第11类RD Sharma解决方案-第32章统计-练习32.6

问题1 计算以下数据的平均值和标准差：

Expenditure (in ₹):	0-10	10-20	20-30	30-40	40-50
Frequency:	14	13	27	21	15

解决方案：

CI	f	x	u = (x – A)/h	fu	u2	fu2
0 – 10	14	5	-2	-28	4	56
10 – 20	13	15	-1	-13	1	13
20 – 30	27	25	0	0	0	0
30 – 40	21	35	1	21	1	21
40 – 50	15	45	2	30	4	60
	90			10		150

Given: 

Number of observations, N = 90 and A = 25

$\sum f_iu_i =10\\ \sum f_iu_i^2 =150$ 

h = 10

Mean = $\overline{x} = A+h(\frac{1}{N}\sum f_iu_i)$

$\overline{x}$ = 25 + 10(10/90) = 26.11 

$var(x) = h^2[\frac{1}{N} \sum f_iu_i^2 - (\frac{1}{N} \sum f_iu_i)^2]$

= 10[(150/90) – (10/90)²]

= 165.4 

Standard Deviation = √var(x) = √165.4 = 12.86  

问题2.计算以下数据的标准差：

Class:	0-30	30-60	60-90	90-120	120-150	150-180	180-210
Frequency:	9	17	43	82	81	44	24

解决方案：

CI	f	x	u = (x – A)/h	f × u	u2	fu2
0 – 30	9	15	-3	-27	9	81
30 – 60	17	45	-2	-34	4	68
60 – 90	43	75	-1	-43	1	43
90 – 120	82	105	0	0	0	0
120 – 150	81	135	1	81	1	81
150 – 180	44	165	2	88	4	176
180 – 210	24	195	3	72	9	216
	300			137		665

Given: N = 300 and A =105

$\sum f_iu_i =137\\ \sum f_iu_i^2 =665$

h = 30

Mean = $\overline{x} = A+h(\frac{1}{N}\sum f_iu_i)$

$\overline{x}$ = 105 + 30(137/300) = 118.7

$var(x) = h^2[\frac{1}{N} \sum f_iu_i^2 - (\frac{1}{N} \sum f_iu_i)^2]$

= 900[(665/300) – (137/300)²] 

= 1807.31 

Standard Deviation = √var(x) = √1807.31 = 42.51

问题3.计算以下分布的AM和SD：

Class:	0-10	10-20	20-30	30-40	40-50	50-60	60-70	70-80
Frequency:	18	16	15	12	10	5	2	1

解决方案：

CI	f	x	u = (x – A)/h	f × u	u2	fu2
0 – 10	18	5	-3	-54	9	162
10 – 20	16	15	-2	-32	4	64
20 – 30	15	25	-1	-15	1	15
30 – 40	12	35	0	0	0	0
40 – 50	10	45	1	10	1	10
50 – 60	5	55	2	10	4	20
60 – 70	2	65	3	6	9	18
70 – 80	1	75	4	4	16	16
	79			-71		305

Given:  N = 79 and A =35

$\sum f_iu_i =-71\\ \sum f_iu_i^2 =305$

h = 10

Mean = $\overline{x} = A+h(\frac{1}{N}\sum f_iu_i)$

$\overline{x}$ = 35 + 10(-71/79) = 26.01

$var(x) = h^2[\frac{1}{N} \sum f_iu_i^2 - (\frac{1}{N} \sum f_iu_i)^2]$

= 100[(305/79) – (-71/79)²] 

= 305.30

Standard Deviation  = √var(x) = √305.30 = 17.47

问题4.一个学生分别获得100个观测值的平均值和标准差，分别为40和5.1。后来发现，一个观测值被错误地复制为50，正确的数字是40。找到正确的均值和SD

解决方案：

According to question, we have, 

n = 100 , $\overline{x} = 40 , \sigma = 5.1 \\ \overline{x} = \frac{1}{n}\sum x_i \\ \sum x_i = n\overline{x} = 4000 \\ Incorrect \sum x_i = 4000$

And, also 

$\sigma = 5.1 \\ \sigma^2 = 26.01 \\ \frac{1}{n}\sum x_i^2 - Mean^2 = 26.01 \\ \frac{1}{100}\sum x_i^2 - 1600 = 26.01 \\$

$\sum x_i^2$ = 1626.01 x 100 

Incorrected $\sum x_i^2$ = 162601

On replacing the incorrect observation of 50 by 40, we get,

Incorrect $\sum x_i$ = 4000 

Corrected $\sum x_i$ = 4000 – 50 + 40 = 3990

Incorrected $\sum x_i^2$= 162601 

Corrected $\sum x_i^2$ = 162601 – 50² + 40² = 161701

Now, we have,

Corrected Mean = 39.90

Corrected Variance = (1/100)(Corrected $\sum x_i^2$) – (Corrected mean)² 

 $= \frac{161701}{100} - (\frac{3990}{100})^2 \\ = \frac{161701 * 100 - (3990)^2}{100^2} \\ = \frac{16170100-15920100}{10000}$

= 25

Corrected standard deviation = √25 = 5

问题5.计算以下分布的平均值，中位数和标准偏差：

Class-interval	31-35	36-40	41-45	46-50	51-55	56-60	61-65	66-70
Frequency:	2	3	8	12	16	5	2	3

