 # NCERT Solutions For Class 10 Maths Chapter 14-Statistics

## NCERT Solutions for Class 10 Maths Chapter 14 – CBSE Free PDF Download

NCERT Solutions for Class 10 Maths Chapter 14 Statistics are provided here, which can be downloaded for free, in PDF format. The solutions are prepared by our subject experts who have mastery in Maths. All the solved questions of Statistics are with respect to the latest updates on the CBSE syllabus and guidelines, to help students solve each exercise question and effectively prepare for the CBSE exam.

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## NCERT Solutions for Class 10 Maths Chapter 14 Statistics

### Access Answers to Maths NCERT Class 10 Chapter 14 – Statistics

Exercise 14.1 Page: 270

1. A survey was conducted by a group of students as a part of their environment awareness program, 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:

To find the mean value, we will use the direct method because the numerical value of fi and xi are small.

Find the midpoint of the given interval using the formula.

Midpoint (xi) = (upper limit + lower limit)/2

 No. of plants (Class interval) No. of houses Frequency (fi) Mid-point (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

The formula to find the mean is:

Mean = x̄ = ∑fi xi /∑fi

= 162/20

= 8.1

Therefore, the mean number of plants per house is 8.1.

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

 Daily wages (in Rs.) 500-520 520-540 540-560 560-580 580-600 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:

Find the midpoint of the given interval using the formula.

Midpoint (xi) = (upper limit + lower limit)/2

In this case, the value of mid-point (xi) is very large, so let us assume the mean value, a = 550.

Class interval (h) = 20

So, ui = (xi – a)/h

ui = (xi – 550)/20

Substitute and find the values as follows:

 Daily wages (Class interval) Number of workers frequency (fi) Mid-point (xi) ui = (xi – 550)/20 fiui 500-520 12 510 -2 -24 520-540 14 530 -1 -14 540-560 8 550 = a 0 0 560-580 6 570 1 6 580-600 10 590 2 20 Total Sum fi = 50 Sum fiui = -12

So, the formula to find out the mean is:

Mean = x̄ = a + h(∑fiui /∑fi ) = 550 + [20 × (-12/50)] = 550 – 4.8 = 545.20

Thus, mean daily wage of the workers = Rs. 545.20

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 c) 11-13 13-15 15-17 17-19 19-21 21-23 23-35 Number of children 7 6 9 13 f 5 4

Solution:

To find out the missing frequency, use the mean formula.

Given, mean x̄ = 18

 Class interval Number of children (fi) Mid-point (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 Total fi = 44+f Sum fixi = 752+20f

The mean formula is

Mean = x̄ = ∑fixi /∑fi = (752 + 20f)/ (44 + f)

Now substitute the values and equate to find the missing frequency (f)

⇒ 18 = (752 + 20f)/ (44 + f)

⇒ 18(44 + f) = (752 + 20f)

⇒ 792 + 18f = 752 + 20f

⇒ 792 + 18f = 752 + 20f

⇒ 792 – 752 = 20f – 18f

⇒ 40 = 2f

⇒ f = 20

So, the missing frequency, f = 20.

4. Thirty women were examined in a hospital by a doctor, and the number of heartbeats per minute were recorded and summarised as follows. Find the mean heartbeats per minute for these women, choosing a suitable method.

 Number of heart beats per minute 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:

From the given data, let us assume the mean as a = 75.5

xi = (Upper limit + Lower limit)/2

Class size (h) = 3

Now, find the ui and fiui as follows:

 Class Interval Number of women (fi) Mid-point (xi) ui = (xi – 75.5)/h fiui 65-68 2 66.5 -3 -6 68-71 4 69.5 -2 -8 71-74 3 72.5 -1 -3 74-77 8 75.5 = a 0 0 77-80 7 78.5 1 7 80-83 4 81.5 2 8 83-86 2 84.5 3 6 Sum fi= 30 Sum fiui = 4

Mean = x̄ = a + h(∑fiui /∑fi )

= 75.5 + 3 × (4/30)

= 75.5 + (4/10)

= 75.5 + 0.4

= 75.9

Therefore, the mean heart beats per minute for these women is 75.9

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 59-61 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:

The given data is not continuous, so we add 0.5 to the upper limit and subtract 0.5 from the lower limit as the gap between two intervals is 1.

