R S AGGARWAL AND V AGGARWAL Solutions for Class 9 Maths Chapter 18 – Mean, Median and Mode of Ungrouped Data
Chapter 18 – Mean, Median and Mode of Ungrouped Data Exercise Ex. 18A
Find the arithmetic mean of:
The first eight natural numbers
first eight natural numbers are:
1,2,3,4,5,6,7and 8
_{}
Find the arithmetic mean of:
The first ten odd numbers
First ten odd numbers are:
1,3,5,7,9,11,13,15, 17, and 19
_{}
Find the arithmetic mean of:
The first seven multiple of 5
First seven multiples of 5 are: 5,10,15, 20, 25, 30, 35
Therefore, Mean =20
Find the arithmetic mean of:
All the factors of 20
Factors of 20 are: 1,2,4,5,10,20
_{ }
Find the mean of:
all prime numbers between 50 and 80.
Prime
numbers between 50 and 80 are as follows:
53,
59, 61, 67, 71, 73, 79
Total
prime numbers between 50 and 80 = 7
The mean of 15 numbers is 27. If each numbers is multiplied by 4, what will be the mean of the new numbers ?
Let the given numbers be x_{1}, x_{2}…….x_{15}
Then the mean of these numbers=27
_{}
The mean of 12 numbers is 40. If each number is divided by 8, what will be the mean of the new numbers?
The mean of 20 number is 18. If 3 is added to each of the first ten numbers , find the mean of the new set of 20 numbers.
The
mean of six numbers is 23 . If one of
the numbers is excluded , the mean of the remaining numbers is 20. Find the
excluded number.
Mean
of 6 numbers = 23
Sum
of 6 numbers =(23×6 )=138
Again
, mean of 5 numbers =20
Sum
of 5 numbers=(20x 5 ) =100
_{} The excluded number= (sum of 6 numbers
)(sum of 5 numbers)
=(138100)
=38
_{} The excluded number=38.
The average height of 30 boys was
calculated to be 150 cm. It was detected later that one value of 165 cm was
wrongly copied as 135 cm for the computation of the mean. Find the correct
mean.
Mean
height of 30 boys = 150 cm
⇒ Total height
of 30 boys = 150 × 30 = 4500 cm
Correct
sum = 4500 – incorrect value + correct value
=
4500 – 135 + 165
=
4530
The mean weight of a class of 34 students is 46.5 kg. If the of the teacher is included, the mean rises by 500 g. Find the weight of the teacher
_{Mean weight of 34 students = 46.5 kg}
_{Total weight of 34 students =(34×46.5)kg =1581 kg}
_{Mean weight of 34 students and the teacher =(46.5+0.5)kg=47kg }
_{Total weight of 34 students and the teacher}
_{=(47×35)kg =1645kg}
_{ Weight of the teacher =(16451581)kg= 64kg}
The mean weight of a class of 36 students is 41 kg. If one of the students leaves the class then the mean is decreased by 200 g. find the weight of the student who left.
_{Mean weight of 36 students = 41 kg}
_{Total weight of 36 students = 41x 36 kg = 1476kg}
_{One student leaves the class mean is decreased by 200 g.}
_{ New mean =(410.2)kg = 40.8 kg }
_{Total weight of 35 students = 40.8×35 kg = 1428 kg.}
_{the weight of the student who left =(14761428)kg =48 kg.}
The average weight of a class of 39 students is 40 kg . When a new student is admitted to the class , the average decreases by 200 g . find the weight of the new student.
_{Mean weight of 39 students =40 kg}
_{Total weight of 39 students = 40x 39) = 1560 kg}
_{One student joins the class mean is decreased by 200 g.}
_{ New mean =(400.2)kg = 39.8 kg }
_{Total weight of 40 students =(39.8×40)kg=1592 kg.}
_{the weight of new student }
_{= Total weight of 40 students – Total weight of 39 students}
_{= 15921560 = 32 kg}
The average weight of 10 oarsmen in a boat is increased by 1.5 kg when one of the crew who weighs 58 kg is replaced by a new man . find the weight of a new man.
The increase in the average of 10 oarsmen = 1.5 kg
_{ Total weight increased =(1.5×10) kg=15 kg}
Since the man weighing 58 kg has been replaced,
_{ Weight of the new man =(58+15)kg =73kg.}
The mean of 8 numbers is 35 . if a number is excluded then the mean is reduced by 3 . find the excluded number.
