第11类RD Sharma解决方案–第32章统计–练习32.3

问题1 计算以下分布的中位数的平均偏差：

Class	0-10	10-20	20-30	30-40	40-50
Frequency	5	10	20	5	10

解决方案：

Calculating the median:

Median is the middle term of the observation in ascending order, X_i,

Here, the middle term is 25.

Therefore, Median = 25

Mean Deviation, $MD = \frac{1}{n} \sum_{i=1}^{n}|d_i|$

MD= 1/50 × 450

= 9

Therefore, mean deviation is 9.

问题2：找到以下数据的均值与均值的偏差：

(一世)

Class Interval	xi	fi	Cumulative Frequency	|di| = |xi – M|	fi |di|
0-10	5	5	5	20	100
10-20	15	10	15	10	100
20-30	25	20	35	0	0
30-40	35	5	91	10	50
40-50	45	10	101	20	200
		Total = 50			Total = 450

(ii)

Classes	0-100	100-200	200-300	300-400	400-500	500-600	600-700	700-800
Frequencies	4	8	9	10	7	5	4	3

解决方案：

(i) $Mean = \frac{\sum f_ix_i}{f_i}$

Mean = 17900/50

= 358

Also, the number of observations, N=50

$MD = \frac{1}{n} \sum_{i=1}^{n}|d_i|$

= 1/50 × 7896

= 157.92

Therefore, mean deviation is 157.92.

(ii) $Mean = \frac{\sum f_ix_i}{f_i}$

Mean = 13630/106

= 128.58

Also, the number of observations, N=106

$MD = \frac{1}{n} \sum_{i=1}^{n}|d_i|$

= 1/106 × 1272.6

= 12.005

问题3。 计算与以下分布的均值的均值偏差：

Classes	95-105	105-115	115-125	125-135	135-145	145-155
Frequencies	9	13	16	26	30	12

解决方案：

$Mean = \frac{\sum f_ix_i}{f_i}$

= 5390/110

= 49

$MD = \frac{1}{n} \sum_{i=1}^{n}|d_i|$

= 1/110 × 1644

= 14.94

Therefore, mean deviation is 14.94.

问题4。 100位寿险保单持有人的年龄分布如下：

Class Interval	xi	fi	Cumulative Frequency	|di| = |xi – M|	fi |di|
0-100	50	4	200	308	1232
100-200	150	8	1200	208	1664
200-300	250	9	2250	108	972
300-400	350	10	3500	8	80
400-500	450	7	3150	92	644
500-600	550	5	2750	192	960
600-700	650	4	2600	292	1168
700-800	750	3	2250	392	1176
		Total = 50	Total = 17900		Total = 7896

计算与中位数年龄的平均偏差。

解决方案：

Number of observations, N = 96

So, N/2 = 96/2 = 48

The cumulative frequency just greater than 48 is 59, and therefore, the corresponding value of x is 38.25

Hence, Median = 38.25

Class Interval	xi	fi	Cumulative Frequency	|di| = |xi – M|	fi |di|
17-19.5	18.25	5	5	20	100
20-25.5	22.75	16	21	15.5	248
36-35.5	30.75	12	33	7.5	90
36-40.5	38.25	26	59	0	0
41-50.5	45.75	14	73	7.5	105
51-55.5	53.25	12	85	15	180
56-60.5	58.25	6	91	20	120
61-70.5	65.75	5	96	27.5	137.5
		Total = 96			Total = 980.5

Number of observations, N = 96.

$MD = \frac{1}{n} \sum_{i=1}^{n}|d_i|$

= 1/96 × 980.5

= 10.21

∴ The mean deviation is 10.21

问题5 找到平均值与以下分布的中位数的平均值偏差：

Class Interval	xi	fi	Cumulative Frequency	|di| = |xi – M|	fi |di|
95-105	100	9	900	28.58	257.22
105-115	110	13	1430	18.58	241.54
115-125	120	16	1920	8.58	137.28
125-135	130	26	3380	1.42	36.92
135-145	140	30	4200	11.42	342.6
145-155	150	12	1800	21.42	257.04
		N = 106	Total = 13630		Total = 1272.6

解决方案：

Number of observations, N = 50

So, N/2 = 50/2 = 25

The cumulative frequency just greater than 25 is 58, and the corresponding value of x is 28.

