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Class 10 NCERT Solutions- Chapter 14 Statistics – Exercise 14.3

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  • Last Updated : 03 Mar, 2021
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Question 1. The following frequency distribution gives the monthly consumption of electricity of 68 consumers 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:

 Total number of consumer n = 68 

n/2 =34

So, the median class is 125-145 with cumulative frequency = 42

Here, l = 125, n = 68, Cf = 22, f = 20, h = 20

Now we find the median:

Median = l +\left(\frac{\frac{n}{2}-c_f}{f}\right)\times h  

=125+\frac{(34−22)}{20} × 20

= 125 + 12 = 137

Hence, the median is 137

Now we find the mode:

Modal class = 125 – 145,

Frequencies are

f1 = 20, f0 = 13, f2 = 14 & h = 20

Mode l+ \left[\frac{(f1-f0)}{(2f1-f0-f2)}\right]×h

On substituting the values in the given formula, we get

Mode = 125 + \frac{(20-13)}{(40-13-14)}×20

= 125 + 140/13 

= 125 + 10.77

= 135.77

Hence, the mode is 135.77

Now we find the mean:

Class Interval fi xi di = xi – a ui = di/h fiui
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

\bar{x} =a+h \frac{∑f_iu_i}{∑f_i}

= 135 + 20(7/68)

= 137.05

Hence, the mean is 137.05

Now, on comparing the median, mean, and mode, we found that mean, median and mode are more/less equal in this distribution.

Question 2. If the median of a distribution given below is 28.5 then, find the value of x & 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:

According to the question

The total number of observations are n = 60

Median of the given data = 28.5

n/2 = 30  

Median class is 20 – 30 with a cumulative frequency = 25 + x

Lower limit of median class, l = 20,

Cf = 5 + x,

f = 20 & h = 10

Now we find the median:

Median = l+\left(\frac{\frac{n}{2}-c_f}{f}\right)\times h

On substituting the values in the given formula, we get

28.5 = 20+\frac{(30−5−x)}{20} × 10

8.5 = (25 – x)/2

17 = 25 – x

Therefore, x = 8

From the cumulative frequency, we can identify the value of x + y as follows:

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

On substituting the values of x, we will find the value of y

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

y = 60 – 53

y = 7

So the value of a is 8 and y is 7

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

Age (in years) 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: 

According to the given question the table is 

Class interval 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

Now we find the median:

Median = l+\left(\frac{\frac{n}{2}-c_f}{f}\right)\times h

On substituting the values in the given formula, we get

Median = 35+\frac{(50-45)}{33} × 5

= 35 + 5(5/33) 

= 35.75

Hence, the median age is 35.75 years.

Question 4. 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:

The data in the given table are not continuous to reduce 0.5 in the lower limit and add 0.5 in the upper limit.

We get a new table:

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

From the given table

n = 40 and n/2 = 20

Median class = 144.5 – 153.5

l = 144.5,

cf = 17, f = 12 & h = 9

Now we find the median:

Median = l+\left(\frac{\frac{n}{2}-c_f}{f}\right)\times h

On substituting the values in the given formula, we get

Median = 144.5+\frac{(20-17)}{12}×9

= 144.5 + 9/4 

= 146.75 mm

Hence, the median length of the leaves is 146.75 mm.

Question 5. 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:

According to the question

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

n = 400 and n/2 = 200

Median class = 3000 – 3500

l = 3000, Cf = 130,

f = 86 & h = 500

Now we find the median:

Median = l+\left(\frac{\frac{n}{2}-c_f}{f}\right)\times h

On substituting the values in the given formula, we get

Median = 3000 + \frac{(200-130)}{86} × 500

= 3000 + 35000/86 = 3000 + 406.97

= 3406.97

Hence, the median lifetime of the lamps is 3406.97 hours

Question 6. 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:

According to the question

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

n = 100 and n/2 = 50

Median class = 7 – 10

Therefore, l = 7, Cf = 36, f = 40 & h = 3

Now we find the median:

Median = l+\left(\frac{\frac{n}{2}-c_f}{f}\right)\times h

On substituting the values in the given formula, we get

Median = 7+\frac{(50-36)}{40} × 3

Median = 7 + 42/40 = 8.05

Hence, the median is 8.05

Now we find the mode:

Modal class = 7 – 10,

Where, l = 7, f1 = 40, f0 = 30, f2 = 16 & h = 3

Mode = l+\left(\frac{f_1-f_0}{2f_1-f_0-f_2}\right)\times h

On substituting the values in the given formula, we get

Mode = 7+\frac{(40-30)}{(2×40-30-16)} × 3

= 7 + 30/34 = 7.88

Hence, the mode is 7.88

Now we find 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 = \bar{x}= \frac{∑f_i x_i }{∑f_i}

= 825/100 = 8.25

Hence, the mean is 8.25

Question 7. 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:

According to the question

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

n = 30 and n/2 = 15

Median class = 55 – 60

l = 55, Cf = 13, f = 6 & h = 5

Now we find the median:

Median = l +\left(\frac{\frac{n}{2}-c_f}{f}\right)\times h

On substituting the values in the given formula, we get

Median = 55+\frac{(15-13)}{6}×5

= 55 + 10/6 = 55 + 1.666

= 56.67

Hence, the median weight of the students is 56.67


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