# NCERT Solutions For Class 10 Maths Chapter 14-Statistics

NCERT Solutions for Class 10 Maths Chapter 14 Statistics are provided here, which can be downloaded for free, in PDF format. The **Solutions for chapter 14 Statistics** are prepared by our experts who have done their specialization in Maths.

All the solved questions of Statistics are with respect to the NCERT syllabus and guidelines, to help students solve each exercise question and prepare for the CBSE Board exam. Using these solutions as a reference tool will be helpful for the students to score good marks. Students can also get the exercise-wise Solutions for Class 10 Maths all chapters and practise solving the problems.

### Access Answers of 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:

In order to find the mean value, we will use 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.) | 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:

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 = 150 and class interval is h = 20.

So, ui = (xi – A)/h = ui = (xi – 150)/20

Substitute and find the values as follows:

Daily wages (Class interval) | Number of workers frequency (fi) | Mid-point (xi) | ui = (xi – 150)/20 | fiui |

100-120 | 12 | 110 | -2 | -24 |

120-140 | 14 | 130 | -1 | -14 |

140-160 | 8 | 150 | 0 | 0 |

160-180 | 6 | 170 | 1 | 6 |

180-200 | 10 | 190 | 2 | 20 |

Total | Sum fi = 50 | Sum fiui = -12 |

So, the formula to find out the mean is:

Mean = x̄ = A + h∑fiui /∑fi =150 + (20 × -12/50) = 150 – 4.8 = 145.20

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

Here, the value of mid-point (xi) 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 = A | 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 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 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 | 0 | 0 |

77-80 | 7 | 78.5 | 1 | 7 |

80-83 | 4 | 81.5 | 3 | 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:

Since, the given data is not continuous so we add 0.5 to the upper limit and subtract 0.45 from the lower limit as the gap between two intervals are 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) | di = xi – A | fidi |

49.5-52.5 | 15 | 51 | -6 | 90 |

52.5-55.5 | 110 | 54 | -3 | -330 |

55.5-58.5 | 135 | 57 = A | 0 | 0 |

58.5-61.5 | 115 | 60 | 3 | 345 |

61.5-64.5 | 25 | 63 | 6 | 150 |

Sum fi = 400 | Sum fidi = 75 |

The formula to find out the Mean is:

Mean = x̄ = A +h ∑fidi /∑fi

= 57 + 3(75/400)

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

Daily expenditure(in c) | 100-150 | 150-200 | 200-250 | 250-300 | 300-350 |

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

Solution:

Find the midpoint of the given interval using the formula.

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

Let is 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 | 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 | 0.20 | 0.40 |

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 air is 0.099 ppm.

**8. A class teacher has the following absentee record of 40 students of a class for the wholeterm. 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 meanliteracy 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 and class interval is h = 10.

So, ui = (xi-A)/h = ui = (xi-70)/10

Substitute and find the values as follows:

Class Interval | Frequency (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 |

So, Mean = x̄ = A+(∑fiui /∑fi)×h

= 70+(-2/35)×10

= 69.42

Therefore, the mean literacy part = 69.42

Exercise 14.2 Page: 275

**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 twomeasures 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,

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

Mode = 35+(20/11) = 35+1.8

Mode = 36.8 year

So the mode of the given data = 36.8 year

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

= 35.37 years

Therefore, the mean of the given data = 35.37 years

**2. The following data gives the information on the observed lifetimes (in hours) of 225electrical 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.

