Ncert Solutions For Class 10 Maths Ex 14.2

Ncert Solutions For Class 10 Maths Chapter 14 Ex 14.2

Question 1:

The age of patients admitted in a hospitals are given in the following table for a complete 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

 

Calculate the mode and mean of the data above. Compare and interpret the two measures of central tendency.

Solution:

Modal class = 35-45, length = 35, height = 10, a1=23,a0=21anda3=14

 

Therefore, Mode = l+a1a02a1a0a2×h

= 35+(23212×232114)×10

= 35+211×10=36.8

Calculating Mean

CLASS INTERVAL ai zi aizi
5-15 6 10 60
15-25 11 20 220
25-35 21 30 630
35-45 21 30 920
45-55 14 50 700
55-65 5 60 300
Sum ai = 80 Sum aizi= 2830

 

p¯=aixiai=283080=35.37

 

 thereforeThe mode of data displays that the maximum number of patients is in the age group of 26.8, whereasthe average age of all the patients is 35.37.

 

Question 2:

The data given below provides the information for 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

  Find the modal lifetimes of the components.

 

Solution:

Modal class =  60-80, l = 60,a1=61,a0=52,a2=38 and h = 20

Mode=l+(a1a02a1a0a2)×h

 

=60+(61522×615238)×20

 

=60+932×20=65.62

 

 Question 3:

The total monthly household expenses of 200 families of a town are given in the following table. Calculate the modal monthly expenses of the families.  Also, calculate the mean monthly expenses.

Expenses 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,a1=40,a0=24,a2=33 and h = 500

Mode=l+(a1a02a1a0a2)×h

 

=1500+(40242×402433)×500

 

=1500+(1623)×500=1847.82

Calculation of mean:

Class Interval ai zi di =zi – b ui= di/h aiui
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
ai = 200 aiui = -35

 

p¯=aixiai×h =275035200×500

 

= 2750 – 87.5 = 2662.50

 

Question 4:

The state – wise teacher- student ratio in a college in India are given in the following table. Calculate the mean and mode of the given 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, a1=10,a0=9,a2=3 and h = 5

 

Mode=l+(a1a02a1a0a2)×h

 

=30(2092×1093)×5

 

=3018×5=30.625

 

Calculating mean:

Number of students per teacher ai zi di= zi –b ui = ui/h aiui
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 ai = 35 sum aiui= -23

 p¯=aixiai×h

 

=30.52335×5

 

=30.5237=29.22

 

Therefore, from the mode we know that the maximum number of states has 30 – 35 students per teacher, and from the mean we know that the average ratio of students per teacher is 29.22

 

Question 5:

Provided table shows the total runs scored by some of the top batsman of the world in one-day cricket matches.

Runs 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

Calculate the mode of the given information.

Solution:

Modal class = 4000-5000, l = 4000, a1=18,a0=4,a2=9 and h = 1000

Mode=l+(a1a02a1a0a2)×h =4000+(1842×1849)×1000 =4000+(1423)×1000=4608.70

 

Question 6:

A complete observation of the number of cars on the roads for 100 periods of 3 minutes are summarized in the following table. Calculate 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, a1=20,a0=12,a2=11 and h = 10

Mode=l+(a1a02a1a0a2)×h

 

=40+(20122×201211)×10

 

=40+(817)×10=44.70

 

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