# RD Sharma Solutions Class 7 Chapter 23 Data Handling Central Values

Read RD Sharma Solutions Class 7 Chapter 23 Data Handling Central Values below, students should study RD Sharma class 7 Mathematics available on Studiestoday.com with solved questions and answers. These chapter wise answers for class 7 Mathematics have been prepared by teacher of Grade 7. These RD Sharma class 7 Solutions have been designed as per the latest NCERT syllabus for class 7 and if practiced thoroughly can help you to score good marks in standard 7 Mathematics class tests and examinations

Exercise 23.1

Question :1. Ashish studies for 4 hours, 5 hours and 3 hours on three consecutive days. How many hours does he study daily on an average?
Solution  1:
On three consecutive days, Ashish studies for 4 hours, 5 hours, and 3 hours.
Average of study hours = (sum of hours)/(number of days)
= (4 + 5 + 3)/3
= 12/3 = 4 hours
Thus, on an average Ashish studies for 4 hours.

Question :2. A cricketer scores the following runs in 8 innings: 58, 76, 40, 35, 48, 45, 0, 100.
Find the mean score.
Solution 2:

Runs in 8 innings: 58, 76, 40, 35, 48, 45, 0, 100
Mean of the score = (total sum of runs)/(number of innings)
= (58 + 76 + 40 + 35 + 48 + 45 + 0 + 100)/8
= 402/8 = 50.25 runs.
Thus, the mean of the score are 50.25 runs.

Question :3. The marks (out of 100) obtained by a group of students in science test are 85, 76, 90, 84, 39, 48, 56, 95, 81 and 75. Find the
(i) Highest and the lowest marks obtained by the students.
(ii) Range of marks obtained.
(iii) Mean marks obtained by the group.
Solution 3:
We must arrange the marks in ascending order in order to find the highest and lowest marks:
39, 48, 56, 75, 76, 81, 84, 85, 90, 95
(i) The highest score is 95, while the lowest is 39.
(ii) The range of the marks gained is: (95 – 39) = 56.
(iii) According to the following data,
Mean of the marks = (Sum of the marks)/(Total number of students)
= (39 + 48 + 56 + 75 + 76 + 81 + 84 + 85 + 90 + 95)/10
= 729/10 = 72.9.
Thus, the mean of the mark of the students is 72.9.

Question :4. The enrolment of a school during six consecutive years was as follows:
1555, 1670, 1750, 2019, 2540, 2820
Find the mean enrolment of the school for this period.
Solution 4:
Enrolment of a school during 6 consecutive years are as follows:
1555, 1670, 1750, 2019, 2540, 2820
The mean of the enrolment = (Sum of the enrolments in each year)/(Total number of years)
= (1555 + 1670 + 1750 + 2019 + 2540 + 2820)/6
= 12354/6 = 2059.
As a result, the school's average enrolment for the specified timeframe is 2059.

Question :5. The rainfall (in mm) in a city on 7 days of a certain week was recorded as follows:
(i) Find the range of the rainfall from the above data.
(ii) Find the mean rainfall for the week.
(iii) On how many days was the rainfall less than the mean rainfall.
Solution 5:
(i) The range of the rainfall = Maximum rainfall – Minimum rainfall
= 20.5 – 0.0 = 20.5 mm.

(ii) The mean of the rainfall = (Sum of the rainfall inthe week)/(Total number of rainfall)
=  (0.0 + 12.2 + 2.1 + 0.0 + 20.5 + 5.3 + 1.0)/7
= 41.1/7 = 5.87 mm.

(iii) There are 5 days (Mon, Wed, Thu, Sat and Sun), when the rainfall was less than the mean, i.e., 5.87 mm.

Question :6. If the heights of 5 persons are 140 cm, 150 cm, 152 cm, 158 cm and 161 cm respectively, find the mean height.
Solution   6:
The mean of the height = (Sum of the heights)/(Total number of persons)
= (140 + 150 + 152 + 158 + 161)/5
= 761/5 = 152.2 cm.