_{Mean of 8 numbers=35}
_{ Total sum of 8 numbers = 35×8 = 280}
_{ Since One number is excluded, }New mean = 35 – 3 = 32
_{Total sum of 7 numbers = 32×7 = 224}
_{the excluded number = Sum of 8 numbers – Sum of 7 numbers}
_{= 280 – 224 = 56}
The
number of children in 10 families of a locality are
2,4,3,4,2,0,3,5,1,6.
Find
the mean number of children per family.
_{}
The mean of 150 items was found to be 60. Later on , it was discovered that the values of two items were misread as 52 and 8 instead of 152 and 88 respectively. Find the correct mean.
_{Mean of 150 items = 60}
_{Total Sum of 150 items = 150×60 = 9000}
_{Correct sum of items =[(sum of 150 items)(sum of wrong items)+(sum of right items)]}
_{= [9000 – (52 + 8) + (152 + 88)]}
_{= [9000(52+8)+(152+88)]}
_{= 9180}
_{ Correct mean =}
The mean of 31 results 60. If the mean of the first 16 results is 58 and that of the last 16 numbers is 62, find the 16^{th} result.
_{Mean of 31 results=60}
_{Total sum of 31 results = 31×60 = 1860}
_{Mean of the first 16 results =16×58=928}
_{Total sum of the first 16 results=16×58=928}
_{Mean of the last 16 results=62}
_{Total sum of the last 16 results=16×62=992}
_{The 16th result = 928 + 992 – 1860}
_{ = 1920 – 1860 = 60}
_{The 16th result = 60.}
The mean of 11 numbers is 42. If the mean of the first 6 numbers is 37 and that of the last 6 numbers is 46 . find the 6^{th} number.
_{Mean of 11 numbers = 42}
_{Total sum of 11 numbers = 42×11 = 462}
_{Mean of the first 6 numbers = 37}
_{Total sum of first 6 numbers = 37×6 = 222}
_{Mean of the last 6 numbers = 46 }
_{Total sum of last 6 numbers = 6×46 = 276}
_{The 6th number= 276 + 222 – 462}
_{ = 498 – 462 = 36}
_{The 6th number = 36}
The mean weight of 25 students of a class is 52 kg . If the mean weight of the first 13 students of the class is 48 kg that of the last 13 students is 55 kg . find the weight of the 13^{th} student.
_{Mean weight of 25 students = 52kg}
_{Total weight of 25 students = 52×25 kg=1300 kg}
_{Mean of the first 13 students = 48 kg}
_{Total weight of the first 13 students = 48×13 kg = 624kg}
_{Mean of the last 13 students = 55 kg}
_{Total weight of the last 13 students = 55×13 kg = 715 kg}
_{The weight of 13th student }
_{= Total weight of the first 13 students + Total weight of the last 13 students – Total weight of 25 students}
_{= 624+7151300 kg}
_{= 39 kg.}
Therefore, the weight of 13th student is 39 kg.
The mean score of 25 observations is 80 and the mean score of another 55 observations is 60. Determine the mean score of the whole set of observations .
_{Mean score of 25 observations = 80}
_{Total score of 25 observations = 80×25 = 2000}
_{Mean score of 55 observations = 60}
_{Total score of 55 observations = 60×55 =3300}
Total no. of observations = 25+55 =80 observations
_{Total score }= 2000+3300 = 5300
_{Mean score =}
Arun scored 36 marks in English , 44 marks in hindi, 75 marks in mathematics and x marks in science . If he has secured an average of 50 marks , find the value of x.
_{ Average marks of 4 subjects = 50}
_{Total marks of 4 subjects = 50×4 = 200}
_{36 + 44 + 75 + x = 200}
_{ 155 + x = 200}
_{ x = 200 – 155 = 45}
_{The value of x = 45}
A ship sails out to an island at the rate of 15 km/h and the sails back to the starting point at 10 km /h . find the average sailing speed for the whole journey .
_{Let the distance of mark from the staring point be x km. }
_{Then , time taken by the ship reaching the marks=}
_{}
_{Time taken by the ship reaching the starting point from the marks =}
_{Total time taken =}
_{Total distance covered =x+x=2x km.}
_{ }
There are 50 students in a class, of which 40 are boys . The average weight of the class is 44 kg and that of the girls is 40 kg . find the average weight of the boys.
