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 plants | 0-2 | 2-4 | 4-6 | 6-8 | 8-10 | 10-12 | 12-14 |

Number of houses | 1 | 2 | 1 | 5 | 6 | 2 | 3 |

Which method did you use for finding the mean, and why?

**Solution:**

Class Interval | fi | xi | fixi |

0-2 | 1 | 1 | 1 |

2-4 | 2 | 3 | 6 |

4-6 | 1 | 5 | 5 |

6-8 | 5 | 7 | 35 |

8-10 | 6 | 9 | 54 |

10-12 | 2 | 11 | 22 |

12-14 | 3 | 13 | 39 |

Sum fi = 20 | Sum fixi = 162 |

Mean can be calculated as follows:

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) | 100-120 | 120-140 | 140-160 | 160-180 | 180-200 |

Number of workers | 12 | 14 | 8 | 6 | 10 |

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 Interval | fi | xi | di = xi – a | fidi |

100-120 | 12 | 110 | -40 | -480 |

120-140 | 14 | 130 | -20 | -280 |

140-160 | 8 | 150 | ||

160-180 | 6 | 170 | 20 | 120 |

180-200 | 10 | 190 | 40 | 400 |

Sum fi = 50 | Sum fidi = -240 |

Now, mean of deviations can be calculated as follows:

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) | 11-13 | 13-15 | 15-17 | 17-19 | 19-21 | 21-23 | 23-25 |

Number of children | 7 | 6 | 9 | 13 | f | 5 | 4 |

**Solution:**

Class Interval | fi | xi | fixi |

11-13 | 7 | 12 | 84 |

13-15 | 6 | 14 | 84 |

15-17 | 9 | 16 | 144 |

17-19 | 13 | 18 | 234 |

19-21 | f | 20 | 20f |

21-23 | 5 | 22 | 110 |

23-25 | 4 | 24 | 96 |

Sum fi = 44 + f | Sum fixi = 752 + 20f |

We have;

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 min | 65-68 | 68-71 | 71-74 | 74-77 | 77-80 | 80-83 | 83-86 |

Number of women | 2 | 4 | 3 | 8 | 7 | 4 | 2 |

**Solution:**

Class Interval | fi | xi | di = xi – a | fidi |

65-68 | 2 | 66.5 | -9 | -18 |

68-71 | 4 | 69.5 | -6 | -24 |

71-74 | 3 | 72.5 | -3 | -9 |

74-77 | 8 | 75.5 | ||

77-80 | 7 | 78.5 | 3 | 21 |

80-83 | 4 | 81.5 | 6 | 24 |

83-86 | 2 | 84.5 | 9 | 18 |

Sum fi = 30 | Sum fidi = 12 |

Now, mean can be calculated as follows:

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 mangoes | 50-52 | 53-55 | 56-58 | 89-61 | 62-64 |

Number of boxes | 15 | 110 | 135 | 115 | 25 |

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

**Solution:**

Class Interval | fi | xi | di = x – a | fidi |

50-52 | 15 | 51 | -6 | 90 |

53-55 | 110 | 54 | -3 | -330 |

56-58 | 135 | 57 | ||

59-61 | 115 | 60 | 3 | 345 |

62-64 | 25 | 63 | 6 | 150 |

Sum fi = 400 | Sum fidi = 75 |

Mean can be calculated as follows:

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) | 100-150 | 150-200 | 200-250 | 250-300 | 300-350 |

Number of households | 4 | 5 | 12 | 2 | 2 |

Find the mean daily expenditure on food by a suitable method.