Here, assumed mean (a) = 57

Class size (h) = 3

Here, the step deviation is used because the frequency values are big.

 Class Interval Number of boxes (fi) Mid-point (xi) ui = (xi – 57)/h fiui 49.5-52.5 15 51 -2 -30 52.5-55.5 110 54 -1 -110 55.5-58.5 135 57 = a 0 0 58.5-61.5 115 60 1 115 61.5-64.5 25 63 2 50 Sum fi = 400 Sum fiui = 25

The formula to find out the Mean is:

Mean = x̄ = a + h(∑fiui /∑fi )

= 57 + 3(25/400)

= 57 + 0.1875

= 57.19

Therefore, the mean number of mangoes kept in a packing box is 57.19

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

 Daily expenditure(in c) 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:

Find the midpoint of the given interval using the formula.

Midpoint (xi) = (upper limit + lower limit)/2

Let us assume the mean (a) = 225

Class size (h) = 50

 Class Interval Number of households (fi) Mid-point (xi) di = xi – A ui = di/50 fiui 100-150 4 125 -100 -2 -8 150-200 5 175 -50 -1 -5 200-250 12 225 = a 0 0 0 250-300 2 275 50 1 2 300-350 2 325 100 2 4 Sum fi = 25 Sum fiui = -7

Mean = x̄ = a + h(∑fiui /∑fi )

= 225 + 50(-7/25)

= 225 – 14

= 211

Therefore, the mean daily expenditure on food is 211.

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:

To find out the mean, first find the midpoint of the given frequencies as follows:

 Concentration of SO2 (in ppm) Frequency (fi) Mid-point (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 Total Sum fi = 30 Sum (fixi) = 2.96

The formula to find out the mean is

Mean = x̄ = ∑fixi /∑fi

= 2.96/30

= 0.099 ppm

Therefore, the mean concentration of SO2 in the air is 0.099 ppm.

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:

Find the midpoint of the given interval using the formula.

Midpoint (xi) = (upper limit + lower limit)/2

 Class interval Frequency (fi) Mid-point (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

The mean formula is,

Mean = x̄ = ∑fixi /∑fi

= 499/40

= 12.48 days

Therefore, the mean number of days a student was absent = 12.48.

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:

Find the midpoint of the given interval using the formula.

Midpoint (xi) = (upper limit + lower limit)/2

In this case, the value of mid-point (xi) is very large, so let us assume the mean value, a = 70.

Class interval (h) = 10

So, ui = (xi – a)/h

ui = (xi – 70)/10

Substitute and find the values as follows:

 Class Interval Frequency (fi) (xi) ui = (xi – 70)/10 fiui 45-55 3 50 -2 -6 55-65 10 60 -1 -10 65-75 11 70 = a 0 0 75-85 8 80 1 8 85-95 3 90 2 6 Sum fi = 35 Sum fiui = -2

So, Mean = x̄ = a + (∑fiui /∑fi) × h

= 70 + (-2/35) × 10

= 69.43

Therefore, the mean literacy part = 69.43%

Exercise 14.2 Page: 275

1. The following table shows the ages of the patients admitted to 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:

To find out the modal class, let us the consider the class interval with high frequency.

Here, the greatest frequency = 23, so the modal class = 35 – 45,

Lower limit of modal class = l = 35,

class width (h) = 10,

fm = 23,

f1 = 21 and f2 = 14

The formula to find the mode is

Mode = l + [(fm – f1)/ (2fm – f1 – f2)] × h

Substitute the values in the formula, we get

Mode = 35+[(23-21)/(46-21-14)]×10

= 35 + (20/11)

= 35 + 1.8

= 36.8 years

So the mode of the given data = 36.8 years

Calculation of Mean:

First find the midpoint using the formula, xi = (upper limit +lower limit)/2

 Class Interval Frequency (fi) Mid-point (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 mean formula is

Mean = x̄ = ∑fixi /∑fi

= 2830/80

= 35.375 years

Therefore, the mean of the given data = 35.375 years

2. The following data gives the information on the observed lifetimes (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:

From the given data the modal class is 60–80.