Question :7. Find the mean of 994, 996, 998, 1002 and 1000.
Solution  7:
Mean = (Sum of the given numbers)/(Total number of given numbers)
Mean = (994 + 996 + 998 + 1002 + 1000)/5
= 4990/5 = 998.

Question :8. Find the mean of first five natural numbers.
Solution   8:
First five natural numbers are 1, 2, 3, 4 and 5
Mean of the first five natural numbers = (Sum of the first five natural numbers)/5
=(1 + 2 + 3 + 4 + 5)/5
= 15/5 = 3

Question :9. Find the mean of all factors of 10.
Solution 9:
Factors of 10 are 1, 2, 5 and 10
Arithmetic mean of all factors of 10 =(Sum of the factors of 10)/(total number of factors)
= ((1 + 2 + 5 + 10))/4
= 18/4 = 4.5

Question :10. Find the mean of first 10 even natural numbers.
Solution 10:
The first ten even natural numbers are 2, 4, 6, 8, 10, 12, 14, 16, 18 and 20.
Mean of first ten even natural numbers =(Sum of the first 10 even natural numbers)/10
= (2 + 4 + 6 + 8 + 10 + 12 + 14 + 16 + 18 + 20)/10
= 110/10 = 11

Question :11. Find the mean of x, x + 2, x + 4, x + 6, x + 8
Solution 11:
Mean =(Sum of observations)/(Number of observations)
= (x + x + 2 + x + 4 + x + 6 + x + 8)/5
= (5x + 20)/5
=(5(x + 4))/5
= x + 4

Question :12. Find the mean of first five multiples of 3.
Solution  12:
The first five multiples of 3 are 3, 6, 9, 12 and 15.
Mean of first five multiples of 3 are = (Sum of first five multiples of 3)/(total number of multiples)
= (3 + 6 + 9 + 12 + 15)/5
= 45/5 = 9

Question :13. Following are the weights (in kg) of 10 new born babies in a hospital on a particular day: 3.4, 3.6, 4.2, 4.5, 3.9, 4.1, 3.8, 4.5, 4.4, 3.6 Find the mean X¯
Solution 13:
Mean of X¯¯¯¯ = (sum of observations)/(number of observations)
= (sum of weights of babies)/(number of babies)
= ((3.4 + 3.6 + 4.2 + 4.5 + 3.9 + 4.1 + 3.8 + 4.5 + 4.4 + 3.6))/10
= 40/10  = 4 kg
Hence, the mean of the weight of babies are 4 kg

Question :14. The percentage of marks obtained by students of a class in mathematics are:
64, 36, 47, 23, 0, 19, 81, 93, 72, 35, 3, 1 Find their mean.
Solution 14:
Mean = (sum of the marks obtained)/(total number of students)
= (64 + 36 + 47 + 23 + 0 + 19 + 81 + 93 + 72 + 35 + 3 + 1)/12
= 474/12 = 39.5%
Hence, the mean of percentage 39.5%.

Question :15. The numbers of children in 10 families of a locality are:
2,  4, 3, 4, 2, 3, 5, 1, 1, 5 Find the mean number of children per family.
Solution 15:
Mean number of children per family = (sum of total number of children)/(total number of families)
= (2 + 4 + 3 + 4 + 2 + 3 + 5 + 1 + 1 + 5)/10
= 30/10 = 3
Thus, on an average there are 3 children per family in the locality.

Question :16. The mean of marks scored by 100 students was found to be 40. Later on it was discovered that a score of 53 was misread as 83. Find the correct mean.
Solution  16:
n = the number of observations = 100
Mean = 40
Mean = (sum of observations)/(total number of observations)
40 = (sum of the observations)/100
Sum of the observations = 40 x 100
Thus, improper sum of the observations = 40 x 100 = 4000.
Now, The proper sum of the observations = Improper sum of the observations – Improper observation + proper observation
The proper sum of observations = 4000 – 83 + 53
The proper sum of observations = 4000 – 30 = 3970
proper mean = (proper sum of the observations)/(number of observations)
= 3970/100 = 39.7

Question :17. The mean of five numbers is 27. If one number is excluded, their mean is 25. Find the excluded number.
Solution  17:
Mean =(sum of five numbers)/5 = 27
So, sum of the 5 numbers = 5 x 27 = 135.
Now, the mean of 4 numbers = (sum of the four numbers)/4 = 25
So, sum of the 4 numbers = 4 x 25 = 100.
Thus, the excluded number = Sum of the 5 number – Sum of the 4 numbers
The excluded number = 135 – 100 = 35.

