第10类NCERT解决方案-第14章统计-练习14.3

问题1.以下频率分布给出了一个地区68位用户的每月用电量。找到数据的中位数，均值和众数并进行比较。

Monthly consumption(in units)	No. of customers
65-85	4
85-105	5
105-125	13
125-145	20
145-165	14
165-185	8
185-205	4

解决方案：

 Total number of consumer n = 68 

n/2 =34

So, the median class is 125-145 with cumulative frequency = 42

Here, l = 125, n = 68, C_f = 22, f = 20, h = 20

Now we find the median:

Median = $l +\left(\frac{\frac{n}{2}-c_f}{f}\right)\times h$ 

$=125+\frac{(34−22)}{20} × 20$

= 125 + 12 = 137

Hence, the median is 137

Now we find the mode:

Modal class = 125 – 145,

Frequencies are

f₁ = 20, f₀ = 13, f₂ = 14 & h = 20

Mode = $l+ \left[\frac{(f1-f0)}{(2f1-f0-f2)}\right]×h$

On substituting the values in the given formula, we get

Mode = $125 + \frac{(20-13)}{(40-13-14)}×20$

= 125 + 140/13 

= 125 + 10.77

= 135.77

Hence, the mode is 135.77

Now we find the mean:

$\bar{x} =a+h \frac{∑f_iu_i}{∑f_i}$

= 135 + 20(7/68)

= 137.05

Hence, the mean is 137.05

Now, on comparing the median, mean, and mode, we found that mean, median and mode are more/less equal in this distribution.

问题2。如果下面给出的分布的中位数为28.5，则找到x和y的值。

Class Interval	fi	xi	di = xi – a	ui = di/h	fiui
65-85	4	75	-60	-3	-12
85-105	5	95	-40	-2	-10
105-125	13	115	-20	-1	-13
125-145	20	135	0	0	0
145-165	14	155	20	1	14
165-185	8	175	40	2	16
185-205	4	195	60	3	12
	Sum fi = 68				Sum fiui = 7

解决方案：

According to the question

The total number of observations are n = 60

Median of the given data = 28.5

n/2 = 30  

Median class is 20 – 30 with a cumulative frequency = 25 + x

Lower limit of median class, l = 20,

C_f = 5 + x,

f = 20 & h = 10

Now we find the median:

Median = $l+\left(\frac{\frac{n}{2}-c_f}{f}\right)\times h$

On substituting the values in the given formula, we get

28.5 = $20+\frac{(30−5−x)}{20} × 10$

8.5 = (25 – x)/2

17 = 25 – x

Therefore, x = 8

From the cumulative frequency, we can identify the value of x + y as follows:

60 = 5 + 20 + 15 + 5 + x + y

On substituting the values of x, we will find the value of y

60 = 5 + 20 + 15 + 5 + 8 + y

y = 60 – 53

y = 7

So the value of a is 8 and y is 7

问题3.人寿保险代理人找到了以下有关100个保单持有人年龄分布的数据。如果仅对18岁以上但小于60岁的人员提供政策，请计算中位数年龄。

Class Interval	Frequency
0-10	5
10-20	x
20-30	20
30-40	15
40-50	y
50-60	5
Total	60

解决方案：

According to the given question the table is 

Given data: n = 100 and n/2 = 50

Median class = 35 – 45

Then, l = 35, c_f = 45, f = 33 & h = 5

Now we find the median:

Median = $l+\left(\frac{\frac{n}{2}-c_f}{f}\right)\times h$

On substituting the values in the given formula, we get

Median = $35+\frac{(50-45)}{33} × 5$

= 35 + 5(5/33) 

= 35.75

Hence, the median age is 35.75 years.

问题4.正确测量植物中40片叶子的长度，精确到最接近的毫米，所获得的数据如下表所示：

Age (in years)	Number of policy holder
Below 20	2
Below 25	6
Below 30	24
Below 35	45
Below 40	78
Below 45	89
Below 50	92
Below 55	98
Below 60	100

找到叶子的中位长度。

解决方案：

The data in the given table are not continuous to reduce 0.5 in the lower limit and add 0.5 in the upper limit.

