Procedure

1. I flipped the 20 coins and counted the number of dropped arms. I repeated it 50 times. The counts recorded in the Table # 2.

Table # 2 The number of dropped arms

№№ №	xi	xi	xi2	№	xi	xi	xi2	№	xi	xi	xi2	№	xi	xi	xi2	№	xi	xi	xi2
1	12	-2	4	11	10	0	0	21	11	-1	1	31	14	-4	16	41	8	2	4
2	6	4	16	12	11	-1	1	22	11	-1	1	32	12	-2	4	42	8	2	4
3	9	1	1	13	8	2	4	23	10	0	0	33	8	2	4	43	12	-2	4
4	11	-1	1	14	10	0	0	24	11	-1	1	34	13	-3	9	44	8	2	4
5	12	-2	4	15	9	1	1	25	13	-3	9	35	8	2	4	45	6	4	16
6	8	2	4	16	10	0	0	26	12	-2	4	36	12	-2	4	46	10	0	0
7	10	0	0	17	9	1	1	27	9	1	1	37	10	0	0	47	13	-3	9
8	15	-5	25	18	11	-1	1	28	12	-2	4	38	12	-2	4	48	11	-1	1
9	10	0	0	19	10	0	0	29	11	-1	1	39	6	4	16	49	10	0	0
10	8	2	4	20	8	2	4	30	9	1	1	40	11	-1	1	50	10	0	0

The number of intervals K will be equal to the quotient of the division

, _{, then K=9, L=1}

_xmin=6 x_max=15

1 - interval (6+L)=> (6 + 1) = 7

2 - interval (6+2L)=> (6+2) = 8

3 - interval (6+3L)=> (6+3) = 9

4 - interval (6+4L)=> (6+4) = 10

5 - interval (6+5L)=> (6+5) = 11

6 - interval (6 + 6L)=> (6+6) = 12

7 - interval (6+7L)=> (6+7) = 13

8 - interval (6+8L)=> (6+8) = 14

9 - interval (6+9L)=> (6+9) = 15

Table#3

The data for the histogram and the Gaussian curve

J	ni	ni/NL			e-	f(x)
1(7)	0	0	-3.5	1.516	0.219	0.0616
2(8)	9	0.18	-2.5	0.7735	0.461	0.1297
3(9)	5	0.1	-1.5	0.2784	0.757	0.213
4(10)	11	0.22	-0.5	0.0309	0.9695	0.2729
5(11)	9	0.18	0.5	0.0309	0.9695	0.2729
6(12)	8	0.16	1.5	0.2784	0.757	0.213
7(13)	3	0.06	2.5	0.7735	0.461	0.1297
8(14)	1	0.02	3.5	1.516	0.219	0.0616
9(15)	1	0.02	4.5	2.506	0.0815	0.0229
0	0	0	-4.5	2. 506	0.0815	0.0229

=198 – sum of absolute square value.

_{= 2.01 - mean square value}

_{e=2.72 – constant.} =4.04 - dispersion.

- Gauss curve.

= 0.287

. = 0.574

х = <x>  x

x=10.160.574

_Conclusion: We have worked hard this lab, and achieved results. we have learned to work with the laws of mistakes. learned to see deferent of error from each other. Probability theory in fact allow us to calculate the number of favorable cases.