_{Total number of students = 50}
_{Total number of girls = 5040 = 10}
_{Average weight of the class = 44 kg }
_{Total weight of 50 students= 44x 50 kg = 2200kg}
_{Average weight of 10 girls = 40 kg}
_{Total weight of 10 girls = 40×10 kg = 400 kg}
_{Total weight of 40 boys = 2200400 kg =1800 kg}
_{the average weight of the boys = }
The aggregate monthly expenditure
of a family was Rs.18720
during the first 3 months, Rs.20340 during the next 4 months and Rs.21708 during the last 5 months of a year. If the total
saving during the year be Rs.35340. Find the average monthly income of the family.
Total
earnings of the year
=
Rs. (3 × 18720 + 4 × 20340 + 5 ×21708 + 35340)
=
Rs. (56160 + 81360 + 108540 + 35340)
=
Rs. 281400
Number
of months = 12
The average weekly payment to 75
workers in a factory is Rs.5680. The mean weekly payment to 25 of them is Rs.5400 and that of 30 others is Rs.5700. Find the mean weekly payment of the remaining workers.
Average
weekly payment of 75 workers = Rs. 5680
⇒ Total weekly
payment of 75 workers = Rs. (75 × 5680) = Rs. 426000
Mean
weekly payment of 25 workers = Rs. 5400
⇒ Total weekly
payment of 25 workers = Rs. (25 × 5400) = Rs. 135000
Mean
weekly payment of 30 workers = Rs. 5700
⇒ Total weekly
payment of 30 workers = Rs. (30 × 5700) = Rs. 171000
Number
of remaining workers = 75 – 25 – 30 = 20
Therefore,
Total weekly payment of remaining 20 workers
=
Rs. (426000 – 135000 – 171000)
=
Rs. 120000
The
following are number of books issued in a school library during a week:
105,
216, 322, 167, 273, 405, and 346.
Find
the average number of books issued per day.
Sol.3
_{}
_{}
The mean marks (out of 100) of
boys and girls in an examination are 70 and 73 respectively. If the mean
marks of all the students in that examination is 71, find the ratio of the
number of boys to the number of girls.
Let
the ratio of number of boys to the number of girls be x : 1.
Then,
Sum
of marks of boys = 70x
Sum
of marks of girls = 73 × 1 = 73
And,
sum of marks of boys and girls = 71 × (x + 1)
⇒ 70x + 73 =
71(x + 1)
⇒ 70x + 73 = 71x
+ 71
⇒ x = 2
Hence,
the ratio of number of boys to the number of girls is 2 :
1.
The average monthly salary of 20
workers in an office is Rs.45900. If the manager’s salary is added, the average
salary becomes Rs.49200
per month. What’s manager’s monthly salary?
Mean
monthly salary of 20 workers = Rs. 45900
⇒ Total monthly
salary of 20 workers = Rs. (20 × 45900) = Rs. 918000
Mean
monthly salary of 20 workers + manager = Rs. 49200
⇒ Total monthly
salary of 20 workers + manager = Rs. (21 × 49200) = Rs. 1033200
Therefore,
manager’s monthly salary = Rs. (1033200 – 918000) =
Rs. 115200
The
daily minimum temperature recorded (in
degree F) at a place during a week was as under:
Monday  Tuesday  Wednesday  Thursday  Friday  Saturday 
35.5  30.8  27.3  32.1  23.8  29.9 
Find
the mean temperature.
_{}
If the mean of five observations
x, x + 2, x + 4, x + 6, x + 8 is 13, find the value of x and hence find the
mean of the last three observations.
Total
numbers of observations = 5
Thus,
last three observations are (9 + 4), (9 + 6) and (9 + 8),
i.e.
13, 15 and 17
The mean weight of 6 boys in a group is 48 kg. The individual weights of five of them are 51 kg, 45 kg, 49 kg, 46 kg and 44 kg. Find the weight of the sixth boy.
Mean weight of the boys =48 kg
Sum of the weight of6 boys =(48×6)kg =288kg
Sum of the weights of 5 boys=(51+45+49+46+44)kg=235kg
Weight of the sixth boy=(sum of the weights of 6 boys ) – (sum of the weights of 5 boys)
=(288235)=53kg.
_{}
The mean of the marks scored by 50 students was found to be 39. Later on it was discovered that a score of 43 was misread as 23. Find the correct mean.