Lower limit of modal class = l = 60,

The frequencies are:

fm = 61, f1 = 52, f2 = 38 and h = 20

The formula to find the mode is

Mode = l+ [(fm – f1)/(2fm – f1 – f2)] × h

Substitute the values in the formula, we get

Mode = 60 + [(61 – 52)/ (122 – 52 – 38)] × 20

Mode = 60 + [(9 × 20)/32]

Mode = 60 + (45/8) = 60 + 5.625

Therefore, modal lifetime of the components = 65.625 hours.

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 (in Rs.) 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:

Given data:

Modal class = 1500-2000,

l = 1500,

Frequencies:

fm = 40 f1 = 24, f2 = 33 and

h = 500

Mode formula:

Mode = l + [(fm – f1)/ (2fm – f1 – f2)] × h

Substitute the values in the formula, we get

Mode = 1500 + [(40 – 24)/ (80 – 24 – 33)] × 500

Mode = 1500 + [(16 × 500)/23]

Mode = 1500 + (8000/23) = 1500 + 347.83

Therefore, modal monthly expenditure of the families = Rupees 1847.83

Calculation for mean:

First find the midpoint using the formula, xi =(upper limit +lower limit)/2

Let us assume a mean, (a) be 2750.

 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 = a 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

The formula to calculate the mean,

Mean = x̄ = a +(∑fiui /∑fi) × h

Substitute the values in the given formula

= 2750 + (-35/200) × 500

= 2750 – 87.50

= 2662.50

So, the mean monthly expenditure of the families = Rs. 2662.50

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

 No of students per teacher Number of states / U.T 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:

Given data:

Modal class = 30 – 35,

l = 30,

Class width (h) = 5,

fm = 10, f1 = 9 and f2 = 3

Mode Formula:

Mode = l + [(fm – f1)/ (2fm – f1 – f2)] × h

Substitute the values in the given formula

Mode = 30 + [(10 – 9)/ (20 – 9 – 3)] × 5

= 30 + (5/8)

= 30 + 0.625

= 30.625

Therefore, the mode of the given data = 30.625

Calculation of mean:

Find the midpoint using the formula, xi =(upper limit +lower limit)/2

 Class Interval Frequency (fi) Mid-point (xi) fixi 15-20 3 17.5 52.5 20-25 8 22.5 180.0 25-30 9 27.5 247.5 30-35 10 32.5 325.0 35-40 3 37.5 112.5 40-45 0 42.5 0 45-50 0 47.5 0 50-55 2 52.5 105.0 Sum fi = 35 Sum fixi = 1022.5

Mean = x̄ = ∑fixi /∑fi

= 1022.5/35

= 29.2 (approx)

Therefore, mean = 29.2

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

 Run Scored Number of Batsman 3000-4000 4 4000-5000 18 5000-6000 9 6000-7000 7 7000-8000 6 8000-9000 3 9000-10000 1 10000-11000 1

Find the mode of the data.

Solution:

Given data:

Modal class = 4000 – 5000,

l = 4000,

class width (h) = 1000,

fm = 18, f1 = 4 and f2 = 9

Mode Formula:

Mode = l + [(fm – f1)/ (2fm – f1 – f2)] × h

Substitute the values

Mode = 4000 + [(18 – 4)/ (36 – 4 – 9)] × 1000

= 4000 + (14000/23)

= 4000 + 608.695

= 4608.695

= 4608.7 (approximately)

Thus, the mode of the given data is 4608.7 runs.

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 Frequency 0-10 7 10-20 14 20-30 13 30-40 12 40-50 20 50-60 11 60-70 15 70-80 8

Solution:

Given Data:

Modal class = 40 – 50, l = 40,

Class width (h) = 10, fm = 20, f1 = 12 and f2 = 11

Mode = l + [(fm – f1)/(2fm – f1 – f2)] × h

Substitute the values

Mode = 40 + [(20 – 12)/ (40 – 12 – 11)] × 10

= 40 + (80/17)

= 40 + 4.7

= 44.7

Thus, the mode of the given data is 44.7 cars.

