# NCERT Solutions for class 10 Maths Chapter 14- Statistics Exercise 14.3

NCERT Solutions for Class 10 Maths Chapter 14- Statistics Exercise 14.3 have been provided here, prepared by our subject experts. These solutions are designed in accordance with NCERT syllabus and guidelines prescribed by CBSE.

NCERT Solutions for Class 10 MathsÂ are in the context of the upcoming session of 2021-2022. These are helpful for students, to do the revision at the time of first and second term exam and also to score well. Go through the solutions presented here for each question of Exercise 14.3.

### Access Other Exercise Solutions of Class 10 Maths Chapter 14- Statistics

Exercise 14.1 Solutions 6 Question (6 long)

Exercise 14.2 Solutions 7 Question (7 long)

Exercise 14.4 Solutions 3 Question (3 long)

### Access Answers to NCERT Class 10 Maths Chapter 14 â€“ Statistics Exercise 14.3

1. The following frequency distribution gives the monthly consumption of 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 henceÂ 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+((34âˆ’22)/20) Ã— 20

=125+12 = 137

Therefore, median = 137

To calculate the mode:

Modal class = 125-145,

f1=20, f0=13, f2=14Â & h = 20

Mode formula:

Mode = l+ [(f1-f0)/(2f1-f0-f2)]Ã—h

Mode = 125 + ((20-13)/(40-13-14))Ã—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 SumÂ fi= 68 SumÂ fiui= 7

xÌ„ =a+h âˆ‘fiui/âˆ‘fiÂ =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/2Â = 30

Median class is 20 â€“ 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+((30âˆ’5âˆ’x)/20) Ã— 10

8.5 = (25 â€“ 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 theÂ 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 andÂ n/2Â = 50

Median class = 35-45

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

Median = 35+((50-45)/33) Ã— 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.Â Â Â Â Â Â Â Â Â Â Â Â Â

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 andÂ n/2Â = 20

Median class = 144.5-153.5

then, l = 144.5,

cf = 17, f = 12 & h = 9

Median = 144.5+((20-17)/12)Ã—9

= 144.5+(9/4)

= 146.75Â 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/2Â = 200

Median class = 3000 â€“ 3500

Therefore, l = 3000, Cf = 130,

f = 86 & h = 500

Median = 3000 + ((200-130)/86) Ã— 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/2Â = 50

Median class = 7-10

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

Median = 7+((50-36)/40) Ã— 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)/(2Ã—40-30-16)) Ã— 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 fiÂ = 100 Sum fixiÂ = 825

Mean = xÌ„ = âˆ‘fi xiÂ /âˆ‘fiÂ

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 andÂ n/2= 15

Median class = 55-60

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

Median = 55+((15-13)/6)Ã—5

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

Median =56.67

Therefore, the median weight of the students = 56.67

In this exercise, students will solve questions based on finding the median of a grouped data, which is a measure of central tendency, which gives the value of the middle-most observation in the data.

A group of collection of data will be provided to the students to find the median for cases having odd and even number of observations. To solve complete problems for Class 10 Maths Chapter 14 exercise-wise, students can visit us any time. Also, get a few more study materials such as notes, books, question papers, etc. to practise well.

The problems in Ex.14.3 are solved as per the guidance and formulas, explained in the example questions before the exercise. NCERT solutions serve as the best materials for students who are looking for solutions for all the subjects class-wise and chapter-wise.