第9类RD Sharma解决方案–第24章集中倾向的度量–练习24.2

问题1.计算以下分布的平均值：

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

解决方案：

Now, mean = $\overline{x} = \frac{\sum f_x}{N}$

= 281/40

= 7.025

问题2.计算以下分布的平均值：

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

解决方案：

Now, mean = $\overline{x} = \frac{\sum f_x}{N}$

= 2650/106

= 25

问题3.以下数据的平均值是20.6。找出p的值。

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

解决方案：

Now, mean = $\overline{x} = \frac{\sum f_x}{N}$

= (25p + 530)/50

Given,

Mean = 20.6 

Solving, we get,

20.6 = (25p + 530)/50

25p + 530 = 1030

25p = 1030 − 530 = 500

that is, p = 20

问题4.如果以下数据的均值为15，则找到p。

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	$\sum f_x = 2650$

解决方案：

Mean = $\overline{x} = \frac{\sum f_x}{N}$

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

Give,

Mean = 15

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

10p + 445 = 15(p + 27)

10p – 15p = 405 – 445 = -40

-5p = -40

that is, p = 8

问题5.为以下平均值为16.6的分布找到p的值。

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

解决方案：

Now, mean = $\overline{x} = \frac{\sum f_x}{N}$

= (24p + 1228)/100

Given,

Mean = 16.6 

Solving, (24p + 1228)/100 = 16.6

24p + 1228 = 1660

24p = 1660 – 1228 = 432

p = 432/24 

= 18

问题6：找到平均值为12.58的以下分布的p的缺失值。

x	f	fx
10	3	30
15	10	150
p	25	25p
25	7	175
35	5	175
	N = 50	$\sum f_x = 25p + 530$

解决方案：

Mean = $\overline{x} = \frac{\sum f_x}{N}$

= (7p + 524)/50

Given, 

Mean = 12.58 

Solving, (7p + 524)/50 = 12.58

7p + 524 = 12.58 x 50

7p + 524 = 629

7p = 629 – 524 = 105

p = 105/7 

= 15

问题7.找到平均值为7.68的以下分布的缺失频率(p)。

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

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 = p +41	$\sum f_x = 9p+303$

Mean = $\overline{x} = \frac{\sum f_x}{N}$

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

Given,

Mean = 7.68

Solving we get, (9p + 303)/(p+41) = 7.68

9p + 303 = 7.68 (p + 41)

9p + 303 = 7.68p + 314.88

9p − 7.68p = 314.88 − 303

1.32p = 11.88

that is, p = (11.881)/(1.32) = 9

问题8。找到平均值为12.58的以下分布的p的缺失值。

x	f	fx
5	6	30
10	p	10p
15	6	90
20	10	200
25	5	125
	N = p+27	$\sum f_x=10p+445$

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	$\sum f_x =7p + 524$

Given,

Mean = 12.58

=> 7p + 524/50 = 12.58

=> 7p + 524 = 629

=> 7p = 629 – 524

Solving for p, we get, 

=> 7p = 105

that is, p = 105/7 = 15

问题9.找到平均值为7.68的以下分布的缺失频率(p)。

x:	8	12	15	p	20	25	30
f:	12	16	20	24	16	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 = p + 41	$\sum f_x =9p + 303$

Mean = $\overline{x} = \frac{\sum f_x}{N}$

Given, 

Mean = 7.68

Now, 

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

Solving, we get, 

9p + 303 = 7.68 + 314.88

9p-7.68p = 314.88 – 303

1.32p = 11.88

p = 9

问题10.如果以下分布的均值为20，则找到p的值。

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	$\sum f_x = 24p + 1228$

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 = 15+5p	$\sum f_x = 295+100p+5p^2$

Given,

Mean = 20

Mean = $\overline{x} = \frac{\sum f_x}{N}$

$\frac{295+100p+5p^2}{15+5p}=20 \\ 295 + 100p + 5p^2 = 300+100p \\ 295 + 100p + 5p^2 - 300 - 100p = 0 \\ 5p^2 - 5 =0 \\ p^2 -1 =0 \\(p+1)(p-1) = 0$

If p+1 = 0 or p-1 =0

p=-1

问题11.四所学校的候选人参加了数学测验。数据如下：

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

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 us assume the number of candidates in school III to be p.

Therefore,

Total number of candidates in all the four schools = 60 + 48 + p + 40 = 148 + p

Average score of four schools = 66

∴Computing total score of the candidates = (148 + p) x 66

Now,

 The mean score of 60 in school I is equivalent to 75 .

Total in school I = 60 x 75 = 4500

The mean score of 48 in school II is equivalent to 80 .

Total in school II = 48 x 80 = 3840

In school III, mean of p = 55

Total in school III= 55 x p = 55p

and in school IV, mean of 40 = 50

Total in school IV = 40 x 50 = 2000

Since, total of the candidates is 148+p.

Also,

Total score = 4500 + 3840 + 55p + 2000 = 10340 + 55p

∴10340 + 55p = (148 + p) x 66 = 9768 + 66p

=> 10340 – 9768 = 66p – 55p

=> 572 = 11p

∴ p = 572/11

Therefore,

The number of candidates in school III = 52

问题12。如果已知频率分布的平均值为50，则在以下频率分布中找到丢失的频率。

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	$\sum f_x = 7p + 524$

解决方案：

x	f	fx
10	17	170
30	f1	30f1
50	32	1600
70	f2	70f2
90	19	1710
	N = 120	$\sum f_x = 3480+30f_1 + 70f_2$

Given,

Mean = 50

$\frac{\sum f_x}{N} = 50 \\ \frac{30f_1 + 70f_2 + 3480}{120} = 50 \\ 30f_1 + 70f_2 + 3480 = 6000 \\ 30f_1 + 70f_2 = 6000- 3480 \\ 30f_1 + 70f_2 = 2520 \\ 3f_! + 7f_2 = 252 ....(i)$

And, given value of N = 120

 $17 + f_1 + 32+ f_2 + 19 = 120 \\ 68 + f_1 + f_2 = 120 \\ f_1 + f_2 = 52 \\ 3f_1+3f_2 = 156 ......(ii)$

Subtracting (ii) from (i) ,

$3f_1+7f_2-3f_1-3f_2 = 252-156 \\ 4f_2 = 96 \\ f_2 = 24$

Substituting f₂  in (i)

$3f_1 + 168 = 252 \\ 3f_1 = 252-168 = 84 \\ f_1 = 28$