Exercise 14.3 Page: 287

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

 Monthly consumption(in units) No. of customers 65-85 4 85-105 5 105-125 13 125-145 20 145-165 14 165-185 8 185-205 4

Solution:

Find the cumulative frequency of the given data as follows:

 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

From the table, it is observed that, N = 68 and hence N/2=34

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

Where, l = 125, N = 68, cf = 22, f = 20, h = 20

Median is calculated as follows:

$$\begin{array}{l}\text{Median}=l+\frac{\frac{N}{2}-cf}{f}\times h\end{array}$$

= 125 + [(34 − 22)/20] × 20

= 125 + 12

= 137

Therefore, median = 137

To calculate the mode:

Modal class = 125-145,

fm or f1 = 20, f0 = 13, f2 = 14 & h = 20

Mode formula:

Mode = l+ [(f1 – f0)/(2f1 – f0 – f2)] × h

Mode = 125 + [(20 – 13)/ (40 – 13 – 14)] × 20

= 125 + (140/13)

= 125 + 10.77

= 135.77

Therefore, mode = 135.77

Calculate the 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 = a 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

x̄ = a + h (∑fiui/∑fi) = 135 + 20 (7/68)

Mean = 137.05

In this case, mean, median and mode are more/less equal in this distribution.

2. If the median of a distribution given below is 28.5, find the value of x & 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:

Given data, n = 60

Median of the given data = 28.5

 CI 0-10 10-20 20-30 30-40 40-50 50-60 Frequency 5 x 20 15 y 5 Cumulative frequency 5 5+x 25+x 40+x 40+x+y 45+x+y

Where, N/2 = 30

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

Lower limit of median class, l = 20,

cf = 5 + x,

f = 20 & h = 10

$$\begin{array}{l}\text{Median}=l+\frac{\frac{N}{2}-cf}{f}\times h\end{array}$$

Substitute the values

28.5 = 20 + [(30 − 5 − x)/20] × 10

8.5 = (25 – x)/2

17 = 25 – x

Therefore, x = 8.

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

Since,

60 = 45 + x + y

Now, substitute the value of x, to find y

60 = 45 + 8 + y

y = 60 – 53

y = 7

Therefore, the value of x = 8 and y = 7.

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

 Age (in years) Number of policy holder 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

Given data: N = 100 and N/2 = 50

Median class = 35-40

Then, l = 35, cf = 45, f = 33 & h = 5

$$\begin{array}{l}\text{Median}=l+\frac{\frac{N}{2}-cf}{f}\times h\end{array}$$

Median = 35 + [(50 – 45)/33] × 5

= 35 + (25/33)

= 35.76

Therefore, the median age = 35.76 years.

4. The lengths of 40 leaves in a plant are measured correctly to the nearest millimeter, and the data obtained is represented as 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 the leaves.

(Hint : The data needs to be converted to continuous classes for finding the median, since the formula assumes continuous classes. The classes then change to 117.5 – 126.5, 126.5 – 135.5, . . ., 171.5 – 180.5.)

Solution:

Since the data are not continuous reduce 0.5 in the lower limit and add 0.5 in the upper limit.

 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

So, the data obtained are:

N = 40 and N/2 = 20

Median class = 144.5-153.5

then, l = 144.5,

cf = 17, f = 12 & h = 9

$$\begin{array}{l}\text{Median}=l+\frac{\frac{N}{2}-cf}{f}\times h\end{array}$$

Median = 144.5 + [(20 – 17)/ 12] × 9

= 144.5 + (9/4)

= 146.75 mm

Therefore, the median length of the leaves = 146.75 mm.

5. The following table gives the distribution of a lifetime 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 lifetime of a lamp.

