NCERT Solutions For Class 10th Maths Chapter 14 : Statistics

CBSE NCERT Solutions For Class 10th Maths Chapter 14 : Statistics. NCERT Solutins For Class 10 Mathematics. Exercise 14.1, Exercise 14.2, Exercise 14.3, Exercise 14.4.


NCERT Solutions for Class X Maths Chapter 14 Statistics – Mathematics CBSE

Exercise 14.1

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:

Class Intervalfixifixi
0-2111
2-4236
4-6155
6-85735
8-106954
10-1221122
12-1431339
Sum fi = 20Sum fixi = 162

Mean can be calculated as follows:

10 math statistics exercise 1 question 1 solution

In this case, the values of fi and xi are small hence direct method has been used.

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

Daily wages (in Rs)100-120120-140140-160160-180180-200
Number of workers12148610

Find the mean daily wages of the workers of the factory by using an appropriate method.

Solution: In this case, value of xi is quite large and hence we should select the assumed mean method.

Let us take assumed mean a = 150

Class Intervalfixidi = xi – afidi
100-12012110-40-480
120-14014130-20-280
140-1608150
160-180617020120
180-2001019040400
Sum fi = 50Sum fidi = -240

Now, mean of deviations can be calculated as follows:

10 math statistics exercise 1 question 2 solution

Mean can be calculated as follows:

x = d + a = -4.8 + 150 = 145.20

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 children76913f54

Solution:

Class Intervalfixifixi
11-1371284
13-1561484
15-17916144
17-191318234
19-21f2020f
21-23522110
23-2542496
Sum fi = 44 + fSum fixi = 752 + 20f

We have;

10 math statistics exercise 1 mean 10 math statistics exercise 1 question 3 solution

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. Find the mean heart beats per minute for these women, choosing a suitable method.

Number of heart beats per min65-6868-7171-7474-7777-8080-8383-86
Number of women2438742

Solution:

Class Intervalfixidi = xi – afidi
65-68266.5-9-18
68-71469.5-6-24
71-74372.5-3-9
74-77875.5
77-80778.5321
80-83481.5624
83-86284.5918
Sum fi = 30Sum fidi = 12

Now, mean can be calculated as follows:

10 math statistics exercise 1 question 4 solution

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 mangoes50-5253-5556-5889-6162-64
Number of boxes1511013511525

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

Solution:

Class Intervalfixidi = x – afidi
50-521551-690
53-5511054-3-330
56-5813557
59-61115603345
62-6425636150
Sum fi = 400Sum fidi = 75

Mean can be calculated as follows:

10 math statistics exercise 1 question 5 solution

In this case, there are wide variations in fi and hence assumed mean method is used.

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:

Class Intervalfixidi = xi – aui = di/hfiui
100-1504125-100-2-8
150-2005175-50-1-5
200-25012225
250-30022755012
300-350232510024
Sum fi = 25Sum fiui = -7

Mean can be calculated as follows:

10 math statistics exercise 1 question 6 solution

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 ppm)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:

Class Intervalfixifixi
0.00-0.0440.020.08
0.04-0.0890.060.54
0.08-0.1290.100.90
0.12-0.1620.140.28
0.16-0.2040.180.72
0.20-0.2420.220.44
Sum fi = 30Sum fixi = 2.96

Mean can be calculated as follows:

10 math statistics exercise 1 question 7 solution

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:

Class Intervalfixifixi
0-611333
6-1010880
10-1471284
14-2041768
20-2842496
28-3833399
38-4013939
Sum fi = 40Sum fixi = 499

Mean can be calculated as follows:

10 math statistics exercise 1 question 8 solution

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

Literacy rate (in %)45-5555-6565-7575-8585-98
Number of cities3101183

Solution:

Class Intervalfixidi = xi – aui = di/hfiui
45-55350-20-2-6
55-651060-10-1-10
65-751170
75-858801018
85-953902026
Sum fi = 35Sum fiui = -2

Mean can be calculated as follows:

10 math statistics exercise 1 question 9 solution

Exercise 14.2

1. The following table shows the ages of the patients admitted in a hospital during a year.

Age (in years)5-1515-2525-3535-4545-5555-65
Number of patients6112123145

Find the mode and the mean of the data given above. Compare and interpret the two measures of central tendency.

Solution: Solution: Modal class = 35 – 45, l = 35, h = 10, f1 = 23, f0 = 21 and f2 = 14

10 statistics exercise 2 question 1 solution

Calculations for Mean:

Class Intervalfixifixi
5-1561060
15-251120220
25-352130630
35-452340920
45-551450700
55-65560300
Sum fi = 80Sum fixi = 2830

The mode of the data shows that maximum number of patients is in the age group of 26.8, while average age of all the patients is 35.37.

2. The following data gives the information on the observed lifetime (in hours) of 225 electrical components:

Lifetime (in hours)0-2020-4040-6060-8080-100100-120
Frequency103552613829

Determine the modal lifetimes of the components.