Calculated mean marks of 50 students =39
_{} calculated sum of these marks=(39x 50)=1950
Corrected sum of these marks
=[1950(wrong number)+(correct number)]
=(195023+43) =1970
_{}correct mean =
The mean of 24 numbers is 35. If 3 is added to each number, what will bethe new mean?
Let the given numbers be x_{1},X_{2……}X_{24}
_{}
The mean of 20 numbers is 43. If 6 is subtracted from each of the numbers , what will be the new mean ?
Let the given numbers be x_{1}, x_{2}…..x_{20}
Then , the mean of these numbers =
_{}
Chapter 18 – Mean, Median and Mode of Ungrouped Data Exercise Ex. 18B
Obtain the mean of the following
distribution:
Variable  4  6  8  10  12 
Frequency  4  8  14  11  3 
_{Find the value of p for the following frequency distribution whose mean is 16.6.}
_{X}  _{8}  _{12}  _{15}  _{p}  _{20}  _{25}  _{30} 
_{F}  _{12}  _{16}  _{20}  _{24}  _{16}  _{8}  _{4} 
We prepare the following frequency distribution table:
(X_{i})  (f_{i})  f_{i}X_{i} 
8 12 15 P 20 25 30  12 16 20 24 16 8 4  96 192 300 24p 320 200 120 

_{}
_{} 1228 + 24p = 1660
_{} 24p = 16601228
_{} 24p = 432
_{}
_{} the value of p =18
Find the missing frequencies in
the following frequency distribution whose mean is 34.
x  10  20  30  40  50  60  Total 
f  4  f_{1}  8  f_{2}  3  4  35 
_{Find the missing frequencies in the following frequency distribution, whose mean is 50.}
_{x}  _{10}  _{30}  _{50}  _{70}  _{90}  _{Total} 
_{f}  _{17}  f_{1}  _{32}  _{f2}  _{19}  _{120} 
Let f_{1 } and f_{2 }be the missing frequencies.
We prepare the following frequency distribution table.
(X_{i})  (f_{i})  f_{i}x_{i} 
10 30 50 70 90  17 f_{1} 32 f_{2} 19  170 30f_{1} 1600 70f_{2} 1710 
Total  120  3480 + 30f_{1 }+ 70f_{2} 
Here,
Thus, …….(1)
Also,
Substituting the value of f_{1} in equation 1, we have,
f_{2}=52 – 28 = 24
Thus, the missing frequencies are f_{1} =28 and f_{2}=24 respectively.
Find the value of p, when the mean
of the following distribution is 20.
x  15  17  19  20  23 
f  2  3  4  5p  6 
The mean of the following
distribution is 50.
x  10  30  50  70  90 
f  17  5a  32  7a  19 
Find the value of a and hence the
frequencies of 30 and 70.
_{The following table shows the weights of 12 workers in a factory :}
_{Weight (in Kg)}  _{60}  _{63}  _{66}  _{69}  _{72} 
_{ No of workers }  _{4}  _{3}  _{2}  _{2}  _{1} 
_{Find the mean weight of the workers.}
For calculating the mean , we prepare the following frequency table :
Weight (in kg) (X_{i})  No of workers (f_{i})  f_{i}X_{i} 
60 63 66 69 72  4 3 2 2 1  240 189 132 138 72 
_{}  771 
The measurements (in mm) of the
diameters of the heads of 50 screws are given below:
Diameter  34  37  40  43  46 
Number  5  10  17  12  6 
Calculate the mean diameter of the
heads of the screws.
_{The following data give the number of boys of a particular age in a class of 40 students.}
_{Age (in years)}  _{15}  _{16}  _{17}  _{18}  _{19}  _{20} 
_{Frequency (f)}  _{3}  _{8}  _{9}  _{11}  _{6}  _{3} 
_{Calculate the mean age of the students}
For calculating the mean , we prepare the following frequency table :
Age (in years) (X_{i})  Frequency (f_{i})  f_{i}X_{i} 
15 16 17 18 19 20  3 8 9 11 6 3  45 128 153 198 114 60 
_{}  698 
_{Find the mean of the following frequency distribution :}
_{Variable (xi)}  _{10}  _{30}  _{50}  _{70}  _{89} 
_{Frequency(fi)}  _{7}  _{8}  _{10}  _{15}  _{10} 
For calculating the mean , we prepare the following frequency table :
Variable (X_{i})  Frequency (f_{i})  f_{i}X_{i} 
10 30 50 70 89  7 8 10 15 10  70 240 500 1050 890 
 _{ } 
Find the mean of daily wages of 40
workers in a factory as per data given below:
Daily  250  300  350  400  450 
Number  8  11  6  10  5 
If the mean of the following data
is 20.2, find the value of p.