**Solution:**

Class Interval | fi | xi | di = xi – a | ui = di/h | fiui |

100-150 | 4 | 125 | -100 | -2 | -8 |

150-200 | 5 | 175 | -50 | -1 | -5 |

200-250 | 12 | 225 | |||

250-300 | 2 | 275 | 50 | 1 | 2 |

300-350 | 2 | 325 | 100 | 2 | 4 |

Sum fi = 25 | Sum fiui = -7 |

Mean can be calculated as follows:

7. To find out the concentration of SO_{2} 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 SO_{2} (in ppm) | Frequency |

0.00-0.04 | 4 |

0.04-0.08 | 9 |

0.08-0.12 | 9 |

0.12-0.16 | 2 |

0.16-0.20 | 4 |

0.20-0.24 | 2 |

Find the mean concentration of SO_{2} in the air.

**Solution:**

Class Interval | fi | xi | fixi |

0.00-0.04 | 4 | 0.02 | 0.08 |

0.04-0.08 | 9 | 0.06 | 0.54 |

0.08-0.12 | 9 | 0.10 | 0.90 |

0.12-0.16 | 2 | 0.14 | 0.28 |

0.16-0.20 | 4 | 0.18 | 0.72 |

0.20-0.24 | 2 | 0.22 | 0.44 |

Sum fi = 30 | Sum fixi = 2.96 |

Mean can be calculated as follows:

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 days | 0-6 | 6-10 | 10-14 | 14-20 | 20-28 | 28-38 | 38-40 |

Number of students | 11 | 10 | 7 | 4 | 4 | 3 | 1 |

**Solution:**

Class Interval | fi | xi | fixi |

0-6 | 11 | 3 | 33 |

6-10 | 10 | 8 | 80 |

10-14 | 7 | 12 | 84 |

14-20 | 4 | 17 | 68 |

20-28 | 4 | 24 | 96 |

28-38 | 3 | 33 | 99 |

38-40 | 1 | 39 | 39 |

Sum fi = 40 | Sum fixi = 499 |

Mean can be calculated as follows:

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

Literacy rate (in %) | 45-55 | 55-65 | 65-75 | 75-85 | 85-98 |

Number of cities | 3 | 10 | 11 | 8 | 3 |

**Solution:**

Class Interval | fi | xi | di = xi – a | ui = di/h | fiui |

45-55 | 3 | 50 | -20 | -2 | -6 |

55-65 | 10 | 60 | -10 | -1 | -10 |

65-75 | 11 | 70 | |||

75-85 | 8 | 80 | 10 | 1 | 8 |

85-95 | 3 | 90 | 20 | 2 | 6 |

Sum fi = 35 | Sum fiui = -2 |

Mean can be calculated as follows:

**Exercise 14.2**

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

Age (in years) | 5-15 | 15-25 | 25-35 | 35-45 | 45-55 | 55-65 |

Number of patients | 6 | 11 | 21 | 23 | 14 | 5 |

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

Calculations for Mean:

Class Interval | fi | 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 |

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) | 0-20 | 20-40 | 40-60 | 60-80 | 80-100 | 100-120 |

Frequency | 10 | 35 | 52 | 61 | 38 | 29 |

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.

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 |

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

Calculations for mean:

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 | |||

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 |

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 teacher | Number of states/UT |

15-20 | 3 |

20-25 | 8 |

25-30 | 9 |

30-35 | 10 |

35-40 | 3 |

40-45 | |

45-50 | |

50-55 | 2 |

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

Calculation for mean:

Class Interval | fi | xi | di = xi – a | ui = di/h | fiui |

15-20 | 3 | 17.5 | -15 | -3 | -9 |

20-25 | 8 | 22.5 | -10 | -2 | -16 |

25-30 | 9 | 27.5 | -5 | -1 | -9 |

30-35 | 10 | 32.5 | |||

35-40 | 3 | 37.5 | 5 | 1 | 3 |

40-45 | 42.5 | 10 | 2 | ||

45-50 | 47.5 | 15 | 3 | ||

50-55 | 2 | 52.5 | 20 | 4 | 8 |

Sum fi = 35 | Sum fiui = -23 |

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 scored | Number of batsmen |

3000-4000 | 4 |

4000-5000 | 18 |

5000-6000 | 9 |

6000-7000 | 7 |

7000-8000 | 6 |

8000-9000 | 3 |

9000-1000 | 1 |

10000-11000 | 1 |

Find the mode of the data.

