第10类NCERT解决方案-第14章统计-练习14.2

问题1.下表显示了一年中住院的患者年龄：

Age (in years)	5-15	15-25	25-35	35-45	45-55	55-65
Number of patients	6	11	21	23	14	5

找到上面给出的数据的众数和均值。比较和解释集中趋势的两种度量。

解决方案：

The greatest frequency in the given table is 23, so the modal class = 35 – 45,

l = 35,

Class width = 10, and the frequencies are

f_m = 23, f₁ = 21 and f₂ = 14

Now, we find the mode using the given formula

Mode = $l+ \left[\frac{(f_m-f_1)}{(2f_m-f_1-f_2)}\right]×h$

On substituting the values in the formula, we get

Mode = $35+\left[\frac{(23-21)}{(46-21-14)}\right]×10$

= 35 + (20/11) = 35 + 1.8

= 36.8

Hence, the mode of the given data is 36.8 year

Now, we find the mean. So for that first we need to find the midpoint.

x_i = (upper limit + lower limit)/2

Mean = $\bar{x}$ = ∑f_ix_i /∑f_i

= 2830/80

= 35.37 years

问题2.以下数据提供了有关225个电气组件的观察到的寿命(以小时为单位)的信息：

Class Interval	Frequency (fi)	Mid-point (xi)	fixi
5-15	6	10	60
15-25	11	20	220
25-35	21	30	630
35-45	23	40	920
45-55	14	50	700
55-65	5	60	300
	Sum fi = 80		Sum fixi = 2830

确定组件的模态寿命。

解决方案：

According to the given question

The modal class is 60 – 80

l = 60, and the frequencies are

f_m = 61, f₁ = 52, f₂ = 38 and h = 20

Now, we find the mode using the given formula

Mode = $l+ \left[\frac{(f_m-f_1)}{(2f_m-f_1-f_2)}\right]×h$

On substituting the values in the formula, we get

Mode = $60+\left[\frac{(61-52)}{(122-52-38)}\right]×20$

= $60+\frac{(9 \times 20)}{32}$

= 60 + 45/8 = 60 + 5.625

Hence, the modal lifetime of the components is 65.625 hours.

问题3.以下数据给出了一个村庄的200个家庭的每月家庭总支出的分布。查找家庭的每月模态支出。此外，找出平均每月支出：

Lifetime (in hours)	0-20	20-40	40-60	60-80	80-100	100-120
Frequency	10	35	52	61	38	29

解决方案：

According to the question

Modal class = 1500-2000,

l = 1500,and the frequencies are

f_m = 40 f₁ = 24, f₂ = 33 and

h = 500

Now, we find the mode using the given formula

Mode = $l+ \left[\frac{(f_m-f_1)}{(2f_m-f_1-f_2)}\right]×h$

On substituting the values in the formula, we get

Mode = $1500+\left[\frac{(40-24)}{(80-24-33)}\right]×500$

= $1500+\frac{16×500}{23}$

= 1500 + 8000/23 = 1500 + 347.83

So, the modal monthly expenditure of the families is 1847.83 Rupees 

Now, we find the mean. So for that first we need to find the midpoint.

x_i = (upper limit + lower limit)/2

Let us considered a mean, A be 2750

Mean = $\overline{x} = a +\frac{∑f_iu_i}{∑f_i}×h$

On substituting the values in the given formula

= $2750+\frac{-35}{200}×500$

= 2750 – 87.50

= 2662.50

Hence, the mean monthly expenditure of the families is  2662.50 Rupees

问题4.以下分布给出了印度高中按州划分的师生比例。查找此数据的模式和均值。解读这两项措施

Expenditure	Number of families
1000-1500	24
1500-2000	40
2000-2500	33
2500-3000	28
3000-3500	30
3500-4000	22
4000-4500	16
4500-5000	7

解决方案：

According to the question

Modal class = 30 – 35,

l = 30,

Class width (h) = 5, and the frequencies are

f_m = 10, f₁ = 9 and f₂ = 3

Now, we find the mode using the given formula

Mode = $l+ \left[\frac{(f_m-f_1)}{(2f_m-f_1-f_2)}\right]×h$

On substituting the values in the formula, we get

Mode = $30+\frac{(10-9)}{(20-9-3)}×5$

= 30 + 5/8 = 30 + 0.625

= 30.625

Hence, the mode of the given data is 30.625

Now, we find the mean. So for that first we need to find the midpoint.

x_i = (upper limit + lower limit)/2

Class Interval	Frequency (fi)	Mid-point (xi)	fixi
15-20	3	17.5	52.5
20-25	8	22.5	180.0
25-30	9	27.5	247.5
30-35	10	32.5	325.0
35-40	3	37.5	112.5
40-45	0	42.5	0
45-50	0	47.5	0
50-55	2	52.5	105.5
	Sum fi = 35		Sum fixi = 1022.5

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

= 1022.5/35 

= 29.2

Hence, the mean is 29.2

问题5.给定的分布情况显示了世界上一些顶尖的板球运动员在一天的国际板球比赛中得分的次数。

Class Interval	fi	xi	di = xi – a	ui = di/h	fiui
1000-1500	24	1250	-1500	-3	-72
1500-2000	40	1750	-1000	-2	-80
2000-2500	33	2250	-500	-1	-33
2500-3000	28	2750	0	0	0
3000-3500	30	3250	500	1	30
3500-4000	22	3750	1000	2	44
4000-4500	16	4250	1500	3	48
4500-5000	7	4750	2000	4	28
	fi = 200				fiui = -35

查找数据的模式。

解决方案：

According to the question

Modal class = 4000 – 5000,

l = 4000,

class width (h) = 1000, and the frequencies are

f_m = 18, f₁ = 4 and f₂ = 9

Now, we find the mode using the given formula

Mode = $l+ \left[\frac{(f_m-f_1)}{(2f_m-f_1-f_2)}\right]×h$

On substituting the values in the formula, we get

Mode = $4000+\frac{(18-4)}{(36-4-9)}×1000$

Mode = 4000 + 14000/23 = 4000 + 608.695

= 4608.695

Hence, the mode of the given data is 4608.7 runs

问题6：一名学生记录了在3分钟内每100分钟通过道路上某个地点的汽车数量，并将其汇总在下表中。查找数据的模式：

No of Students per teacher	Number of states / U.T
15-20	3
20-25	8
25-30	9
30-35	10
35-40	3
40-45	0
45-50	0
50-55	2

解决方案：

According to the question

Modal class = 40 – 50, l = 40,

Class width (h) = 10, and the frequencies are

f_m = 20, f₁ = 12 and f₂ = 11

Now, we find the mode using the given formula

Mode = $l+ \left[\frac{(f_m-f_1)}{(2f_m-f_1-f_2)}\right]×h$

On substituting the values in the formula, we get

Mode = $40+\frac{(20-12)}{(40-12-11)}×10$

Mode = 40 + 80/17 = 40 + 4.7 = 44.7

Hence, the mode of the given data is 44.7 cars