Variable  10  15  20  25  30 
Frequency  6  8  p  10  6 
_{If the mean of the following frequency distribution is 8, find the value of p.}
_{X}  _{3}  _{5}  _{7}  _{9}  _{11}  _{13} 
_{F}  _{6}  _{8}  _{15}  _{p}  _{8}  _{4} 
We prepare the following frequency table :
(X_{i})  (f_{i})  f_{i}X_{i} 
3 5 7 9 11 13  6 8 15 P 8 4  18 40 105 9P 88 52 
_{} 
_{}
_{} 303 + 9p = 8(41+p)
_{ }303 + 9p= 328 + 8p
_{ }9p – 8p = 328 303
_{ }P=25
_{} the value of P=25
_{Find the missing frequency p for the following frequency distribution whose mean is 28.25.}
_{X}  _{15}  _{20}  _{25}  _{30}  _{35}  _{40} 
_{F}  _{8}  _{7}  _{p}  _{14}  _{15}  _{6} 
We prepare the following frequency distribution table:
(X_{i})  (f_{i})  f_{i}X_{i} 
15 20 25 30 35 40  8 7 P 14 15 6  120 140 25p 420 525 240 
_{} 
_{}
_{} 1445 + 25p = (28.25)(50+p)
_{} 1445 + 25p = 1412.50 + 28.25p
_{} 28.25p + 25p = 1445 + 1412.50
_{} 3.25p = 32.5
_{}
_{} the value of p=10
Chapter 18 – Mean, Median and Mode of Ungrouped Data Exercise Ex. 18C
Find the median of:
2,10, 9, 9, 5, 2, 3, 7, 11
Arranging the data in accending order, we have
2,2,3, 5, 7, 9, 9, 10, 11
Here n = 9, which is odd
_{}
Find the median of:
15, 6, 16, 8, 22, 21, 9, 18, 25
Arranging the data in ascending order , we have
6, 8, 9, 15, 16, 18, 21, 22, 25
Here n = 9, which is odd
Find the median of
20, 13, 18, 25, 6, 15, 21, 9, 16, 8, 22
Arranging data in ascending order:
6, 8, 9, 13, 15, 16, 18, 20, 21, 22, 25
Here n = 11 odd
_{}
Find the median of:
7, 4, 2, 5, 1, 4, 0, 10, 3, 8, 5, 9, 2
Arranging the data in ascending order , we have
0, 1, 2, 2, 3, 4, 4, 5, 5, 7, 8, 9, 10
Here n = 13, which is odd
_{}
Find the median of the data
46, 41, 77, 58, 35, 64, 87, 92,
33, 55, 90.
In the above data, if 41 and 55
are replaced by 61 and 75 respectively, what will be the new median?
Total
number of observations = n = 11 (odd)
Arranging
data in ascending order, we have
33,
35, 41, 46, 55, 58, 64, 77, 87, 90, 92
Now,
41 and 55 are replaced by 61 and
75 respectively.
Arranging
new data in ascending order, we have
33,
35, 46, 58, 61, 64, 75, 77, 87, 90, 92
Find the median of:
17, 19, 32, 10, 22, 21, 9, 35
Arranging the data in ascending order , we have
9, 10, 17, 19, 21, 22, 32, 35
Here n = 8, which is even
_{ }
Find the median of:
72, 63, 29, 51, 35, 60, 55, 91, 85, 82
Arranging the data in ascending order , we have
29, 35, 51, 55, 60, 63, 72, 82, 85, 91
Here n = 10, which is even
_{}
Find the median of:
10, 75, 3, 15, 9, 47, 12, 48, 4, 81, 17, 27
Arranging the data in ascending order , we have
_{3, 4, 9, 10, 12, 15, 17, 27, 47, 48, 75, 81 Here n = 12, which is even}
_{The marks of 15 students in an examination are :}
25,19,17,24,23,29,31,40,19,20,22,26,17,35,21
Find the median score.
_{Arranging the data in ascending order , we have}
_{17, 17, 19, 19, 20, 21, 22, 23, 24, 25, 26, 29, 31, 35, 40}
_{Here n = 15, which is odd}
_{}
Thus, the median score is 23.
