITUD DEL INTERVALO
C=RANGO=35/7=5

INTERVALOS
Yj-Yj

Yj

Fi

30.1-35
35.1-40
40.1-45
45.1-50
50.1-55
55.1-6060.1-65

8
6
2
9
4
10
1

8
14
16
25
29
39
40

NUMERO DE DATOS 40/2 =20
1.

Md=Yj-1+0 [

nj+1 ]
Nj-1+nj-1

40.1+5= [ 125 ]14+25
Md =40.1+5=[ 25]
39
Md =40.1+5 x 0.64
Md =5*0.64=3.2
Md =40.1+3.2
Md = 43.3
2. Md =Yj-1+c [nj-nj-1 ]
( j-nj-1)+(nj-nj+1)

[

Md =16-14
(16-14)+(16-25)

[

Md =40.1+
Md =

[

2]
-7

]

2]
2+(9)

= -0.28

Md=40.1=5*(-0.28)
5 *(-0.28)=1.4
Md=40.1+(1.4)+38.6

[

3. Md= Yji – 1 + c

Md= 40.1 + 5

nj+1 - nj-1
2(2nj – nj-1 – nj+1)

[

nj+1 - nj-1
2(32 – 14 – 25)

Nota 2x16= 32
Md= 40.1 + 5[

Md= 40.1 + 5

[

]

11
2(-7)
5
-14

] = -0.78

Md = 40.1 + 5 x 0.78
Md = 5 x 0.78 = 40.1 + (-3.9)
Md = 36.2

 MEDIANAMe=Yji –1+c n/2 – Nj - 1
______________________

n/2=20

]
]

Me= 40.1+5 20 – 14
___________ = 0.3
20
0.3 ˟ 5 = 1.5 + 40.1= 41.6
 MEDIA
Xi___Xi ˟ fi_____

n
 INTERVALOS

___332.85 x 171___ = 28458.63

40

___30.1 + 35__ = 32.55

2
___35.1 + 40___ = 37.55

2
___40.1 +45__ = 42.55

2
___45.1 + 50__ = 47.55

2
___50.1 + 55__ = 52.55

2
___55.1 + 60___ = 57.55

2
___60.1 + 65___ = 62.55

2

332.85