We get a new table:

From the given table

n = 40 and n/2 = 20

Median class = 144.5 – 153.5

l = 144.5,

cf = 17, f = 12 & h = 9

Now we find the median:

Median = $l+\left(\frac{\frac{n}{2}-c_f}{f}\right)\times h$

On substituting the values in the given formula, we get

Median = $144.5+\frac{(20-17)}{12}×9$

= 144.5 + 9/4 

= 146.75 mm

Hence, the median length of the leaves is 146.75 mm.

问题5.下表给出了400盏霓虹灯的寿命分布。

Class interval	Frequency	Cumulative frequency
15-20	2	2
20-25	4	6
25-30	18	24
30-35	21	45
35-40	33	78
40-45	11	89
45-50	3	92
50-55	6	98
55-60	2	100

求出灯的平均寿命。

解决方案：

According to the question

Class Interval	Frequency	Cumulative
1500-2000	14	14
2000-2500	56	70
2500-3000	60	130
3000-3500	86	216
3500-4000	74	290
4000-4500	62	352
4500-5000	48	400

n = 400 and n/2 = 200

Median class = 3000 – 3500

l = 3000, C_f = 130,

f = 86 & h = 500

Now we find the median:

Median = $l+\left(\frac{\frac{n}{2}-c_f}{f}\right)\times h$

On substituting the values in the given formula, we get

Median = $3000 + \frac{(200-130)}{86} × 500$

= 3000 + 35000/86 = 3000 + 406.97

= 3406.97

Hence, the median lifetime of the lamps is 3406.97 hours

问题6.在这100个姓氏中，是从本地电话簿中随机抽取的，姓氏中英文字母的字母频率分布如下：

Length (in mm)	Number of leaves
118-126	3
127-135	5
136-144	9
145-153	12
154-162	5
163-171	4
172-180	2

确定姓氏中的中间字母的数量。在姓氏中找到均数字母的数量，并在姓氏中找到情态字的大小。

解决方案：

According to the question

Class Interval	Frequency	Cumulative Frequency
1-4	6	6
4-7	30	36
7-10	40	76
10-13	16	92
13-16	4	96
16-19	4	100

n = 100 and n/2 = 50

Median class = 7 – 10

Therefore, l = 7, C_f = 36, f = 40 & h = 3

Now we find the median:

Median = $l+\left(\frac{\frac{n}{2}-c_f}{f}\right)\times h$

On substituting the values in the given formula, we get

Median = $7+\frac{(50-36)}{40} × 3$

Median = 7 + 42/40 = 8.05

Hence, the median is 8.05

Now we find the mode:

Modal class = 7 – 10,

Where, l = 7, f₁ = 40, f₀ = 30, f₂ = 16 & h = 3

Mode = $l+\left(\frac{f_1-f_0}{2f_1-f_0-f_2}\right)\times h$

On substituting the values in the given formula, we get

Mode = $7+\frac{(40-30)}{(2×40-30-16)} × 3$

= 7 + 30/34 = 7.88

Hence, the mode is 7.88

Now we find the mean:

Class Interval	fi	xi	fixi
1-4	6	2.5	15
4-7	30	5.5	165
7-10	40	8.5	340
10-13	16	11.5	184
13-16	4	14.5	51
16-19	4	17.5	70
	Sum fi = 100		Sum fixi = 825

Mean = $\bar{x}= \frac{∑f_i x_i }{∑f_i}$

= 825/100 = 8.25

Hence, the mean is 8.25

问题7.下面的分布分配了一个班级的30名学生的权重。查找学生的体重中位数。

Class Interval	Frequency	Cumulative frequency
117.5-126.5	3	3
126.5-135.5	5	8
135.5-144.5	9	17
144.5-153.5	12	29
153.5-162.5	5	34
162.5-171.5	4	38
171.5-180.5	2	40

解决方案：

According to the question

Class Interval	Frequency	Cumulative frequency
40-45	2	2
45-50	3	5
50-55	8	13
55-60	6	19
60-65	6	25
65-70	3	28
70-75	2	30

n = 30 and n/2 = 15

Median class = 55 – 60

l = 55, C_f = 13, f = 6 & h = 5

Now we find the median:

Median = $l +\left(\frac{\frac{n}{2}-c_f}{f}\right)\times h$

On substituting the values in the given formula, we get

Median = $55+\frac{(15-13)}{6}×5$

= 55 + 10/6 = 55 + 1.666

= 56.67

Hence, the median weight of the students is 56.67