10类RD Sharma解决方案–第7章统计–练习7.1 |套装1

问题1：计算以下分布的平均值：

x:	5	6	7	8	9
f:	4	8	14	11	3

解决方案：

x	f	fx
5	4	20
6	8	48
7	14	98
8	11	88
9	3	27
	N = 40	∑ fx = 281

We know that, Mean = ∑fx/ N = 281/40 = 7.025

问题2：找到以下数据的平均值：

x:	19	21	23	25	27	29	31
f:	13	15	16	18	16	15	13

解决方案：

x	f	fx
19	13	247
21	15	315
23	16	368
25	18	450
27	16	432
29	15	435
31	13	403
	N = 106	∑ fx = 2650

We know that, Mean = ∑fx/ N = 2650/106 = 25

问题3：如果以下数据的平均值为20.6。找出p的值。

x:	10	15	p	25	35
f:	3	10	25	7	5

解决方案：

x	f	fx
10	3	30
15	10	150
p	25	25p
25	7	175
35	5	175
	N = 50	∑ fx = 530 + 25p

Given,

Mean = 20.6

We know that,

Mean = ∑fx/ N = (530 + 25p)/50

Now,

20.6 = (530 + 25p)/ 50

(20.6 × 50) – 530 = 25p

p = 500/25

So, p = 20

问题4：如果以下数据的平均值为15，则找到p。

x:	5	10	15	20	25
f:	6	p	6	10	5

解决方案：

x	f	fx
5	6	30
10	p	10p
15	6	90
20	10	200
25	5	125
	N = p + 27	∑ fx = 445 + 10p

Given,

Mean = 15

We know that,

Mean = ∑fx/ N = (445 + 10p)/(p + 27)

Now,

15 = (445 + 10p)/(p + 27)

15p + 405 = 445 + 10p

5p = 40

So, p = 8

问题5：找到平均值为16.6的以下分布的p值

x:	8	12	15	p	20	25	30
f:	12	16	20	24	16	8	4

解决方案：

x	f	fx
8	12	96
12	16	192
15	20	300
p	24	24p
20	16	320
25	8	200
30	4	120
	N = 100	∑ fx = 1228 + 24p

Given,

Mean = 16.6

We know that,

Mean = ∑fx/ N = (1228 + 24p)/ 100

Now,

16.6 = (1228 + 24p)/ 100

1660 = 1228 + 24p

24p = 432

So, p = 18

问题6：找到平均值为12.58的以下分布的p的缺失值

x:	5	8	10	12	p	20	25
f:	2	5	8	22	7	4	2

解决方案：

x	f	fx
5	2	10
8	5	40
10	8	80
12	22	264
p	7	7p
20	4	80
25	2	50
	N = 50	∑ fx = 524 + 7p

Given,

Mean = 12.58

We know that,

Mean = ∑fx/ N = (524 + 7p)/ 50

Now, 

12.58 = (524 + 7p)/ 50

629 = 524 + 7p

7p = 105

So, p = 15

问题7：找到平均值为7.68的以下分布的缺失频率(p)

x:	3	5	7	9	11	13
f:	6	8	15	p	8	4

解决方案：

x	f	fx
3	6	18
5	8	40
7	15	105
9	p	9p
11	8	88
13	4	52
	N = 41 + p	∑ fx = 303 + 9p

Given,

Mean = 7.68

We know that,

Mean = ∑fx/ N = (303 + 9p)/(41 + p)

Now,

7.68 = (303 + 9p)/(41 + p)

7.68(41 + p) = 303 + 9p

7.68p + 314.88 = 303 + 9p

1.32p = 11.88

So, p = 9

问题8：如果以下分布的均值为20，则找到p的值。

x:	15	17	19	20 + p	23
f:	2	3	4	5p	6

解决方案：

x	f	fx
15	2	30
17	3	51
19	4	76
20 + p	5p	100p + 5p2
23	6	138
	N = 5p + 15	∑ fx = 295 + 100p + 5p2

Given,

Mean = 20

We know that,

Mean = ∑fx/ N = (295 + 100p + 5p²)/(5p + 15)

Now,

20 = (295 + 100p + 5p²)/(5p + 15)

20(5p + 15) = 295 + 100p + 5p²

100p + 300 = 295 + 100p + 5p²

5p² – 5 = 0

5(p² – 1) = 0

p² – 1 = 0

(p + 1)(p – 1) = 0

So, p = 1 or -1

Here, p = -1 (Reject as frequency of a number cannot be negative)

So, p = 1

问题9：下表列出了每班40名学生中特定年龄的男孩的数量。计算学生的平均年龄。

Age (in years):	15	16	17	18	19	20
No. of students:	3	8	10	10	5	4

解决方案：

Age (in years)          (x)	No. of students            (f)	fx
15	3	45
16	8	128
17	10	170
18	10	180
19	5	95
20	4	80
	N = 40	∑ fx = 698

We know that, Mean = ∑fx/ N = 698/40 = 17.45

So, the mean age of the students is 17.45 years.

问题10：四所学校的候选人参加了数学测验。数据如下：

Schools	No. of Candidates	Average Score
I	60	75
II	48	80
III	NA	55
IV	40	50

如果所有四所学校的候选人的平均分数是66，请找到从学校III出现的候选人的数量。

解决方案：

Let the number of candidates that appeared from school III = p

Schools	No. of Candidates                (f)	Average Score          (x)	fx
I	60	75	4500
II	48	80	3840
III	p	55	55p
IV	40	50	2000
	N = 148 + p		∑ fx = 10340 + 55p

Given,

Average score of the candidates of all the four schools = Mean = 66

We know that,

Mean = ∑fx/ N = (10340 + 55p)/(148 + p)

Now,

66 = (10340 + 55p)/(148 + p)

66(148 + p) = 10340 + 55p

66p + 9768 = 10340 + 55p

11p = 572

So, p = 52

So, the number of candidates that appeared from school III are 52