Генетика. Часть 5. Статистические методы изучения изменчивости. Крюков В.И. - 36 стр.

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. .  5. . 
. –: , 2006. –208 .
36
 12


  
 ()





(f)
-

()
f·a f·a
2
1,51
–
1,60
1,61–1,70
1,71–1,80
1,81–1,90
1,91–2,00
2,01–2,10
2,11–2,20
2,21–2,30
1,66
1,76
1,86=
1,96
2,06
2,16
2,26
3
11
21
25
20
13
4
1
–
3
–2
–1
0
1
2
3
4
–
9
–2
–21
0
20
26
12
4
27
44
21
0
20
52
36
16
k
=0,1

n
=98
fa
=10
f·a
2
= 216
855,1
2
90,181,1
A 1,86
x
=1,86+0,1·(10 : 98) = 1,86+0,1·0,1 = 1,86 + 0,01 = 1,87 
01,02,21,0
98
10
98
216
1,0
2
15,048,11,019,21,0 .
%8%10008,0%100
87,1
15,0
%100
x
C
v
   -
, 
 
 .     -
. -
-
 ()     
     , , 
.     
       
     .  
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                                                                                              12


                                                             -
                 ( )                                                f·a        f·a2
                                           (f)        ( )
1,51–1,60               1,56               3          –3         –9            27
1,61–1,70               1,66               11         –2         –2            44
1,71–1,80               1,76               21         –1         –21           21
1,81–1,90               1,86=              25         0          0             0
1,91–2,00               1,96               20         1          20            20
2,01–2,10               2,06               13         2          26            52
2,11–2,20               2,16               4          3          12            36
2,21–2,30               2,26               1          4          4             16
k=0,1                                      n=98                    fa=10         f·a2= 216

                                  1,81 1,90
                             A              1,855 1,86
                                      2
            x =1,86+0,1·(10 : 98) = 1,86+0,1·0,1 = 1,86 + 0,01 = 1,87
                                                  2
                                 216        10
                       0,1                                 0,1   2,2 0,01
                                 98         98
                           0,1 2,19 0,1 1,48 0,15 .
                                 0,15
            Cv           100%         100% 0,08 100%                           8%
                       x         1,87

                                                                                               -
                                                                               ,

                                       .                                                       -
                                                                           .                   -
                                                                                               -
    (             )
                                                                           ,              ,
        .

                                                                                      .

————————————————————————————————————————————
     .    .   5.                               .
                    .– :        , 2006. –208 .

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