Ncert Solutions For Class 10 Maths 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, $a_{1} = 23, a_{0} = 21\, and\, a_{3} = 14$

 

Therefore, Mode = $l +\frac{a_{1}-a_{0}}{2a_{1}-a_{0}-a_{2}}\times h$

= $35+\left ( \frac{23-21}{2\times 23-21-14} \right )\times 10$

= $35+\frac{2}{11}\times 10=36.8$

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

 

$\bar{p}=\frac{\sum a_{i}x_{i}}{\sum a_{i}}=\frac{2830}{80}=35.37$

 

$\ therefore$The 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,$a_{1} = 61, a_{0} = 52, a_{2} = 38$ and h = 20

$Mode = l + \left ( \frac{a_{1}-a_{0}}{2a_{1}-a_{0}-a_{2}} \right )\times h$

 

$= 60 + \left ( \frac{61 – 52}{2 \times 61 – 52 – 38 } \right ) \times 20$

 

$= 60 + \frac{9}{32} \times 20 = 65.62$

 

 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,$a_{1} = 40, a_{0} = 24, a_{2} = 33$ and h = 500

$Mode = l + \left ( \frac{a_{1}-a_{0}}{2a_{1}-a_{0}-a_{2}} \right )\times h$

 

$= 1500 + \left ( \frac{40 – 24}{2 \times 40 – 24 – 33} \right ) \times 500$

 

$= 1500 + \left ( \frac{16}{23} \right ) \times 500 = 1847.82$

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

 

$\bar{p}=\frac{\sum a_{i}x_{i}}{\sum a{i}}\times h$ $= 2750 – \frac{35}{200} \times 500$

 

= 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, $a_{1} = 10, a_{0} = 9, a_{2} = 3$ and h = 5

 

$Mode = l +\left ( \frac{a_{1}-a_{0}}{2a_{1}-a_{0}-a_{2}} \right )\times h$

 

$= 30 – \left ( \frac{20 – 9}{2 \times 10 – 9 – 3 } \right ) \times 5$

 

$= 30 – \frac{1}{8} \times 5 = 30.625$

 

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

 $\bar{p}=\frac{\sum a_{i}x_{i}}{\sum a_{i}}\times h$

 

$= 30.5 – \frac{23}{35}\times 5$

 

$= 30.5 – \frac{23}{7} = 29.22$

 

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, $a_{1} = 18, a_{0} = 4, a_{2} = 9$ and h = 1000

$Mode = l + \left ( \frac{a_{1}-a_{0}}{2a_{1}-a_{0}-a_{2}} \right )\times h$ $= 4000 + \left ( \frac{18 – 4}{2 \times 18 – 4 -9} \right )\times 1000$ $= 4000 + \left ( \frac{14}{23} \right )\times 1000 = 4608.70$

 

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, $a_{1} = 20, a_{0} = 12, a_{2} = 11$ and h = 10

$Mode = l + \left ( \frac{a_{1}-a_{0}}{2a_{1}-a_{0}-a_{2}} \right )\times h$

 

$= 40 + \left ( \frac{20 – 12}{2 \times 20 – 12 – 11} \right )\times 10$

 

$= 40 + \left ( \frac{8}{17} \right )\times 10 = 44.70$

 

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