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NCERT Solutions for Class 10 Maths Chapter 14 Statistics Ex 14.1

January 17, 2019 by Veerendra

NCERT Solutions for Class 10 Maths Chapter 14 Statistics Ex 14.1 are part of NCERT Solutions for Class 10 Maths. Here are we have given Chapter 14 Statistics Class 10 NCERT Solutions Ex 14.1.

Students find NCERT Solutions for Class 10 Maths Chapter 14 Statistics quite helpful for their CBSE Board Preparation. Refer to our Textbook Solutions any time, while doing your homework or while preparing for the exam. All questions and answers from the NCERT Book of class 10 Math Chapter 14 are provided here for you for free.

  • Statistics Class 10 Ex 14.2
  • Statistics Class 10 Ex 14.3
  • Statistics Class 10 Ex 14.4
BoardCBSE
TextbookNCERT
ClassClass 10
SubjectMaths
ChapterChapter 14
Chapter NameStatistics
ExerciseEx 14.1
Number of Questions Solved9
CategoryNCERT Solutions

NCERT Solutions for Class 10 Maths Chapter 14 Statistics Ex 14.1

NCERT Solutions for Class 10 Maths

Question 1.
A survey was conducted by a group of students as a part of their environment awareness programme, in which they collected the following data regarding the number of plants in 20 houses in a locality. Find the mean number of plants per house.

Number of plants0 – 22 – 44 – 66 – 88 – 1010 – 1212 – 14
Number of houses1215623

Which method did you use for finding the mean, and why?
Solution:
Let us find class marks (xi) for each interval by using the relation.
NCERT Solutions for Class 10 Maths Chapter 14
Now we may compute xi and fixi as following

Number of plantsNumber of houses (fi)xifixi
0 – 2111— 1 = 1
2 – 4232 — 3 = 6
4 – 6151 — 5 = 5
6 – 8575 — 7 = 35
8 – 10696 — 9 = 54
10 – 122112 —11 = 22
12 – 143133 — 13 = 39
Total20162

From the table we may observe that
NCERT Solutions for Class 10 Maths Chapter 14 Statistics 1s1
So, mean number of plants per house is 8.1.
We have used here direct method as values of class marks (xi) and fi are small.

Question 2.
Consider the following distribution of daily wages of 50 worker of a factory.

Daily wages (in Rs)100 – 120120 – 140140 -160160 – 180180 – 200
Number of workers1214 8610

Solution:
Let us find class mark for each interval by using the relation.

NCERT Solutions for Class 10 Maths Chapter 14 Statistics 2s
Class size (h) of this data = 20
Now taking 150 as assured mean (a) we may calculate di, ui and fiui as following.

Daily wages (in Rs)Number of workers (fi)xidi = xi – 150NCERT Solutions for Class 10 Maths Chapter 14 Statistics 2s.1fiui
100 -12012110– 40-2– 24
120 – 14014130– 20-1– 14
140 – 1608150000
160 -18061702016
180 – 2001019040220
Total50-12

From the table we may observe that
NCERT Solutions for Class 10 Maths Chapter 14 Statistics 2s1
So mean daily wages of the workers of the factory is Rs.145.20

Question 3.
The following distribution shows the daily pocket allowance of children of a locality. The mean pocket allowance is Rs.18. Find the missing frequency f.

Daily pocket allowance (in Rs)11 – 1313 – 1515 -1717 – 1919 – 2121 – 2323 – 25
Number of workers76913f54

Solution:
We may find class mark (xi) for each interval by using the relation.

NCERT Solutions for Class 10 Maths Chapter 14 Statistics 3s
Given that mean pocket allowance \(\overline { x } \) = Rs.18
Now taking 18 as assured mean (a) we may calculate di and fidi as following.

Daily pocket allowance (in Rs.)Number of children fiClass mark xidi = xi – 18fidi
11 – 13712– 6– 42
13 – 15614– 4– 24
15 – 17916– 2– 18
17 – 19131800
19 – 21f2022 f
21 – 23522420
23 – 25424624
TotalNCERT Solutions for Class 10 Maths Chapter 14 Statistics 3s.12f – 40

From the table we may obtain
NCERT Solutions for Class 10 Maths Chapter 14 Statistics 3s.2
Hence the missing frequency f is 20.

Question 4.
Thirty women were examined in a hospital by a doctor and the number of heart beats per minute were recorded and summarized as follows. Fine the mean heart beats per minute for these women, choosing a suitable method.

Number of heart beats per minute65 – 6868 – 7171-7474 – 7777 – 8080 – 8383 – 86
Number of women2438742

Solution:
We may find class mark of each interval (xi) by using the relation.
NCERT Solutions for Class 10 Maths Chapter 14 Statistics 4s
Class size h of this data = 3
Now taking 75.5 as assumed mean (a) we may calculate di, ui, fiui as following.

Number of heart beats per minuteNumber of women fixidi = xi -75.5NCERT Solutions for Class 10 Maths Chapter 14 Statistics 4s.1fiui
65 – 68266.5– 9– 3– 6
68 – 71469.5– 6– 2– 8
71 – 74372.5– 3– 1– 3
74 – 77875.5000
77 – 80778.5317
80 – 83481.5628
83 – 86284.5936
Total304

Now we may observe from table that
NCERT Solutions for Class 10 Maths Chapter 14 Statistics 4s.2
So mean heart beats per minute for these women are 75.9 beats per minute.