The heights (in cm) of 9 students
of a class are
148, 144, 152, 155, 160, 147, 150,
149, 145.
Find the median height
Total
number of students = n = 9 (odd)
Arranging
heights (in cm) in ascending order, we have
144,
145, 147, 148, 149, 150, 152, 155, 160
_{The weights (in kg ) of 8 children are:}
_{ 13.4, 10.6, 12.7, 17.2, 14.3, 15, 16.5, 9.8}
_{Find the median weight.}
_{Arranging the weights of 8 children in ascending order, we have }
_{9.8, 10.6, 12.7, 13.4, 14.3, 15, 16.5, 17.2}
_{Here , n= 8 , which is even }
_{}
_{The ages (in years ) of 10 teachers in a school are:}
_{32, 44, 53, 47, 37, 54, 34, 36, 40, 50}
_{Fid the median age.}
_{Arranging the ages of teachers in ascending order , we have}
_{32, 34, 36, 37, 40, 44, 47, 50, 53, 54}
_{Here, n =10, which is even}
_{ }
_{If 10, 13, 15, 18, x+1, x+3, 30, 32, 35, 41 are ten observations in an ascending order with median 24, find the value of x.}
_{ The ten observations in ascending order:}
_{10, 13, 15, 18, x+1, x+3, 30, 32, 35, 41}
_{Here, n =10, which is even}
_{}
The following observations are
arranged in ascending order:
26, 29, 42, 53, x, x + 2, 70, 75,
82, 93.
If the median is 65, find the
value of x.
Total
number of observations = n = 10 (even)
Median
= 65
The numbers 50, 42, 35, (2x + 10),
(2x – 8), 12, 11, 8 have been written in a descending order. If their median
is 25, find the value of x.
Total
number of observations = n = 8 (even)
Median
= 25
Chapter 18 – Mean, Median and Mode of Ungrouped Data Exercise Ex. 18D
_{Find the mode of the following items.}
_{0, 6, 5, 1, 6, 4, 3, 0, 2, 6, 5, 6}
_{Arrange the given data in ascending order we have}
_{0, 0, 1, 2, 3, 4, 5, 5, 6, 6, 6, 6}
Let us prepare the following table:
_{Observations(x)}  _{0}  _{1}  _{2}  _{3}  _{4}  _{5}  _{6} 
_{Frequency}  _{2}  _{1}  _{1}  _{1}  _{1}  _{2}  _{4} 
_{As 6 ocurs the maximum number of times i.e. 4, mode = 6}
_{Determine the mode of the following values of a variable.}
_{23, 15, 25, 40, 27, 25, 22, 25, 20}
_{Arranging the given data in ascending order , we have:}
_{15, 20, 22, 23, 25, 25, 25, 27, 40}
_{The frequency table of the data is :}
_{Observations(x)}  _{15}  _{20}  _{22}  _{23}  _{25}  _{27}  _{40} 
_{Frequency}  _{1}  _{1}  _{1}  _{1}  _{3}  _{1}  _{1} 
_{As 25 ocurs the maximum number of times i.e. 3, mode = 25}
_{Calculate the mode of the following sizes of shoes by a shop on a particular day}
_{5, 9, 8, 6, 9, 4, 3, 9, 1, 6, 3, 9, 7, 1, 2, 5, 9}
_{Arranging the given data in ascending order , we have:}
_{1, 1, 2, 3, 3, 4, 5, 5, 6, 6, 7, 8, 9, 9, 9, 9, 9, }
_{The frequency table of the data is :}
_{Observations(x)}  _{1}  _{2}  _{3}  _{4}  _{5}  _{6}  _{7}  _{8}  _{9} 
_{Frequency}  _{2}  _{1}  _{2}  _{1}  _{2}  _{2}  _{1}  _{1}  _{5} 
_{As 9, occurs the maximum number of times i.e. 5, mode = 9}
_{A cricket player scored the following runs in 12 oneday matches:}
_{50, 30, 9, 32, 60, 50, 28, 50, 19, 50, 27, 35.}
_{Find his modal score.}
_{Arranging the given data in ascending order , we have:}
_{9, 19, 27, 28, 30, 32, 35, 50, 50, 50, 50, 60}
_{The frequency table of the data is :}
_{Observations(x)}  _{9}  _{19}  _{27}  _{28}  _{30}  _{32}  _{35}  _{50}  _{60} 
_{Frequency}  _{1}  _{1}  _{1}  _{1}  _{1}  _{1}  _{1}  _{4}  _{1} 
_{As 50, ocurs the maximum number of times i.e. 4, mode = 50}
Thus, the modal score of the cricket player is 50.
