1) Primero ordenemos de menor a mayor
12, 3 - 12, 5 - 12, 6 - 12, 7 - 12, 8 - 13, 1 - 13, 5 - 13, 8 - 13, 8 - 13, 9
14, 2 - 14, 2 - 14, 4 - 14, 5 - 14, 8 - 14, 9 - 15, 2 - 15, 2 - 15, 3 - 15, 3
15, 5 - 15, 6 - 15, 8 - 15.
9 - 16, 4 - 16, 4 - 16, 5 - 16, 5 - 16, 7 - 16, 7
16, 8 - 16, 9 - 17, 0 - 17, 2 - 17, 3 - 17, 5 - 17, 5 - 17, 8 - 17, 9 - 17, 9
2) Hallemos el ancho del intervalo utilizando la siguiente fórmula <img src="https://tex.z-dn.net/?f=w%3D%5Cdfrac%7BR%7D%7Bk%7D" />
donde R es la diferencia entre el mayor y el menor : R = 17.
9 - 12.
3 = 5.
6
y k viene dada por la fórmula de Sturges <img src="https://tex.z-dn.net/?f=k%3D1%2B3.3%5Clog%20n" />
en este caso n es la cantidad de datos : n = 40, entonces k = {5, 6, 7}, tomemos k = 5
Entonces <img src="https://tex.z-dn.net/?f=w%3D%5Cdfrac%7B5.6%7D%7B5%7D%3D1.12" />
3) pongamos en intervalos
Para hallar los intervalos, sumamos al elemento menor del intervalos másw
INTERVALOS - - - - - - - - - - - - - - - FRECUENCIA
[12.
30 ; 13.
42> .
6
[13.
42 ; 14.
54> .
8
[14.
54 ; 15.
66>. 8
[15.
66 ; 16.
78>. 8
[16.
78 ; 17.
90>. 10 = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = =
TOTAL.
40 - - - - - - - -
INTERVALOS - - - - - - - - - - - - - - - xi (marcas de clase) - - - - - (hi)Freq rel
[12.
30 ; 13.
42> .
12. 86 .
0. 15
[13.
42 ; 14.
54> .
13. 98.
0. 20
[14.
54 ; 15.
66>. 15.
10. 0.
20
[15.
66 ; 16.
78>. 16.
22. 0.
20
[16.
78 ; 17.
90>. 17.
34. 0.
25
Promedio <img src="https://tex.z-dn.net/?f=%5Cdisplaystyle%0A%5Coverline%20x%3D%5Csum_%7Bi%3D1%7D%5E%7B5%7Dx_i%5Ccdot%20h_i" /> <img src="https://tex.z-dn.net/?f=%5Coverline%20x%20%3D%2012.86%280.15%29%2B13.98%280.20%29%2B15.10%280.20%29%2B16.22%280.20%29%2B17.34%280.25%29%5C%5C%20%5C%5C%0A%5Cboxed%7B%5Coverline%20x%20%3D15.324%7D" />.