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 | fi | xi | Di = xi – a | Ui = di /h | Fi/ui |
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 fi = 68 Sum fiui=
7
X= a + h fiui/fi
= 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