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 x 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 200families of a village. Find the modal monthly expenditure of the families. Also, find themean 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:

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 | 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 = Rupees 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

Mode = 30+(5/8) = 30+0.625

Mode = 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 | 105.5 |

Sum fi = 35 | Sum fixi = 1022.5 |

Mean = x̄ = ∑fixi /∑fi

= 1022.5/35

= 29.2

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

Mode = 4000+(14000/23) = 4000+608.695

Mode = 4608.695

Mode = 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

Mode = 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:

=125+((34−22)/20) × 20

=125+12 = 137

Therefore, median = 137

To calculate the mode:

Modal class = 125-145,

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 | 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 then, 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

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

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=5+20+15+5+x+y

Now, substitute the value of x, to find y

60 = 5+20+15+5+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 = 35-45

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

Median = 35+((50-45)/33) × 5

= 35 + (5/33)5

= 35.75

Therefore, the median age = 35.75 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 leaves. **

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

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 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 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

Median = 3000 + ((200-130)/86) × 500

= 3000 + (35000/86)

= 3000 + 406.97

= 3406.97

Therefore, the median life time of the lamps = 3406.97 hours

**6. In this 100 surnames were randomly picked up from a local telephone directory and the frequency distribution of the number of letters in 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 number of median letters in the surnames. Find the number of mean letters in the surnames and also, find the size of modal in 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

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 | 51 |

16-19 | 4 | 17.5 | 70 |

Sum fi = 100 | Sum fixi = 825 |

Mean = x̄ = ∑fi xi /∑fi

Mean = 825/100 = 8.25

Therefore, mean = 8.25

**7. The distributions of below give a weight of 30 students of a class. Find the median weight of a student.**

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

Median = 55+((15-13)/6)×5

Median=55 + (10/6) = 55+1.666

Median =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 if a factory. Convert the distribution above to a less than type cumulative frequency distribution and draw its ogive.**

Daily income in Rupees | 100-120 | 120-140 | 140-160 | 160-180 | 180-200 |

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

Solution

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

Daily income | Frequency | Cumulative Frequency |

Less than 120 | 12 | 12 |

Less than 140 | 14 | 26 |

Less than 160 | 8 | 34 |

Less than 180 | 6 | 40 |

Less than 200 | 10 | 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 an 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 mode by making a table.

Class interval | Number of students(Frequency) | Cumulative Frequency |

Less than 38 | 0 | 0 |

Less than 40 | 3-0=3 | 3 |

Less than 42 | 5-3=2 | 8 |

Less than 44 | 9-5=4 | 9 |

Less than 46 | 14-9=5 | 14 |

Less than 48 | 28-14=14 | 28 |

Less than 50 | 32-28=4 | 32 |

Less than 52 | 35-22=3 | 35 |

The class 46 – 48 has the maximum frequency, therefore, this is modal class

Here, *l *= 46, h = 2, f1= 14, f0= 5 and f2 = 4

The mode formula is given as:

Now, Mode =

= 46 + 0.95 = 46.95

Thus, mode is verified.

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

Production Yield | 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

this graph paper. The graph obtained is known as more than type ogive curve.

Class 10 Maths Chapter 14, Statistics, is one of the most important of all the chapter present in the textbook. The weightage of this chapter in the final 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;

- 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 Question ( 9 long)

Exercise 14.2 Solutions 6 Question ( 6 long)

Exercise 14.3 Solutions 7 Question ( 7 long)

Exercise 14.4 Solutions 3 Question ( 3 long)

NCERT solutions for Class 10 Maths Chapter 14- Statistics are made available for students who want 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 or 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 board 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 the statistics chapter.
- Students can prepare and do the revision for chapter 14 with this source.
- The questions of statistics are solved by subject experts.
- The content of the material is as per the CBSE syllabus and guidelines.

- Chapter 1 Real Numbers
- Chapter 2 Polynomials
- Chapter 3 Pair of Linear Equations in Two Variables
- Chapter 4 Quadratic Equations
- Chapter 5 Arithmetic Progressions
- Chapter 6 Triangles
- Chapter 7 Coordinate Geometry
- Chapter 8 Introduction to Trigonometry
- Chapter 9 Some Applications of Trigonometry
- Chapter 10 Circles
- Chapter 11 Constructions
- Chapter 12 Areas Related to Circles
- Chapter 13 Surface Areas and Volumes
- Chapter 14 Statistics
- Chapter 15 Probability