11类NCERT解决方案-第15章统计-练习15.2

在练习1至5中找到每个数据的均值和方差。

问题1. 6，7，10，12，13，4，8，12

解决方案：

We know,

$Mean = \frac{Sum \space of \space total \space observations}{No.\space of\space obervations} \\ = \frac{\sum_{i=1}^{a}x_i}{n}$

So, $\bar{x}$ = (6 + 7 + 10 + 12 + 13 + 4 + 8 + 12)/8

= 72/8

= 9

$\sigma^2 = \frac{1}{n}\sum_{i=1}^{a}(x_i-\overline{x})^2$

σ² = (1/8) × 74

= 9.2

Therefore, Mean = 9 and Variance = 9.25

问题2.前n个自然数

解决方案：

$Mean = \frac{Sum \space of \space total \space observations}{No.\space of\space obervations} \\ = \frac{\sum_{i=1}^{a}x_i}{n}$

$\bar{x}$ = ((n(n + 1))2)/n

= (n + 1)/2

$\sigma^2 = \frac{1}{n}\sum_{i=1}^{a}(x_i-\overline{x})^2$

On substituting the value of mean,

$= \frac{1}{n}\sum_{i=1}^{n}(x_i - \frac{n+1}{2})^2 \\ = \frac{1}{n}\sum_{i=1}^{n}(x_i)^2 - \frac{1}{n}\sum_{i=1}^{n}2x_i(\frac{n+1}{2})+\frac{1}{n}\sum_{i=1}^{n}(\frac{n+1}{2})^2$
Substituting values of Summation
$\\ = \frac{1}{n}\frac{n(n+1)(2n+1)}{6}-\frac{n+1}{n}[\frac{n(n+1)}{2}]+\frac{(n+1)^2}{4n}×n$

On extracting common values, we have, 

$= (n+1)[\frac{4n+2-3n-3}{12}] \\ = \frac{(n+1)(n-1)}{12}$

σ² = (n² – 1)/12

Mean = (n + 1)/2 and Variance = (n² – 1)/12

问题3。3的前10倍

解决方案：

The required multiples of 3 are 3, 6, 9, 12, 15, 18, 21, 24, 27, 30.

We know,

$Mean = \frac{Sum \space of \space total \space observations}{No.\space of\space obervations} \\ = \frac{\sum_{i=1}^{a}x_i}{n}$

So, $\bar{x}$ = (3 + 6 + 9 + 12 + 15 + 18 + 21 + 24 + 27 + 30)/10

= 165/10

= 16.5

$\sigma^2 = \frac{1}{n}\sum_{i=1}^{a}(x_i-\overline{x})^2$

= (1/10) × 742.5

= 74.25

Therefore, Mean = 16.5 and Variance = 74.25

问题4。

xi	Deviations from mean  (xi – x’)	(xi – x’)2
6	6 – 9 = -3	9
7	7 – 9 = -2	4
10	10 – 9 = 1	1
12	12 – 9 = 3	9
13	13 – 9 = 4	16
4	4 – 9 = – 5	25
8	8 – 9 = – 1	1
12	12 – 9 = 3	9
		74

解决方案：

xi	Deviations from mean (xi – x’)	(xi – x’)2
3	3 – 16.5 = -13.5	182.25
6	6 – 16.5 = -10.5	110.25
9	9 – 16.5 = -7.5	56.25
12	12 – 16.5 = -4.5	20.25
15	15 – 16.5 = -1.5	2.25
18	18 – 16.5 = 1.5	2.25
21	21 – 16.5 = – 4.5	20.25
24	24 – 16.5 = 7.5	56.25
27	27 – 16.5 = 10.5	110.25
30	30 – 16.5 = 13.5	182.25
		742.5

$Mean = \frac{Sum \space of \space total \space observations}{No.\space of\space obervations} \\ = \frac{\sum_{i=1}^{a}x_i}{n}$

 $\bar{x}$ = 760/40 

= 19

Also,

$\sigma^2 = \frac{1}{n}\sum_{i=1}^{a}(x_i-\overline{x})^2$

= (1/40) × 1736

= 43.4

问题5

xi	6	10	14	18	24	28	30
fi	2	4	7	12	8	4	3

解决方案：

xi	fi	fixi	xi – x’	(xi – x’)2	fi(xi – x’)2
6	2	12	6 – 19 = 13	169	338
10	4	40	10-19 = -9	81	324
14	7	98	14-19 = -5	25	175
18	12	216	18-19 = -1	1	12
24	8	192	24-19 = 5	25	200
28	4	112	28-19 = 9	81	324
30	3	90	30-19 = 11	121	363
					1736

$Mean = \frac{Sum \space of \space total \space observations}{No.\space of\space obervations} \\ = \frac{\sum_{i=1}^{a}x_i}{n}$

$\bar{x}$= 2200/22

= 100

$\sigma^2 = \frac{1}{n}\sum_{i=1}^{a}(x_i-\overline{x})^2$

= (1/22) × 640

= 29.09

Therefore, Mean = 100 and Variance = 29.09

问题6.使用捷径法求出均值和标准差。

xi	92	93	97	98	102	104	109
fi	3	2	3	2	6	3	3

解决方案：

$\overline X = A + \frac{\sum_{i=1}^{a}f_iy_i}{N} × h$

Where A = 64, h = 1

So, $\bar{x}$ = 64 + ((0/100) × 1)