Question 5.
In a retail market, fruit vendors were selling mangoes kept in packing boxes. These boxes contained varying number of mangoes. The following was the distribution of mangoes according to the number of boxes.

Number of mangoes50 – 5253 – 5556 – 5859 – 6162 – 64
Number of boxes1511013511525

Find the mean number of mangoes kept in a packing box. Which method of finding the mean did you choose?
Solution:

Number of mangoesNumber of boxes
fi
50 – 5215
53 – 55110
56 – 58135
59 – 61115
62 – 6425

We may observe that class intervals are not continuous. There is a gap of 1 between two class intervals. So we have to add 1/2 to upper class limit and subtract 1/2 from lower class limit of each interval.
And class mark (xi) may be obtained by using the relation
NCERT Solutions for Class 10 Maths Chapter 14 Statistics 5s
Class size (h) of this data = 3

Now taking 57 as assumed mean (a) we may calculate di, ui, fiui as following –

Class intervalfixidi = xi – 57NCERT Solutions for Class 10 Maths Chapter 14 Statistics 5s.1fiui
49.5 – 52.51551-6-2-30
52.5 – 55.511054-3-1-110
55.5 – 58.513557000
58.5 – 61.51156031115
61.5 – 64.525636250
Total40025

Now we may observe that

NCERT Solutions for Class 10 Maths Chapter 14 Statistics 5s.2

Clearly, mean number of mangoes kept in a packing box is 57.19.
We have chosen step deviation method here as values of fi, di are big and also there is a common multiple between all di.

Question 6.
The table below shows the daily expenditure on food of 25 households in a locality.

Daily expenditure (in Rs)100 – 150150 – 200200 – 250250 – 300300 – 350
Number of households451222

Find the mean daily expenditure on food by a suitable method.
Solution:
We may calculate cla

mark (xi) for each interval by using the relation

NCERT Solutions for Class 10 Maths Chapter 14 Statistics 6s
Class size = 50

Now taking 225 as assumed mean (a) we may calculate di, ui, fiui as following

Daily expenditure (in Rs)fixidi = xi – 225NCERT Solutions for Class 10 Maths Chapter 14 Statistics 6s.1fiui
100 – 1504125-100-2-8
150 – 2005175-50-1-5
200 – 25012225000
250 – 30022755012
300 – 350232510024
Total 25-7

Now we may observe that –

NCERT Solutions for Class 10 Maths Chapter 14 Statistics 6s.2

Question 7.
To find out the concentration of SO2 in the air (in parts per million, i.e., ppm), the data was collected for 30 localities in a certain city and is presented below:

concentration of SO2 (in pmm)Frequency
0.00 – 0.044
0.04 – 0.089
0.08 – 0.129
0.12 – 0.162
0.16 – 0.204
0.20 – 0.242

Find the mean concentration of SO2 in the air.

Solution:

NCERT Solutions for Class 10 Maths Chapter 14 Statistics 7s

Concentration of SO2 (in ppm)FrequencyClass mark xidi = xi – 0.14NCERT Solutions for Class 10 Maths Chapter 14 Statistics 7s.1fiui
0.00 – 0.0440.02-0.12-3-12
0.04 – 0.0890.06-0.08-2-18
0.08 – 0.1290.10-0.04-1-9
0.12 – 0.1620.14000
0.16 – 0.2040.180.0414
0.20 – 0.2420.220.0824
Total30-31

NCERT Solutions for Class 10 Maths Chapter 14 Statistics 7s.2

Question 8.
A class teacher has the following absentee record of 40 students of a class for the whole term. Find the mean number of days a student was absent.

Number of days0 – 66 – 1010 – 1414 – 2020 – 2828 – 3838 –  40
Number of students111074431

Solution:
We may find class mark of each interval by using the relation

NCERT Solutions for Class 10 Maths Chapter 14 Statistics 8s
Now taking 16 as assumed mean (a) we may calculate di and fidi as following

Number of daysNumber of students
fi
xidi= xi – 16fidi
0 – 6113-13-143
6 -10108-8-80
10 – 14712-4-28
14 – 2041600
20 – 28424832
28 – 383331751
38 – 401392323
Total40-145

Now we may observe that

NCERT Solutions for Class 10 Maths Chapter 14 Statistics 8s.1
So, mean number of days is 12.38 days, for which a student was absent.

Question 9.
The following table gives the literacy rate (in percentage) of 35 cities. Find the mean literacy rate.

Literacy rate
(in %)
45 – 5555 – 6565 – 7575 – 8585 – 95
Number of cities3101183

Solution:
We may find class marks by using the relation

NCERT Solutions for Class 10 Maths Chapter 14 Statistics 9s
Class size (h) for this data = 10

Now taking 70 as assumed mean (a) we may calculate di, ui, and fiui as following

Literacy rate 
(in %)
Number of cities 
fi
xidi= xi – 70uifiui
45 – 55350-20-2-6
55 – 651060-10-1-10
65 – 751170000
75 – 858801018
85 – 953902026
Total35-2

Now we may observe that
NCERT Solutions for Class 10 Maths Chapter 14 Statistics 9s.1
So, mean literacy rate is 69.43%.

 

We hope the NCERT Solutions for Class 10 Maths Chapter 14 Statistics Ex 14.1 help you. If you have any query regarding NCERT Solutions for Class 10 Maths Chapter 14 Statistics Ex 14.1, drop a comment below and we will get back to you at the earliest.

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Filed Under: CBSE Tagged With: NCERT Solutions for Class 10 Maths

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