Solution:

 Class Interval Frequency Cumulative 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

Data:

N = 400 & N/2 = 200

Median class = 3000 – 3500

Therefore, l = 3000, cf = 130,

f = 86 & h = 500

$$\begin{array}{l}\text{Median}=l+\frac{\frac{N}{2}-cf}{f}\times h\end{array}$$

Median = 3000 + [(200 – 130)/86] × 500

= 3000 + (35000/86)

= 3000 + 406.98

= 3406.98

Therefore, the median lifetime of the lamps = 3406.98 hours

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:

To calculate 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

Given:

N = 100 & N/2 = 50

Median class = 7-10

Therefore, l = 7, cf = 36, f = 40 & h = 3

$$\begin{array}{l}\text{Median}=l+\frac{\frac{N}{2}-cf}{f}\times h\end{array}$$

Median = 7 + [(50 – 36)/40] × 3

Median = 7 + (42/40)

Median = 8.05

Calculate the Mode:

Modal class = 7-10,

Where, l = 7, f1 = 40, f0 = 30, f2 = 16 & h = 3 Mode = 7 + [(40 – 30)/(2 × 40 – 30 – 16)] × 3

= 7 + (30/34)

= 7.88

Therefore mode = 7.88

Calculate the 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 58 16-19 4 17.5 70 Sum fi = 100 Sum fixi = 832

Mean = x̄ = ∑fi xi /∑fi

Mean = 832/100 = 8.32

Therefore, mean = 8.32

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

Given: N = 30 and N/2= 15

Median class = 55-60

l = 55, Cf = 13, f = 6 & h = 5

$$\begin{array}{l}\text{Median}=l+\frac{\frac{N}{2}-cf}{f}\times h\end{array}$$

Median = 55 + [(15 – 13)/6] × 5

= 55 + (10/6)

= 55 + 1.666

= 56.67

Therefore, the median weight of the students = 56.67

Exercise 14.4 Page: 293

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

Convert the given distribution table to a less than type cumulative frequency distribution, and we get

 Daily Income Cumulative Frequency (or) Number of workers Less than 120 12 Less than 140 26 Less than 160 34 Less than 180 40 Less than 200 50

From the table plot the points corresponding to the ordered pairs such as (120, 12), (140, 26), (160, 34), (180, 40) and (200, 50) on graph paper and the plotted points are joined to get a smooth curve and the obtained curve is known as less than type ogive curve 2. During the medical check-up 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:

From the given data, to represent the table in the form of graph, choose the upper limits of the class intervals are in x-axis and frequencies on y-axis by choosing the convenient scale. Now plot the points corresponding to the ordered pairs given by (38, 0), (40, 3), (42, 5), (44, 9),(46, 14), (48, 28), (50, 32) and (52, 35) on a graph paper and join them to get a smooth curve. The curve obtained is known as less than type ogive. Locate the point 17.5 on the y-axis and draw a line parallel to the x-axis cutting the curve at a point. From the point, draw a perpendicular line to the x-axis. The intersection point perpendicular to x-axis is the median of the given data. Now, to find the median by making a table.

 Class interval Number of students(Frequency) Cumulative Frequency Less than 38 0 – 38 0 0 Less than 40 38 – 40 3 – 0 = 3 3 Less than 42 40 – 42 5 – 3 = 2 5 Less than 44 42 – 44 9 – 5 = 4 9 Less than 46 44 – 46 14 – 9 = 5 14 Less than 48 46 – 48 28 – 14 = 14 28 Less than 50 48 – 50 32 – 28 = 4 32 Less than 52 50 – 52 35 – 22 = 3 35

Here, N = 35 and N/2 = 35/2 = 17.5

Median class = 46 – 48

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

The mode formula is given as:

$$\begin{array}{l}\text{Median}=l+\frac{\frac{N}{2}-cf}{f}\times h\end{array}$$

= 46 + [(17.5 – 14)/ 14] × 2

= 46 + 0.5

= 46 + 0.5 = 46.5

Thus, median is verified.