解决方案：

CI	Freq	Mid Value	ui	fiui	fiui2
31 – 35	2	33	-4	-8	32
36 – 40	3	38	-3	-9	27
41 – 45	8	43	-2	-16	32
46 – 50	12	48	-1	-12	12
51 – 55	16	53	0	0	0
56 – 60	5	58	1	5	5
61 – 65	2	63	2	4	8
66 – 70	2	68	3	6	18
	N = 50			Total = – 30	Total = 134

Now, using the given values, we have

Mean = 53 + 5 x (-30/50) 

= 50

Variance = 25 x ((134/50) – (9/25)

= 58

Standard Deviation = √58 

= 7.62

问题6.找到下面给出的频率分布的均值和方差：

xi	1 ≤ x < 3	3 ≤ x < 5	5 ≤ x < 7	7 ≤ x < 9
fi	6	4	5	1

解决方案：

The data can be converted to a continuous frequency distribution by subtracting 0.5 from lower limit and adding 0.5 to upper limit of each of the class interval. 

Given: N = 16 and A = 5.5

$\sum f_iu_i =-30\\ \sum f_iu_i^2 =116$ and h=1

Mean = $\overline{x} = A+h(\frac{1}{N}\sum f_iu_i)$

$\overline{x}$ = 5.5 + 1((1/6) x (-30))

= 3.625 

$var(x) = h^2[\frac{1}{N} \sum f_iu_i^2 - (\frac{1}{N} \sum f_iu_i)^2]$

= 1 [((1/16) x 116) – ((1/16) x (-30)²] 

= 3.74

问题7.下表显示了70罐咖啡的重量：

Class Interval	fi	xi	ui	fiui	ui2	fiui2
1 – 2	6	1.5	-4	-24	16	96
3 – 4	4	3.5	-2	-8	4	16
5 – 6	5	5.5	0	0	0	0
7 – 8	1	7.5	2	2	4	4
	N = 16			Total = -30		Total = 116

计算均值，方差和标准差。

解决方案：

Weight (in grams)	200-201	201-202	202-203	203-204	204-205	205-206
Frequency	13	27	18	10	1	1

Now, using the given values, we have

Mean = 203.5 + 2 x (-54/70)

= 201.9

Variance = 4 x (62/70) – (-54/70)

= 0.98

Standard Deviation = √0.98

= 0.099

问题8：100个观测值的平均值和标准偏差分别为40和10。如果在计算时错误地将两个观测值分别取为30和70而不是3和27，则找到正确的标准偏差。

解决方案：

Mean = 40

Standard Deviation = 10

n = 100

$\sum x_i = 40 * 100 = 4000$

Corrected Sum = 4000 – 30 +70 + 3 + 27 = 3930

Corrected mean = 39.3

Variance = 100

$100 = \frac{\sum x_i^2}{100} -(40)^2$

Incorrect \sum x_i^2 = 170000 

So, Corrected \sum x_i^2 = Incorrect $\sum x_i^2$ – (Sum of squares of incorrect values) + 

                                                                                    (Sum of squares of corrected values) 

Corrected$\sum x_i^2$ = 170000 – (900 + 4900) + (9+729)  

= 164938

$Corrected \space \sigma = \sqrt{\frac{Corrected \space \sum x_i^2}{n}-(Corrected \space Mean)^2} \\ = \sqrt{\frac{164938}{100}-(39.3)^2} \\$

 = 10.24

问题9.在计算10个读数的均值和方差时，一名学生错误地将52的读数用于正确的读数25。他获得的均值和方差分别为45和16。找到正确的均值和方差。

解决方案：

Mean = 45

Variance = 16

n = 10

$\sum x_i = 450$

So, Corrected Sum = 450 – 52 + 25 = 423

Corrected mean = 42.3

Variance = 16

1$16 = \frac{\sum x_i^2}{10} -(45)^2 \\ Incorrect \sum x_i^2 = 20410$

Corrected $\sum x_i^2$= Incorrect  $\sum x_i^2$– (Sum of squares of incorrect values) + 

                                                                             (Sum of squares of corrected values) 

Corrected $\sum x_i^2$  = 20410 – 2704 + 625 = 18331

$Corrected \space \sigma = \sqrt{\frac{Corrected \space \sum x_i^2}{n}-(Corrected \space Mean)^2} \\ = \sqrt{\frac{18331}{10}-(42.3)^2}$

= 6.62

So, Corrected variance = 6.62 * 6.62 = 43.82

问题10.计算以下频率分布的均值，方差和标准偏差：

CI	xi	fi	ui	fiui	fiui2
200 – 201	200.5	13	-15	-19.5	29.25
201 – 202	201.5	27	-1	-27	27
202 – 203	202.5	18	-0.5	-9	4.5
203 – 204	203.5	10	0	0	0
204 – 205	204.5	1	0.5	0.5	0.25
205 – 206	205.5	1	1	1	1
		N = 70		Total = – 54	Total = 62

解决方案：

Class	0-10	10-20	20-30	30-40	40-50	50-60
Frequency	11	29	18	4	5	3

Given:

Number of observations, N = 70 and A = 35

$\sum f_iu_i =-98\\ \sum f_iu_i^2 =250$

h = 10

Mean = $\overline{x} = A+h(\frac{1}{N}\sum f_iu_i)$

$\overline{x}$ = 35 + 10(-98/70) = -21

$var(x) = h^2[\frac{1}{N} \sum f_iu_i^2 - (\frac{1}{N} \sum f_iu_i)^2]$

= 100[(1/70) x 250 – (1/70) x (-98)²]

= 161 

Standard Deviation = √var(x) 

= √161

= 12.7