Therefore, Median = 28

Now,

$Mean = \frac{\sum f_ix_i}{f_i}$

= 1350/50

= 27

Class Interval	xi	fi	Cumulative Frequency	|di| = |xi – Median|	fi |di|	FiXi	|Xi – Mean|	Fi |Xi – Mean|
0-10	5	5	5	23	115	25	22	110
10-20	15	8	13	13	104	120	12	96
20-30	25	15	28	3	45	375	2	30
30-40	35	16	44	7	112	560	8	128
40-50	45	6	50	17	102	270	18	108
		N = 50			Total = 478	Total = 1350		Total = 472

Mean deviation from the median of observation = 478/50 = 9.56 

And, Mean deviation from mean of observation = 472/50 = 9.44 

∴ The mean deviation from the median is 9.56 and from the mean is 9.44.

问题6.计算下面给出的100个人的年龄分布的平均年龄的平均偏差：

Marks	10-20	20-30	30-40	40-50	50-60	60-70	70-80	80-90
No. of students	8	10	15	25	20	18	9	5

解决方案：

Converting the given data into continuous frequency distribution by subtracting 0.5 from the lower limit and adding 0.5 to the upper limit of each class interval.

Age	xi	fi	Cumulative Frequency	|di| = |xi –38|	fi |di|
15.5-20.5	18	5	5	20	100
20.5-25.5	23	6	11	15	90
25.5-30.5	28	12	23	10	120
30.5-35.5	33	14	37	5	70
35.5-40.5	38	26	63	0	0
40.5-45.5	43	12	75	5	60
45.5-50.5	48	16	91	10	160
50.5-55.5	53	9	100	15	135
		N = 100			Total = 735

We have, N = 100 

So, N/2 = 100/2 = 50

The cumulative frequency just greater than N/2 is 63 and the corresponding class is 35.5-40.5.

l=35.5, f=26, h= 5, F =37

Therefore, Median = l + (N/2 – F)/f * h  = 35.5 + 50-37/26 * 5 =38

$MD = \frac{1}{n}\sum f_i|d_i| = \frac{735}{100} = 73.5$

问题7.计算以下数据的中位数的平均偏差：

Class Interval	xi	fi	Cumulative Frequency	|di| = |xi – M|	fi |di|
10-20	15	8	120	34	272
20-30	25	10	250	24	240
30-40	35	15	525	14	210
40-50	45	25	1125	4	100
50-60	55	20	1100	6	120
60-70	65	18	1170	16	288
70-80	75	9	675	26	234
80-90	85	5	425	36	180
		N = 110	Total = 5390		Total = 1644

解决方案：

	fi	xi	fixi	|xi-9.2|	fi|xi-9.2|
0-4	4	2	8	7.2	28.8
4-8	6	6	36	3.2	19.2
8-12	8	10	80	0.8	6.4
12-16	5	14	70	4.8	24.0
16-20	2	18	36	8.8	17.6
	N=25		Total=230		total = 96.0

Mean = 230/25 =9.2

Mean Deviation=96/25 = 3.84

问题8.计算以下数据的中位数的平均偏差：

Age (on nearest birthday)	17-19.5	20-25.5	26-35.5	36-40.5	41-50.5	51-55.5	56-60.5	61-70.5
No. of persons	5	16	12	26	14	12	6	5

解决方案：

	fi	xi	fixi	|xi-14.1|	fi|xi-14.1|
0-6	4	3	12	11.1	44.4
6-12	5	9	45	5.1	25.5
12-18	3	15	45	0.9	2.7
18-24	6	21	126	6.9	41.4
24-30	2	27	54	12.9	25.8
	N=20		Total=282		total = 139.8

Mean = 282/20 =14.1

Mean Deviation= 139.8/20 = 6.99