Solution: Modal class = 60-80, l = 60, f1 = 61, f0 = 52, f2 = 38 and h = 20

3. The following data gives the distribution of total monthly household expenditure of 200 families of a village. Find the modal monthly expenditure of the families. Also, find the mean monthly expenditure.

ExpenditureNumber of families
1000-150024
1500-200040
2000-250033
2500-300028
3000-350030
3500-400022
4000-450016
4500-50007

Solution: Modal class = 1500-2000, l = 1500, f1 = 40, f0 = 24, f2 = 33 and h = 500

10 statistics exercise 2 question 3 solution

Calculations for mean:

Class Intervalfixidi = xi – aui = di/hfiui
1000-1500241250-1500-3-72
1500-2000401750-1000-2-80
2000-2500332250-500-1-33
2500-3000282750
3000-3500303250500130
3500-40002237501000244
4000-45001642501500348
4500-5000747502000428
fi = 200fiui = -35

10 statistics exercise 2 question 3 solution 4. The following distribution gives the state-wise teacher-student ratio in higher secondary schools of India. Find the mode and mean of this data. Interpret the two measures.

Number of students per teacherNumber of states/UT
15-203
20-258
25-309
30-3510
35-403
40-45
45-50
50-552

Solution: Modal class = 30-35, l = 30, f1 = 10, f0 = 9, f2 = 3 and h = 5

10 statistics exercise 2 question 4 solution

Calculation for mean:

Class Intervalfixidi = xi – aui = di/hfiui
15-20317.5-15-3-9
20-25822.5-10-2-16
25-30927.5-5-1-9
30-351032.5
35-40337.5513
40-4542.5102
45-5047.5153
50-55252.52048
Sum fi = 35Sum fiui = -23

10 statistics exercise 2 question 4 solution

The mode shows that maximum number of states has 30-35 students per teacher. The mean shows that average ratio of students per teacher is 29.22

5. The given distribution shows the number of runs scored by some top batsmen of the world in one-day international cricket matches.

Runs scoredNumber of batsmen
3000-40004
4000-500018
5000-60009
6000-70007
7000-80006
8000-90003
9000-10001
10000-110001

Find the mode of the data.

Solution: Modal class = 4000-5000, l = 4000, f1 = 18, f0 = 4, f2 = 9 and h = 1000

10 statistics exercise 2 question 5 solution

6. A student noted the number of cars passing through a spot on a road for 100 periods each of 3 minutes and summarized it in the table given below. Find the mode of the data.

Number of cars0-1010-2020-3030-4040-5050-6060-7070-80
Frequency71413122011158

Solution: Modal class = 40 – 50, l = 40, f1 = 20, f0 = 12, f2 = 11 and h = 10

10 statistics exercise 2 question 6 solution

Exercise 14.3(NCERT)

1. The following frequency distribution gives the monthly consumption of electricity of 68 consumers of a locality. Find the median, mean and mode of the data and compare them.

Monthly consumption (in units)Number of customers
65-854
85-1055
105-12513
125-14520
145-16514
165-1858
185-2054

Solution:

Class IntervalFrequencyCumulative frequency
65-8544
85-10559
105-1251322
125-1452042
145-1651456
165-185864
185-205468
N = 68

Here; n = 68 and hence n/2 = 34

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

now, l = 125, n = 68, cf = 22, f = 20, h = 20

Median can be calculated as follows:

10 statistics exercise 3 question 1 solution

Calculations for Mode:

Modal class = 125-145, f1 = 20, f0 = 13, f2 = 14 and h = 20

10 statistics exercise 3 question 1 solution

Calculations for Mean:

Class Intervalfixidi = xi – aui = di/hfiui
65-85475-60-3-12
85-105595-40-2-10
105-12513115-20-1-13
125-14520135
145-1651415520114
165-185817540216
185-205419560312
Sum fi = 68Sum fiui = 7

10 statistics exercise 3 question 1 solution

Mean, median and mode are more or less equal in this distribution.

2. If the median of the distribution given below is 28.5, find the value of x and y.

Class IntervalFrequency
0-105
10-20x
20-3020
30-4015
40-50y
50-605
Total60

Solution: n = 60 and hence n/2 = 30

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

lower limit of median class = 20, cf = 5 + x , f = 20 and h = 10

10 statistics exercise 3 question 2 solution

Now, from cumulative frequency, we can find the value of x + y as follows:

10 statistics exercise 3 question 2 solution

Hence, x = 8 and y = 7

3. A life insurance agent found the following data for distribution of ages of 100 policy holders. Calculate the median age, if policies are given only to persons having age 18 years onwards but less than 60 years.

