# NCERT Solutions For Class 10th Maths Chapter 14 : Statistics

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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 0 0 160-180 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 0 0 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 0 0 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 0 0 0 250-300 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 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 0.00-0.04 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 SO2 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 0 0 0 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 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

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 0 45-50 0 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 0 0 0 35-40 3 37.5 5 1 3 40-45 0 42.5 10 2 0 45-50 0 47.5 15 3 0 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 0 0 0 145-165 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 0 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 0 0 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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