NCERT Solutions For Class 10 MATH CHAPTER – 14 EXERCISE – 14.3

NCERT Solutions For Class 10 MATH Chapter – 14 Exercise – 14.3

Q1. The following frequency distribution gives the monthly consumption of electricity of 68 consumer 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:-

Median = l + (n/2 – cf/f) x h

= 125 + (34-22)/20) x 20

= 125 + 12 = 137

Median = 137

Mode = l + [(f₁-f₀)/(2f₁– f₀ – f₂)] x h

Mode = 125 + (20-13)/(40-13-14)x 20

Mode = 135.77

Calculate the Mean

Class intervals	f_i	x_i	D_i = x_i – a	U_i = d_i /h	F_i/u_i
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 f_i = 68 Sum f_iu_i=

X= a + h f_iu_i/f_i

= 135 + 20(7/68)

= Mean = 137.05

Q2. If the median of a distribution given below is 28.5 then , find the value 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:-

Median = l + (n/2 – cf/f) x h

28.5 = 20+(30-5-x)/20) x 10

17 = 25 – x

X = 8

60 = 5 + 20 + 15 + 5 + x +y

Now substitute the value of x , to find y

60 = 5 + 20 + 15 + 5 + 8 + y

Y = 7

Q3. The life insurance agent found the following data for the following data for the distribution of ages of policy 100 policy holders Calculate the median age , if policies are given only to the person whose age is 18 years onward but less than the 60 years.

Age	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 intervals	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.

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.

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

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

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