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

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 cars | 0-10 | 10-20 | 20-30 | 30-40 | 40-50 | 50-60 | 60-70 | 70-80 |

Frequency | 7 | 14 | 13 | 12 | 20 | 11 | 15 | 8 |

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

**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 |

65-85 | 4 |

85-105 | 5 |

105-125 | 13 |

125-145 | 20 |

145-165 | 14 |

165-185 | 8 |

185-205 | 4 |

**Solution:**

Class Interval | Frequency | Cumulative frequency |

65-85 | 4 | 4 |

85-105 | 5 | 9 |

105-125 | 13 | 22 |

125-145 | 20 | 42 |

145-165 | 14 | 56 |

165-185 | 8 | 64 |

185-205 | 4 | 68 |

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:

Calculations for Mode:

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

Calculations for Mean:

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 | |||

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 |

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 Interval | Frequency |

0-10 | 5 |

10-20 | x |

20-30 | 20 |

30-40 | 15 |

40-50 | y |

50-60 | 5 |

Total | 60 |

**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

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

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 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 |

**Solution:**

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 |

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

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

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 |

118-126 | 3 |

127-135 | 5 |

136-144 | 9 |

145-153 | 12 |

154-162 | 5 |

163-171 | 4 |

172-180 | 2 |

Find the median length of leaves.

**Solution:**

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 |

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

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

Lifetime (in hours) | Number of lamps |

1500-2000 | 14 |

2000-2500 | 56 |

2500-3000 | 60 |

3000-3500 | 86 |

3500-4000 | 74 |

4000-4500 | 62 |

4500-5000 | 48 |

Find the median life time of a lamp.

**Solution:**

Class Interval | Frequency | Cumulative Frequency |

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 |

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

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

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 letters | 1-4 | 4-7 | 7-10 | 10-13 | 13-16 | 16-19 |

Number of surnames | 6 | 30 | 40 | 16 | 4 | 4 |

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 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 |

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

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

Calculations for Mode:

Modal class = 7-10,

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

Calculations for 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 |

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

Weight (in kg) | 40-45 | 45-50 | 50-55 | 55-60 | 60-65 | 65-70 | 70-75 |

Number of students | 2 | 3 | 8 | 6 | 6 | 3 | 2 |

**Solution:**

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 |

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

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

**Exercise 14.4(NCERT)**

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

Daily income (in Rs) | 100-120 | 120-140 | 140-160 | 160-180 | 180-200 |

Number of workers | 12 | 14 | 8 | 6 | 10 |

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

**Solution:**

Daily income | Cumulative frequency |

Less than 120 | 12 |

Less than 140 | 26 |

Less than 160 | 34 |

Less than 180 | 40 |

Less than 200 | 50 |

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 40 | 3 |

Less than 42 | 5 |

Less than 44 | 9 |

Less than 46 | 14 |

Less than 48 | 28 |

Less than 50 | 32 |

Less than 52 | 35 |

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:**

Weight (in kg) | Frequency | Cumulative Frequency |

36-38 | ||

38-40 | 3 | 3 |

40-42 | 2 | 5 |

42-44 | 4 | 9 |

44-46 | 5 | 14 |

46-48 | 14 | 28 |

48-50 | 4 | 32 |

50-52 | 3 | 35 |

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:

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) | 50-55 | 55-60 | 60-65 | 65-70 | 70-75 | 75-80 |

Number of farms | 2 | 8 | 12 | 24 | 38 | 16 |

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

**Solution:**

Production yield | Cumulative frequency |

More than 50 | 100 |

More than 55 | 98 |

More than 60 | 90 |

More than 65 | 78 |

More than 70 | 54 |

More than 75 | 16 |

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