Question :18. The mean weight per student in a group of 7 students is 55 kg. The individual weights of 6 of them (in kg) are 52, 54, 55, 53, 56 and 54. Find the weight of the seventh student.
Solution   18:
Mean = (sum of weights of students)/(number of students)
Let be x kg the weight of the 7th student.
Mean = (52 + 54 + 55 + 53 + 56 + 54 + x)/7
55 = (52 + 54 + 55 + 53 + 56 + 54 + x)/7
55 x 7 = 324 + x
385 = 324 + x
x = 385 – 324
x = 61 kg.
Thus, weight of 7th student is 61kg.

Question :19. The mean weight of 8 numbers is 15 kg. If each number is multiplied by 2, what will be the new mean?
Solution  19:
Let x1, x2, x3…x8 be the eight numbers whose mean is 15 kg.
15 = (x1 + x2 + x3+⋯…+ x8)/8
x1 + x2 + x3 + …+ x8 = 15 × 8
x1 + x2 + x3 +…+ x8 = 120
so, the new numbers be 2x1, 2x2, 2x3 …2x8.
Let M be the arithmetic mean of the new numbers.
M = (2x1 + 2x2 + 2x3 +⋯+ 2x8)/8
= (2(x1 + x2 + x3 + …+ x8))/8
= ((2 × 120))/8 = 30

Question :20. The mean of 5 numbers is 18. If one number is excluded, their mean is 16. Find the excluded number.
Solution  20:
Let x1, x2, x3, x4 and x5 be 5 numbers whose mean is 18.
18 = (Sum of five numbers)/5
So, sum of 5 numbers = 18 × 5 = 90
If one of the numbers is removed, the mean is 16.
16 = (Sum of four numbers )/4
Thus, sum of 4 numbers = 16 × 4 = 64.
The omitted number = Sum of 5 observations – Sum of 4 observations
The omitted number = 90 – 64 = 26
Thus, The omitted number is 26.

Question :21. The mean of 200 items was 50. Later on, it was discovered that the two items were misread as 92 and 8 instead of 192 and 88. Find the correct mean.
Solution  21:
n = Number of observations = 200
Mean = (sum of observations)/(number of observations)
50 = (sum of observations)/200
Sum of the observations = 50 x 200 = 10,000.
Thus, the improper sum of the observations = 50 x 200
Now,
The proper sum of the observations = Improper sum of the observations – Improper observations + proper observations
Proper sum of the observations = 10,000 – (92 + 8) + (192 + 88)
= 10,000 – 100 + 280
= 9900 + 280
= 10,180.
Thus, proper mean = (proper sum of the observations)/(number of observations)
= 10180/200 = 50.9

Question :22. The mean of 5 numbers is 27. If one more number is included, then the mean is 25. Find the included number.
Solution  22:
Mean = Sum of 5 numbers ÷ 5
Sum of the 5 numbers = 27 × 5 = 135.
New mean = 25
25 = Sum of 6 numbers ÷ 6
Sum of the 6 numbers = 25 × 6 = 150.
The included number = Sum of the six numbers – Sum of the five numbers
The included number = 150 – 135 =15
Thus, 15 is the included number.

Question :23. The mean of 75 numbers is 35. If each number is multiplied by 4, find the new mean.
Solution  23:
Let x1, x2, x3…x75 be 75 numbers with their mean equal to 35. Then,
35 = (x1 + x2 + x3 + …..+ x75)/75
x1 + x2 + x3 + …..+ x75 = 35 × 75
x1 + x2 + x3 + …..x x75 = 2625
=4 x 1, 4 x 2, 4 x 3…4 x 75 are new number.
Let M be the arithmetic mean of the new numbers.
M = (4x1 + 4x2 + 4x3 +⋯+ 4 x75)/75
= (4 (x1 + x2 + x3 + …+ x75))/75
= ((4 × 2625))/75 =140
So, the new mean is 140.