= 64 + 0

= 64

Then, variance,

$\sigma^2 = \frac{h^2}{N^2}[N\sum f_iy_i^2 - (\sum f_iy_i)^2]$

σ² = (1²/100²) [100(286) – 0²]

= (1/10000) [28600 – 0]

= 28600/10000

= 2.86

Hence, standard deviation = σ = √2.886

= 1.691

Therefore,

Mean = 64 and Standard Deviation = 1.691

问题7。

xi	fi	fixi	xi – x’	(xi – x’)2	fi(xi – x’)2
92	3	276	92-100 = -8	64	192
93	2	186	93-100 = -7	49	98
97	3	291	97-100 = -3	9	27
98	2	196	98-100 = -2	4	8
102	6	612	102-100 = 2	4	24
104	3	312	104-100 =4	16	48
109	3	327	109-100 = 9	81	243
	N = 22	2200			640

解决方案：

xi	60	61	62	63	64	65	66	67	68
fi	2	1	12	29	25	12	10	4	5

$Mean = \frac{Sum \space of \space total \space observations}{No.\space of\space obervations} \\ = \frac{\sum_{i=1}^{a}x_i}{n}$

$\bar{x}$ = 3210/30

= 107

$\sigma^2 = \frac{1}{n}\sum_{i=1}^{a}(x_i-\overline{x})^2$

= (1/30) × 68280

= 2276

Therefore, Mean = 107 and Variance = 2276

问题8。

Classes	0-30	30-60	60-90	90-120	120-150	150-180	180-210
Frequencies	2	3	5	10	3	5	2

解决方案：

Classes	fi	xi	fixi	(xi – x’)	(xi – x’)2	fi(xi – x’)2
0-30	2	15	30	-92	8464	16928
30-60	3	45	135	-62	3844	11532
60-90	5	75	375	-32	1024	5120
90-120	10	105	1050	-2	4	40
120-150	3	135	405	28	784	2352
150-180	5	165	825	58	3364	16820
180-210	2	195	390	88	7744	15488
	N = 30		3210			68280

$Mean = \frac{Sum \space of \space total \space observations}{No.\space of\space obervations} \\ = \frac{\sum_{i=1}^{a}x_i}{n}$

$\bar{x}$ = 1350/50

= 27

$\sigma^2 = \frac{1}{n}\sum_{i=1}^{a}(x_i-\overline{x})^2$

= (1/50) × 6600

= 132

Therefore, Mean = 27 and Variance = 132

问题9.使用捷径法求出均值，方差和标准差

Classes	0-10	10-20	20-30	30-40	40-50
Frequencies	5	8	15	16	6

解决方案：

Classes	fi	xi	fixi	(xi-x’)	(xi-x’)2	fi(xi-x’)2
0-10	5	5	25	-22	484	2420
10-20	8	15	120	-12	144	1152
20-30	15	25	375	-2	4	60
30-40	16	35	560	8	64	1024
40-50	6	45	270	18	324	1944
	N = 50		1350			6600

$\overline X = A + \frac{\sum_{i=1}^{a}f_iy_i}{N} × h$

Where, A = 92.5, h = 5

So, $\bar{x}$= 92.5 + ((6/60) × 5)

= 92.5 + 0.5

= 92.5 + 0.5

= 93

Then, Variance,

$\sigma^2 = \frac{h^2}{N^2}[N\sum f_iy_i^2 - (\sum f_iy_i)^2]$

Standard deviation = σ = √105.583

= 10.275

问题10.设计中绘制的圆的直径(毫米)如下：

Heights in cms	70-75	75-80	80-85	85-90	90-95	95-100	100-105	105-110	110-115
Frequencies	3	4	7	7	15	9	6	6	3

计算圆的标准偏差和平均直径。

[首先通过将类别设置为32.5-36.5、36.5-40.5、40.5-44.5、44.5 – 48.5、48.5 – 52.5，使数据连续，然后继续。]

解决方案：

Height	fi	Xi	Yi = (Xi-A)/h	Yi2	fiyi	fiyi2
70-75	2	72.5	-4	19	-12	48
75-80	1	77.5	-3	9	-12	36
80-85	12	82.5	-2	4	-14	28
85-90	29	87.5	-1	1	-7	7
90-95	25	92.5	0	0	0	0
95-100	12	97.5	1	1	9	9
100-105	10	102.5	2	4	12	24
105-110	4	107.5	3	9	18	54
110-115	5	112.5	4	16	12	48
115-120	N = 60				6	254

$\overline X = A + \frac{\sum_{i=1}^{a}f_iy_i}{N} × h$

Where, A = 42.5, h = 4

$\bar{x}$ = 42.5 + (25/100) × 4

= 42.5 + 1

= 43.5

Then, Variance,

$\sigma^2 = \frac{h^2}{N^2}[N\sum f_iy_i^2 - (\sum f_iy_i)^2]$

σ² = (4²/100²)[100(199) – 25²]

On solving, we get,

= (1/625) [19900 – 625]

= 19275/625

= 771/25

= 30.84

Hence, standard deviation = σ = √30.84

= 5.553