If the mean of the data 3, 21, 25,
17, (x + 3), 19, (x – 4) is 18, find the value of x. Using this value of x,
find the mode of the data.
Total
number of observations = n = 7
Mean
= 18
Thus,
data is as follows:
3,
21, 25, 17, 24, 19, 17
The
most occurring value is 17.
Hence,
the mode of the data is 17.
The numbers 52, 53, 54, 54, (2x +
1), 55, 55, 56, 57 have been arranged in an ascending order and their median
is 55. Find the value of x and hence find the mode of the given data.
Number
of values = n = 9 (odd)
Numbers
in ascending order:
52, 53, 54, 54, (2x + 1), 55, 55,
56, 57
Thus,
we have
52, 53, 54, 54, 55, 55, 55, 56, 57
The most occurring number is 55.
Hence, the mode of the data is 55.
For what value of x is the mode of
the data 24, 15, 40, 23, 27, 26, 22, 25, 20, x + 3 found 25? Using this value
of x, find the median.
Mode
of the data = 25
So,
we should have the value 25 occurring maximum number of times in the given
data.
That
means, x + 3 = 25
⇒ x = 22
Thus,
we have 24, 15, 40, 23, 27, 26, 22, 25, 20,
25.
Arranging
data in ascending order, we have
15,
20, 22, 23, 24, 25, 25, 26, 27, 40
Number
of observations = 10 (even)
The numbers 42, 43, 44, 44, (2x +
3), 45, 45, 46, 47 have been arranged in an ascending order and their median
is 45. Find the value of x. Hence, find the mode of the above data.
Total
number of observations = n = 9 (odd)
Median
= 45
Thus, we have
42, 43, 44, 44, 45, 45, 45, 46, 47
The most occurring value is 45.
Hence, the mode of the data is 45.
Chapter 18 – Mean, Median and Mode of Ungrouped Data Exercise MCQ
If the mean of x, x+2, x+4, x+6 and x+8 is 11, the value of
x is
The mean of the following data is 8
X  3  5  7  9  11  13 
Y  6  8  15  P  8  4 
The value of p is
 23
 24
 25
 21
The runs scored by 11 members of a cricket team are
15, 34, 56, 27, 43, 29, 31, 13, 50, 20, 0
The median score is
 27
 29
 31
 20
The weight of 10 students (in kgs)
are
55,40,35,52,60,38,36,45,31,44
The median weight is
 40 kg
 41 kg
 42 kg
 44 kg
The median of the numbers 4,4,5,7,6,7,7,12,3 is
 4
 5
 6
 7
The median of the numbers 84,78,54,56,68,22,34,45,39,54 is
 45
 49.5
 54
 56
Mode of the data
15,17,15,19,14,18,15,14,16,15,14,20,19,14,15 is
 14
 15
 16
 17
The median of the data arranged in ascending order
8, 9, 12, 18, (x +2), (x + 4), 30, 31, 34, 39 is 24. The
value of x is.
 22
 21
 20
 24
If the mean of x, x + 3, x + 5, x + 10 is 9, the mean of
the last three observations is
If each observation of the data is decreased by 8 then
their mean
(a) remains
the same
(b) is
decreased by 8
(c) is
increased by 5
(d) becomes
8 times the original mean
Correct
option: (b)
If each observation of the data is decreased by 8 then their mean
is also decreased by 8.
The
mean weight of six boys in a group is 48 kg. The individual weights of five
them are 51 kg, 45 kg, 48 kg and 44 kg. The weight of 6^{th} boy is
 52 kg
 52.8 kg
 53 kg
 47 kg
The mean of the marks scored by 50 students was found to be
39. Later on it was discovered that a score of 43 was misread as 23. The
correct mean is
 38.6
 39.4
 39.8
 39.2
The mean of 100 items was found to be 64. Later on it was
discovered that two items were misread as 26 and 9 instead of 36 and 90
respectively. The correct mean is
 64.86
 65.31
 64.91
 64.61
The mean of 100 observations is 50. If one of the
observations 50 is replaced by 150, the resulting mean will be
 50.5
 51
 51.5
 52