Exercise 23.2

Question :1. A die was thrown 20 times and the following scores were recorded:
5, 2, 1, 3, 4, 4, 5, 6, 2, 2, 4, 5, 5, 6, 2, 2, 4, 5, 5, 1
Prepare the frequency table of the scores on the upper face of the die and find the mean score.
Solution 1:
To calculate the arithmetic mean, prepare the following table:

Mean score = Σfixi/Σfi
= 73/20 = 3.65
Thus, the mean score of given frequency is 3.65.

Question :2. The daily wages (in Rs) of 15 workers in a factory are given below:
200, 180, 150, 150, 130, 180, 180, 200, 150, 130, 180, 180, 200, 150, 180
Prepare the frequency table and find the mean wage.

Solution 2:

Mean score = Σfixi/Σfi
= 2540/15 = 169.33
Thus, the mean score of given frequency is 169.33.

Question :3. The following table shows the weights (in kg) of 15 workers in a factory:

Calculate the mean weight.

Solution 3:
Calculation of mean:
Mean of weight = Σfixi/Σfi
= 975/15 = 65 kg
Thus, the mean of weight of given frequency is 65kg.

Question :4. The ages (in years) of 50 students of a class in a school are given below:
Solution 4:
Calculation of mean:

Mean of age in year = Σfixi/Σfi
= 770/50 = 15.4 years
Thus, the mean of age in year is 65kg.

Question :5. Calculate the mean for the following distribution:

Solution 5:
Mean score = Σfixi/Σfi
= 281/40 = 7.025
Thus, the mean score of given frequency is 7.025.

Question :6. Find the mean of the following data:

Solution 6:

Mean score = Σfixi/Σfi
= 2650/106  = 25
Thus, the mean score of given frequency is 25.

Question :7. The mean of the following data is 20.6. Find the value of p.

Solution 7:

Mean score = Σfixi/Σfi
= 2650/106  = 25
20.6 = (530+25p)/50
530 + 25 p = 20.6 x 50
25 p = 1030 – 530
p = 500/25 = 20
Thus, the value of p is 20.

Question :8. If the mean of the following data is 15, find p.

Solution 8:

Mean score = Σfixi/Σfi
= 2650/106  = 25
15 = (445+10p)/(27+p)
445 + 10 p = 405 + 15p
5 p = 445 – 405
p = 40/5 = 8
Thus, the value of p is 8.

Question :9. Find the value of p for the following distribution whose mean is 16.6

Solution 9:

Mean score = Σfixi/Σfi
16.6 = (1228+24p)/100
1228 + 24 p = 16.6 x 100
24 p = 1660 – 1228
p = 432/24 = 18
Thus, the value of p is 18.

Question :10. Find the missing value of p for the following distribution whose mean is 12.58
Solution 10:

Mean score = Σfixi/Σfi
12.58 = (524+7p)/50
524 + 7 p = 12.58 x 50
7 p = 629 – 524
p = 105/7 = 15
Thus, the value of p is 15.

Question :11. Find the missing frequency (p) for the following distribution whose mean is 7.68

Solution 11:

Mean score = Σfixi/Σfi
7.68 = 303 + 9p/41 + p
303 + 9 p = 314.88 + 7.68p
1.32 p = 314.88 – 303
p = 11.88/1.32 = 9
Thus, the value of p is 9.

