10类RD Sharma解决方案–第7章统计–练习7.1 |套装2

问题11：五枚硬币同时被抛掷1000次，每次抛掷都观察到正面数。下表显示了获得0、1、2、3、4和5头的抛掷次数。求出每一次抛掷的平均头数。

No. of heads per toss	No. of tosses
0	38
1	144
2	342
3	287
4	164
5	25
Total	1000

解决方案：

No. of heads per toss                (x)	No. of tosses         (f)	fx
0	38	0
1	144	144
2	342	684
3	287	861
4	164	656
5	25	125
	Total (N) = 1000	∑ fx = 2470

We know that, Mean = ∑fx/ N = 2470/1000 = 2.47

So, the mean number of heads per toss are 2.47

问题12：如果知道分布的平均值为50，则在下面的频率分布中找到丢失的频率。

x:	10	30	50	70	90
f:	17	f1	32	f2	19	Total = 120

解决方案：

x	f	fx
10	17	170
30	f1	30f1
50	32	1600
70	f2	70f2
90	19	1710
	N = 68 + f1 + f2 = 120	∑ fx = 30f1 + 70f2 + 3480

Given,

Mean = 50, N = 120 

We know that,

Mean = ∑fx/ N = (30f₁ + 70f₂ + 3480)/(68 + f₁ + f₂)

Now, 

50 = (30f₁ + 70f₂ + 3480)/(68 + f₁ + f₂)

Also , 68 + f₁ + f₂ = 120 

 f₁ = 52 – f₂     ……. (i)

Now,

50 = (30f₁ + 70f₂ + 3480)/ 120

30f₁ + 70f₂ = 6000 – 3480

Now, putting the value of f₁ from equation (i) –

30(52 – f₂) + 70f₂ = 2520

1560 – 30f₂ + 70f₂ = 2520

40f₂ = 960 

So, f₂ = 24 

and f₁ = 52 – f₂ = 52 – 24 = 28

Thus, f₁ = 28 and f₂ = 24

问题13：以下数据的算术平均值为14，求k的值。

xi :	5	10	15	20	15
fi :	7	k	8	4	5

解决方案：

x	f	fx
5	7	35
10	k	10k
15	8	120
20	4	80
25	5	125
	N = k + 24	∑ fx = 360 + 10k

Given,

Mean = 14

We know that,

Mean = ∑fx/ N = (360 + 10k)/(k + 24)

Now, 

14(k + 24) = 360 + 10k

14k + 336 = 360 + 10k

4k = 24

So, k = 6 

问题14：以下数据的算术平均值为25，求出k的值。

xi :	5	15	25	35	45
fi :	3	k	3	6	2

解决方案：

x	f	fx
5	3	15
15	k	15k
25	3	75
35	6	210
45	2	90
	N = 14 + k	∑ fx = 390 + 15k

Given,

Mean = 25

We know that,

Mean = ∑fx/ N = (390 + 15k)/(k + 14)

Now, 

25(k + 14) = 390 + 15k

25k + 350 = 390 + 15k

10k = 40

So, k = 4 

问题15：如果以下数据的平均值为18.75。找出p的值。

xi :	10	15	p	25	30
fi :	5	10	7	8	2

解决方案：

x	f	fx
10	5	50
15	10	150
p	7	7p
25	8	200
30	2	60
	N = 32	∑ fx = 460 + 7p

Given,

Mean = 18.75

We know that,

Mean = ∑fx/ N = (460 + 7p)/ 32

Now, 

18.75 × 32 = 460 + 7p

600 = 460 + 7p

7p = 140

So, p = 20