问题11.以下是有关某城市30天空气中二氧化硫浓度(百万分之几)的频率分布表。
Conc. of SO2 |
0.00-0.04 |
0.04-0.08 |
0.08-0.12 |
0.12-0.16 |
0.16-0.20 |
0.20-0.24 |
No of days |
4 |
8 |
9 |
2 |
4 |
3 |
在这些天中的任何一天中，求出二氧化硫浓度在0.12-0.16之间的概率。
解决方案：
Total number of days = 30
Probability of concentration of SO2 in the internal 0.12 – 0.16 =
= Favorable Outcome / Total outcome
= 2/30 = 0.06
问题12：一个组织随机选择了2400个家庭，并对其进行了调查，以确定收入水平与家庭中车辆数量之间的关系。下表中列出了收集的信息：
每个家庭的车辆
Monthly Income(in Rs) |
0 |
1 |
2 |
Above 2 |
Less than 7000 |
10 |
160 |
25 |
0 |
7000-10000 |
0 |
305 |
27 |
2 |
10000-13000 |
1 |
535 |
29 |
1 |
13000-16000 |
2 |
469 |
59 |
25 |
16000 or more |
1 |
579 |
82 |
88 |
假设选择了一个家庭，则找到选择的家庭的概率为
(i)每月收入10000 − 13000卢比，恰好拥有2辆车。
(ii)每月赚取16000卢比或以上，并且正好拥有1辆车。
(iii)每月收入少于7000卢比，并且不拥有任何车辆。
(iv)每月收入13000卢比至16000卢比，拥有2辆以上的汽车。
(v)拥有不超过1辆车。
(vi)拥有至少一辆车。
解决方案：
Total numbers of families selected by the organization to Survey= 2400. -(According to Question)
(i) Let E1 be the event of selecting of family earning ₹(10000 -13000)
per month and owning exactly two vehicles.
Numbers of families earning ₹10000 –13000
per month and owning exactly 2 vehicles = 29
Required probability P(E1) = 29/2400
(ii) Let E2 be the event of selecting of family earning ₹16000 or
more per month and owning exactly 1 vehicle.
Number of families earning ₹16000 or
more per month and owning exactly 1 vehicle = 579
Required probability,P(E2) = 579/2400
(iii) Let E3 be the event of selecting of family earning than ₹ 7000 per month and
doesn’t own any vehicle.
Number of families earning but ₹7000 per month and
doesn’t own any vehicle = 10
Required probability, P(E3)= 10/2400 = 1/240
(iv) Let E4 be the event of selecting a family earning ₹(13000 -16000) per month and
owning quite 2 vehicles.
Number of families earning ₹13000-16000 per month and
owning quite 2 vehicles = 25
Required probability, P(E4) = 25/2400 = 1/96
(v) Let E5 be the event of selecting a family owning less than 1 vehicle.
Number of families owning less than 1 vehicle i.e. the number
of families owning 0 vehicle and 1 vehicle = 10+160+0+305+1+535+2+469+1+579 = 2069
Required probability, P(E5) = 2062/2400 = 1031/1200
(vi) Let P(E6) is the probability that the family of owning atleast one vehicle
P(E6) = (160+305+535+469+579+25+27+29+29+82+0+2+1+25+88)/2400
= 2356/2400 = 589/600
问题13.下表列出了400盏霓虹灯的寿命：
Lifetime |
300-400 |
400-500 |
500-600 |
600-700 |
700-800 |
800-900 |
900-1000 |
Number of lambs |
14 |
56 |
60 |
86 |
74 |
62 |
48 |
随机选择一个灯泡。找出所选灯泡的寿命为
(i)少于400?
(ii)300至800之间?
(iii)至少700小时?
解决方案：
(i) The probability that the lifetime of the selected bulb is less than 400
= Favorable outcomes / Total outcome
= 14/400 = 7/400
(ii) The probability that the lifetime of the selected bulb is between 300 – 800 hours
= Favorable outcomes / Total outcome
= (14 +56 +60 +86 +74) / 400
= 29/40
(iii) The probability that the lifetime of the selected bulb is at least 700 hours
= Favorable outcomes / Total outcome
= (74 +62+ 48)/400 = 23/50
问题14.以下是某工厂中30名工人的工资频率分布(以卢比为单位)：
Wages(in Rs) |
110-130 |
130-150 |
150-170 |
170-190 |
190-210 |
210-230 |
230-250 |
No. of workers |
3 |
4 |
5 |
6 |
5 |
4 |
3 |
随机选择一个工人。找出
(i)少于150卢比
(ii)Atleast Rs210
(iii)大于或等于150卢比但小于210卢比。
解决方案：
(i) The probability that his wages are less than Rs 150 =
= Favorable outcomes / Total outcome
=(3 + 4) / 30 = 7 / 30
(ii)The probability that his wages are at least Rs 210
= Favorable outcomes / Total outcome
= (3 + 4) / 30 = 7 / 30
(iii) The probability that his wages are more than or equal to 150 but less than Rs 200
= Favorable outcomes / Total outcome
= (5 + 6 + 5) / 30 = 16 / 30 = 8 / 15