Question :12. Find the value of p, if the mean of the following distribution is 20

Solution 12:

Mean score = Σfixi/Σfi
20 = ([(295 + (20 + p) 5p)])/(15 + 5p)
295 + 100 p + 5p2 = 300 + 100p
5p2 = 300 – 295
5p2= 5
p2 = 1
p = 1
thus, the value of P is 1.
Exercise 23.3

Find the median of the following data (1 – 8)
Question :1. 83, 37, 70, 29, 45, 63, 41, 70, 34, 54
Solution  1:
Firstly, arrange the given data into ascending order,
29, 34, 37, 41, 45, 54, 63, 70, 70, 83
Now, the number of observations is 10. So, n = 10 (even)
Thus, median = (n/2)th term + (((n+1))/2)th term
Median = (value of 5th term + value of 6th term)/2
= (45+54)/2 = 49.5
Thus, median for given data is 49.5

Question :2. 133, 73, 89, 108, 94,104, 94, 85, 100, 120
Solution  2:
Firstly, arrange the given data into ascending order,
73, 85, 89, 94, 100, 104, 108, 120, 133
Now, the number of observations, n = 10 (even)
Thus, median = (n/2)th term + (((n+1))/2)th term
Median = ((value of 5th term + value of 6th term))/2
= ((94+100))/2 = 97
So, the median for given data is 97

Question :3. 31, 38, 27, 28, 36, 25, 35, 40
Solution  3:
Firstly, arrange the given data into ascending order
25, 27, 28, 31, 35, 36, 38, 40
And number of observations, n = 8 (even)
Thus, median = (n/2)th term + (((n+1))/2)th term
Median = ((value of 4th term + value of 5th term))/2
= ((31+35))/2 = 33
So, the median for given data is 33

Question :4. 15, 6, 16, 8, 22, 21, 9, 18, 25
Solution  4:
Firstly, arrange the given data into ascending order
6, 8, 9, 15, 16, 18, 21, 22, 25
And number of observations, n = 9 (odd)
Thus, median = (((n+1))/2)th term
Median = value of 5th term
= 16
So, the median for given number is 16.

Question :5. 41, 43,127, 99, 71, 92, 71, 58, 57
Solution  5:
Firstly, arrange the given data into ascending order
41, 43, 57, 58, 71, 71, 92, 99, 127
and number of observations, n = 9 (odd)
Thus, median = (((n+1))/2)th term
Median = value of 5th term
= 71
So, the median for given number is 71.

Question :6. 25, 34, 31, 23, 22, 26, 35, 29, 20, 32
Solution  6:
Firstly, arrange the given data into ascending order,
20, 22, 23, 25, 26, 29, 31, 32, 34, 35
And number of observations, n = 10 (even)
Thus, median = (n/2)th term + (((n+1))/2)th term
Median = ((value of 5th term + value of 6th term))/2
= ((26+29))/2 = 27.5
So, the median for given data is 27.5

Question :7. 12, 17, 3, 14, 5, 8, 7, 15
Solution  7:
Firstly, arrange the given data into ascending order,
3, 5, 7, 8, 12, 14, 15, 17
and number of observations, n = 8 (even)
Thus, median = (n/2)th term + (((n+1))/2)th term
Median = ((value of 4th term + value of 5th term))/2
= ((8+12))/2 = 10
So, the median for given data is 10

Question :8. 92, 35, 67, 85, 72, 81, 56, 51, 42, 69
Solution  8:
Firstly, arrange the given data into ascending order,
35, 42, 51, 56, 67, 69, 72, 81, 85, 92
And number of observations, n = 10 (even)
Thus, median = (n/2)th term + (((n+1))/2)th term
Median = ((value of 5th term + value of 6th term))/2
= ((67+69))/2 = 68
So, the median for given data is 68

Question :9. Numbers 50, 42, 35, 2x +10, 2x – 8, 12, 11, 8, 6 are written in descending order and their median is 25, find x.
Solution   9:
The number of observations n is 9.
The median is the n+12th observation, which is the 5th observation, since n is odd.
We begin with the last number because the numbers are ordered in ascending order.
Median = 5th observation.
=> 25 = 2x – 8
=> 2x = 25 + 8
=> 2x = 33
=> x = 33/2
x = 16.5

Question :10. Find the median of the following observations: 46, 64, 87, 41, 58, 77, 35, 90, 55, 92, 33. If 92 is replaced by 99 and 41 by 43 in the above data, find the new median?
Solution   10:
Arranging the given data in ascending order:
33, 35, 41, 46, 55, 58, 64, 77, 87, 90, 92
The number of observations (n) is 11 in this case (odd).
Since the number of observations is odd, therefore,
Thus, median = (((n+1))/2)th term
Median = value of the 5th term
= 58.
So, the median is 58.
If 92 is replaced by 99 and 41 by 43, then the new observations arranged in ascending order are:
33, 35, 43, 46, 55, 58, 64, 77, 87, 90, 99
New median = Value of the 6th term
= 58.