Age (in years)Number of policyt hodlers
Below 202
Below 256
Below 3024
Below 3545
Below 4078
Below 4589
Below 5092
Below 5598
Below 60100

Solution:

Class intervalFrequencyCumulative frequency
15-2022
20-2546
25-301824
30-352145
35-403378
40-451189
45-50392
50-55698
55-602100

Here; n = 100 and n/2 = 50, hence median class = 35-45

In this case; l = 35, cf = 45, f = 33 and h = 5

10 statistics exercise 3 question 3 solution

4. The lengths of 40 leaves of a plant are measured correct to the nearest millimeter, and the data obtained is represented in the following table:

Length (in mm)Number of leaves
118-1263
127-1355
136-1449
145-15312
154-1625
163-1714
172-1802

Find the median length of leaves.

Solution:

Class IntervalFrequencyCumulative frequency
117.5-126.533
126.5-135.558
135.5-144.5917
144.5-153.51229
153.5-162.5534
162.5-171.5438
171.5-180.5240

We have; n = 40 and n/2 = 20 hence median class = 144.5-153.5

Thus, l = 144.5, cf = 17, f = 12 and h = 9

10 statistics exercise 3 question 4 solution

5. The following table gives distribution of the life time of 400 neon lamps.

Lifetime (in hours)Number of lamps
1500-200014
2000-250056
2500-300060
3000-350086
3500-400074
4000-450062
4500-500048

Find the median life time of a lamp.

Solution:

Class IntervalFrequencyCumulative Frequency
1500-20001414
2000-25005670
2500-300060130
3000-350086216
3500-400074290
4000-450062352
4500-500048400

We have; n = 400 and n/2 = 200 hence median class = 3000 – 3500

So, l = 3000, cf = 130, f = 86 and h = 500

10 statistics exercise 3 question 5 solution

6. 100 surnames were randomly picked up from a local telephone directory and the frequency distribution of the number of letters in the English alphabets in the surnames was obtained as follows:

Number of letters1-44-77-1010-1313-1616-19
Number of surnames630401644

Determine the median number of letters in the surnames. Find the mean number of letters in the surnames. Also, find the modal size of the surnames.

Solution: Calculations for median:

Class IntervalFrequencyCumulative Frequency
1-466
4-73036
7-104076
10-131692
13-16496
16-194100

Here; n = 100 and n/2 = 50 hence median class = 7-10

So, l = 7, cf = 36, f = 40 and h = 3

10 statistics exercise 3 question 6 solution

Calculations for Mode:

Modal class = 7-10,

Here; l = 7, f1 = 40, f0 = 30, f2 = 16 and h = 3

10 statistics exercise 3 question 6 solution

Calculations for Mean:

Class intervalfixifixi
1-462.515
4-7305.5165
7-10408.5340
10-131611.5184
13-16414.551
16-19417.570
Sum fi = 100Sum fixi = 825

10 statistics exercise 3 question 6 solution

7. The distribution below gives the weights of 30 students of a class. Find the median weight of the students.

Weight (in kg)40-4545-5050-5555-6060-6565-7070-75
Number of students2386632

Solution:

Class IntervalFrequencyCumulative frequency
40-4522
45-5035
50-55813
55-60619
60-65625
65-70328
70-75230

 

We have; n = 30 and n/2 = 15 hence median class = 55-60

So, l = 55, cf = 13, f = 6 and h = 5

10 statistics exercise 3 question 7 solution

Exercise 14.4(NCERT)

1. The following distribution gives the daily income of 50 workers of a factory.

Daily income (in Rs)100-120120-140140-160160-180180-200
Number of workers12148610

Convert the distribution above to a less than type cumulative frequency distribution, and draw its ogive.

Solution:

Daily incomeCumulative frequency
Less than 12012
Less than 14026
Less than 16034
Less than 18040
Less than 20050

statistics exercise 4 question 1 ogive 2. During the medical checkup of 35 students of a class, their weights were recorded as follows:

Weight (in kg)Number of students
Less than 38
Less than 403
Less than 425
Less than 449
Less than 4614
Less than 4828
Less than 5032
Less than 5235

Draw a less than type ogive for the given data. Hence obtain the median weight from the graph and verify the result by using the formula.

Solution:

statistics exercise 4 question 2 ogive

Weight (in kg)FrequencyCumulative Frequency
36-38
38-4033
40-4225
42-4449
44-46514
46-481428
48-50432
50-52335

Since N = 35 and n/2 = 17.5 hence median class = Less than 46-48

Here; l = 46, cf = 14, f = 14 and h = 2

Median can be calculated as follows:

statistics exercise 4 question 2 solution

This value of median verifies the median shown in ogive.

3. The following table gives production yield per hectare of wheat of 100 farms of a village.

Production yield (in kg)50-5555-6060-6565-7070-7575-80
Number of farms2812243816

Change this distribution to a more than type distribution, and draw its ogive.

Solution:

Production yieldCumulative frequency
More than 50100
More than 5598
More than 6090
More than 6578
More than 7054
More than 7516

statistics exercise 4 question 3 ogive


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