NCERT Solutions Maths Class 10 Exercise 14.3

NCERT Solutions Maths Class 10 Exercise 14.3

 

Maths Class 10 Exercise 14.3

 

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 consumers

65 – 85

4

85 – 105

5

105 – 125

13

125 – 145

20

145 – 165

14

165 – 185

8

185 – 205

4

 

Solution: To calculate median:


Here, Σfi = n = 68, therefore n/2 = 68/2 = 34, which lies in interval 125 – 145.

Therefore, the median class = 125 – 145

So, l = 125, n = 68, f = 20, cf = 22 and h = 20

Now, Median = 

= 125 + (34 – 22)/20 × 20  

= 125 + 12

= 137


To calculate mean:


From given data, assumed mean (a) = 135

Width of the class (h) = 20

Therefore,  

= 7/68

= 0.102 

Using formula, Mean   

= 135 + 20(0.102)

= 135 + 2.04

= 137.04


To calculate mode:

In the given data, maximum frequency is 20 and it corresponds to the class interval 125 – 145.

Therefore, the modal class = 125 – 145

And l = 125, f1 = 20, f0 = 13, f2 = 14 and h = 20

Since, Mode = 

=

= 

= 125 + 140/13 

= 125 + 10.7692

= 125 + 10.77

= 135.77

Hence, median, mean and mode of the given data is 137 units, 137.04 units and 135.77 units respectively.

 

2. If the median of the distribution given below is 28.5, then find the values 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:


Here, Σfi = n = 60, therefore n/2 = 60/2 = 30

It is given that the median of the distribution is 28.5, which lies in the interval 20 – 30.

Therefore, the median class = 20 – 30

So, l =20, n = 60, f = 20, cf = 5 + x and h = 10

Since, 45 + x + y = 60 

x + y = 15           ………. (i)

Now, Median = 

 

 

 

2(28.5) = 65 – x

 57 = 65 – x

 x = 65 – 57

 x = 8

Putting the value of x in equation (i), we get

8 + y = 15

 y = 7

Hence, the value of x and y is 8 and 7 respectively.

 

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


Age (in years)

Number of policy holders

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:


Here, Σfi = n = 100, therefore n/2 = 100/2 = 50, which lies in interval 35 – 40.

So, Median class = 35 – 40

So, l = 35, n = 100, f = 33, cf = 45 and h = 5

Now, Median = 

= 

= 

= 35 + 25/33

= 35 + 0.7575

= 35 + 0.76 (approx.)

= 35.76

Hence, median age of given data is 35.76 years.

 

4. The lengths of 40 leaves of a plant are measured correct to the nearest millimetre, and data obtained is represented in the following table. Find the median length of the leaves.


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

 

Solution: Since the frequency distribution is not continuous, so first of all, we shall change it into continuous distribution.


Here, Σfi = n = 40, therefore n/2 = 40/2 = 20, which lies in interval 144.5 – 153.5.

Therefore, Median class = 144.5 – 153.5

So, l = 144.5, n = 40, f = 12, cf = 17 and h = 9

Now, Median = 

= 

= 

= 144.5 + 2.25

= 146.75

Hence, median length of the leaves is 146.75 mm.

 

5. The following table gives the distribution of the life time of 400 neon lamps. Find the median life time of a lamp.

 

Life time (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

 

Solution:  


Here, Σfi = n = 400, therefore n/2 = 400/2 = 200, which lies in interval 3000 – 3500.

Therefore, Median class = 3000 – 3500

So, l = 3000, n = 400, f = 86, cf = 130 and h = 500

Now, Median = 

= 

= 

= 3000 + 406.9767441

= 3000 + 406.98 (approx.)

= 3406.98

Hence, median life time of a lamp is 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:


Here, Σfi = n = 100, therefore n/2 = 100/2 = 50, which lies in interval 7 – 10.

Therefore, Median class = 7 – 10

So, l = 7, n = 100, f = 40, cf = 36 and h = 3

Now, Median = 

= 

= 

= 7 + 1.05

= 8.05


To calculate mean:


From given data, assumed mean (a) = 8.5

Width of the class (h) = 3

Therefore,    

= –6/100

= –0.06

Using formula, Mean   

= 8.5 + 3(– 0.06)

= 8.5 – 0.18

= 8.32


To calculate mode:

In the given data, maximum frequency is 40 and it corresponds to the class interval 7 – 10.

Therefore, the modal class = 7 – 10

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

Since, Mode = 

= 

= 

= 7 + 30/34 

= 7 + 0.88 (approx.)

= 7.88

Hence, median, mean and mode of the given data is 8.05 letters, 8.32 letters and 7.88 letters, respectively.

 

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

 

Weight (in kg)

Number of students

40 – 45

2

45 – 50

3

50 – 55

8

55 – 60

6

60 – 65

6

65 – 70

3

70 – 75

2

 

Solution:


Here, Σfi = n = 30, therefore n/2 = 30/2 = 15, which lies in interval 55 – 60.

Therefore, the median class = 55 – 60

So, l = 55, n = 30, f = 6, cf = 13 and h = 5

Now, Median = 

= 

= 55 + 10/6 

= 55 + 1.66666

= 55 + 1.67 (approx.)

= 56.67

Hence, median weight of the students is 56.67 kg.

 

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