Question :11. Find the median of the following data: 41, 43, 127, 99, 61, 92, 71, 58, 57, If 58 is replaced by 85, what will be the new median?
Solution   11:
Arranging the given data in ascending order,
41, 43, 57, 58, 61, 71, 92, 99,127
The number of observations (n) is 9 in this case (odd).
Thus, median = (((n+1))/2)th term
Median = value of 5th term
So, the median = 61.
If 58 is replaced by 85, then the new observations arranged in ascending order are:
41, 43, 57, 61, 71, 85, 92, 99, 12
New median = Value of the 5th term = 71.

Question :12. The weights (in kg) of 15 students are: 31, 35, 27, 29, 32, 43, 37, 41, 34, 28, 36, 44, 45, 42, 30. Find the median. If the weight 44 kg is replaced by 46 kg and 27 kg by 25 kg, find the new median.
Solution  12:
Arranging the given data in ascending order, we have:
27, 28, 29, 30, 31, 32, 34, 35, 36, 37, 41, 42, 43, 44, 45
The number of observations (n) is 15 in this case (odd).
The number of observations is odd,
Thus, median = (((n+1))/2)th term
Median = value of 8th term
= 35 kg.
So, the median is 35 kg.
If 44 kg is replaced by 46 kg and 27 kg by 25 kg, then the new observations arranged in ascending order are:
25, 28, 29, 30, 31, 32, 34, 35, 36, 37, 41, 42, 43, 45, 46
∴ New median = Value of the 8th observation = 35 kg.

Question :13. The following observations have been arranged in ascending order. If the median of the data is 63, find the value of x: 29, 32, 48, 50, x, x + 2, 72, 78, 84, 95
Solution  13:
The number of observations n is 10. Since n is even,
Thus, median = (n/2)th term + (((n+1))/2)th term
Median = ((value of 5th term + value of 6th term))/2
63 = x + ((x + 2))/2
63 = ((2x + 2))/2
63 = (2(x + 1))/2
63 = x + 1
x = 63 – 1 = 62
so, the value of x is 62

Exercise 23.4

Question :1. Find the mode and median of the data: 13, 16, 12, 14, 19, 12, 14, 13, 14
By using the empirical relation also find the mean.
Solution 1:
Arrange the data in ascending order such that similar values are grouped together, as follows:
12, 12, 13, 13, 14, 14, 14, 16, 19
The number of observations n is 9. Since n is odd,
Thus, median = (((n+1))/2)th term
Median = value of 5th term
= 14
So, 14 occurs the maximum number of times, i.e., three times.
Thus, 14 is the mode of the data.
Now,
Mode = 3 Median – 2 Mean
14 = 3 x 14 – 2 Mean
2 Mean = 42 – 14 = 28
Mean = 28/2 = 14.
So, the mean of the data is 14,
Mode of the data is 14,
And median of the data is 14.

Question :2. Find the median and mode of the data: 35, 32, 35, 42, 38, 32, 34
Solution   2:
Arrange the data in ascending order such that similar values are grouped together, as follows:32, 32, 34, 35, 35, 38, 42
The number of observations n is 7. Since n is odd,
Therefore median = (((n+1))/2)^th term
Median = value of 4th term
Median = 35
So, 32 and 35, both occur twice. Therefore, there are two modes 32 and 35.