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

 Production Yield (in kg/ha) 50-55 55-60 60-65 65-70 70-75 75-80 Number of Farms 2 8 12 24 38 16

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

Solution:

Converting the given distribution to a more than type distribution, we get

 Production Yield (kg/ha) Number of farms More than or equal to 50 100 More than or equal to 55 100 – 2 = 98 More than or equal to 60 98 – 8 = 90 More than or equal to 65 90 – 12 = 78 More than or equal to 70 78 – 24 = 54 More than or equal to 75 54 – 38 = 16

From the table obtained draw the ogive by plotting the corresponding points where the upper limits in x-axis and the frequencies obtained in the y-axis are (50, 100), (55, 98), (60, 90), (65, 78), (70, 54) and (75, 16) on the graph paper. The graph obtained is known as more than type ogive curve. ## NCERT Solutions for Class 10 Maths Chapter 14 Statistics

Class 10 Maths Chapter 14, Statistics, is one of the most important chapters present in the textbook. The weightage of this chapter in the CBSE exam is around 11 to 12 marks. On average, there will be 3 questions which could be asked from this chapter and marks will be distributed in a manner of 3+4+4 (it could vary as per question).

Topics covered in Chapter 14, Statistics are as follows:

• Mean of Grouped Data
• Mode of Grouped Data
• Median of Grouped Data
• Graphical Representation of Cumulative Frequency Distribution

List of Exercises in Class 10 Maths Chapter 14 :
Exercise 14.1 Solutions 9 Questions ( 9 long)
Exercise 14.2 Solutions 6 Questions ( 6 long)
Exercise 14.3 Solutions 7 Questions ( 7 long)
Exercise 14.4 Solutions 3 Questions ( 3 long)

NCERT solutions for Class 10 Maths Chapter 14 – Statistics are made available for students aiming to obtain good marks in this chapter. The methods and procedure to solve the questions have been explained clearly in these NCERT Solutions, such that, students find it easy to understand the fundamentals quickly.

The world is highly data-oriented, in fact, each and every field has a group of data, which represents the relevant information. Statistics is the branch of mathematics which deals with the representation of data in a meaningful way.

You will face many real-life scenarios where the fundamentals of statistics are used to represent a set of data in tabular form, in graphs or in pie charts. There are a number of methods you will learn from this chapter such as, step deviation methods, finding mode and median of grouped data, converting frequency distribution and the relation between mode, mean and median methods, etc. 10th Class NCERT solutions are the best study materials to prepare for the CBSE exam.

### Key Features of NCERT Solutions for Class 10 Maths Chapter 14 – Statistics

• The solutions for the statistics chapter works as a reference for the students.
• It will help students to score marks against the questions asked from this chapter.
• Students can prepare and do the revision for Chapter 14 with this source.
• The questions of statistics have been solved by subject experts.
• The content of the material is as per the CBSE Syllabus (2023-24) and guidelines.

Statistics can also be understood in a more effective way by using the other solutions which are provided at BYJU’S. The solutions are prepared to help students perform well in the CBSE exams.

Disclaimer –

Dropped Topics – 14.5 Graphical representation of cumulative frequency distribution

## Frequently Asked Questions on NCERT Solutions for Class 10 Maths Chapter 14

Q1

### Does NCERT Solutions for Class 10 Maths Chapter 14 help you to clear CBSE exams?

Yes, NCERT Solutions for Class 10 Maths Chapter 14 is one of the important chapters of Class 10 Maths. These solutions are focused on learning various Mathematics tricks and shortcuts for quick and easy calculations. This makes them learn and clear the Maths subject in CBSE exams.
Q2

### How do the NCERT Solutions for Class 10 Maths Chapter 14 help in learning areas and volumes of geometrical shapes?

NCERT Solutions for Class 10 Maths Chapter 14 is created by our faculty in an interactive manner to make it easy for the students to understand the concepts. Students can refer to the PDF of solutions as a major study material to improve their speed in solving problems accurately.
Q3

### How many questions are present in the NCERT Solutions for Class 10 Maths Chapter 14?

The 14th Chapter of NCERT Solutions for Class 10 Maths has 4 exercises. The first exercise contains 9 questions, the second exercise contains 6 questions, the third exercise has 7 questions and the last or fourth exercise has 3 questions. So, there are a total of 25 long questions that are present in the NCERT Solutions for Class 10 Maths Chapter 14.