Question :3. Find the mode of the data: 2, 6, 5, 3, 0, 3, 4, 3, 2, 4, 5, 2, 4
Solution  3:
Arrange the data in ascending order such that similar values are grouped together, as follows:
0, 2, 2, 2, 3, 3, 3, 4, 4, 4, 5, 5, 6
Here, 2, 3 and 4 occur three times each. Therefore, there are three modes 2, 3 and 4.

Question :4. The runs scored in a cricket match by 11 players are as follows:

6, 15, 120, 50, 100, 80, 10, 15, 8, 10, 10

Find the mean, mode and median of this data.

Solution 4:

Arrange the data in ascending order such that similar values are grouped together, as follows:

6, 8, 10, 10, 15, 15, 50, 80, 100, 120

The number of observations n is 11. Since n is odd,

Thus, median = (((n+1))/2)th term

Median = value of 6th term

= 15

Here, 10 occur three times. Therefore, 10 is the mode of the given data.

Now,

Mode = 3 Median – 2 Mean

10 = 3 x 15 – 2 Mean

2 Mean = 45 – 10 = 35

Mean = 35/2 = 17.5

The value of mean: 17.5

Mode: 10

Median: 15

Question :5. Find the mode of the following data:
12, 14, 16, 12, 14, 14, 16, 14, 10, 14, 18, 14
Solution 5:
Arrange the data in ascending order such that similar values are grouped together, as follows:
10, 12, 12, 14, 14, 14, 14, 14, 14, 16, 18
Here, 14 occurs the maximum number of times.
Thus, 14 is the mode of the data.

Question :6. Heights of 25 children (in cm) in a school are as given below:
168, 165, 163, 160, 163, 161, 162, 164, 163, 162, 164, 163, 160, 163, 163, 164, 163, 160, 165, 163, 162
What is the mode of heights?
Also, find the mean and median.
Solution 6:
Arranging the data in tabular form, we get:

Thus, the median = (((n+1))/2)th term
Median = value of 13th term
= 163 cm
Here, 163 cm occurs the most number of times. Thus, the mode of the data is 163 cm.
Mode = 3 Median – 2 Mean
163 = 3 x 163 – 2 Mean
2 Mean = 326
Mean = 163 cm.
So, the value of mean = 163; mode = 163; and median = 163.

Question :7. The scores in mathematics test (out of 25) of 15 students are as follows:
19, 25, 23, 20, 9, 20, 15, 10, 5, 16, 25, 20, 24, 12, 20
Find the mode and median of this data. Are they same?
Solution 7:
Arrange the data in ascending order such that similar values are grouped together, as follows:
5, 9, 10, 12, 15, 16, 19, 20, 20, 20, 20, 23, 24, 25, 25
The number of observations n is 15. Since n is odd,
Thus, median = (((n+1))/2)^th term
Median = value of 8th term
= 20
Here, 20 occurs the greatest number of times, i.e., 4 times.
Thus, the mode of the given data is 20.
Yes, the median and mode of the given data are the same.

Question :8. Calculate the mean and median for the following data:

Using empirical formula, find its mode.
Solution 8:
Calculation of mean
Mean = Σfi xi/Σfi
= 332/25 = 13.28
The number of observations n is 25. Since n is odd,
Therefore median = (((n+1))/2)th term
Median = value of 13th term
= 13
Now, by using empirical formula,
Mode = 3Median – 2 Mean
Mode = 3 (13) – 2 (13.28)
Mode = 39 – 26.56
Mode = 12.44.
So, the value of mean is  13.28,
Mode is 12.44,
Median is 13.

Question :9. The following table shows the weights of 12 persons.

Find the median and mean weights. Using empirical relation, calculate its mode.

Solution 9:

Calculation of mean
Mean = Σfi xi/Σfi
= 612/12 = 51 kg
Here number of observation = 12
Therefore median = (n/2)th term + (((n+1))/2)th term
Median = ((value of 6th term + value of 7th term))/2
= ((50 + 50))/2 = 50
Now by empirical formula we have,
Mode = 3 Median – 2 Mean
Mode = 3 x 50 – 2 x 51
Mode = 150 – 102
Mode = 48 kg.
Thus, Mean = 51 kg, Median = 50 kg and Mode = 48 kg.