Chapter 14: Statistics – Chapter-Wise Important Questions and Answers | Class 10 Mathematics CBSE Solutions

CBSE Class 10 Mathematics Chapter 14: Statistics Important Questions with Solutions and Answers

Chapter 14: STATISTICS

              SECTION = A

Question:  The mode of the following data: 15, 8, 26, 24, 15, 18, 20, 24, 15, 19, 15 is:
(a) 24        (b) 15          (c) 20           (d) 18

Answer:  (b)  15

[Since 15 has the maximum frequency (4 times). So, the mode is 15 .]

Question: In a moderately skewed distribution, if Mean = 50 and Mode = 40, find the Median.
 (a) 45.67                 (b) 46.67                (c) 47.67              (d) 48.67

Answer: (b) 46.67
[ We have,  3 Median = Mode + 2 Mean
                     3 Median = 40 + 2(50) = 140
                       Median    ]

Question: If the Mean of a distribution is 25 and Median is 28, find the Mode using the empirical formula.
        (a) 30                 (b) 32                   (c) 34              (d) 36

Answer: (c) 34
[ We have,  3 Median = Mode + 2 Mean
        3(28) = Mode + 2(25)
           84 = Mode + 50
             Mode = 34   ]

Question: For a moderately skewed data, if Median = 30 and Mode = 24, what is the Mean?
             (a) 31                     (b) 32                   (c) 33                    (d) 34

Answer:   (c) 33
[ We have,  3(30) = 24 + 2 Mean
     90 = 24 + 2 Mean
     2 Mean = 66
     Mean = 33

Question:  Which of the following is true for a perfectly symmetrical distribution?
(a) Mean = Median = Mode       (b) Mean > Median > Mode          (c) Mean < Median < Mode     (d) 3 Median = Mode + 2 Mean

Answer:  (a) Mean = Median = Mode
[ In a perfectly symmetrical distribution, Mean, Median, and Mode are all equal. ]

Question:  If Mode = 60 and Mean = 45, then using the empirical relationship, the Median is:
             (a) 48               (b) 50                (c) 52                (d) 54

Answer: (b) 50
[ We have, 3 Median = 60 + 2(45) = 60 + 90 = 150
                       Median    ]

Question: For a moderately skewed distribution, the empirical relationship between the measures of central tendency is given by:

(a) Mode = 3 Median – 2 Mean      (b) Mode = 2 Mean – 3 Median         (c) Mode = 3 Mean – 2 Median             (d) Mode = 2 Median – 3 Mean

Answer:  (a) Mode = 3 Median – 2 Mean

Question: For a moderately skewed distribution, which of the following empirical relationship holds true?

 (a) Mean – Mode = 3(Mean – Median)      (b) Mode = 3 Median – 2 Mean      (c)        (d) All of the above

Answer:  (d) All of the above

Question: The arithmetic mean of 1 , 2 , 3 , ……….. , n is :

           (a)                   (b)                      (c)                  (d) 

Answer:  (a)              

We have,    ]

Assertion-Reasoning 

Assertion (A): For the data: 3, 4, 5, 6, 7, 8, the mode is not defined.

Reason (R): Mode is the value that occurs most frequently in a dataset.

Which of the following is correct?
(a) Both A and R are true, and R is the correct explanation of A.
(b) Both A and R are true, but R is not the correct explanation of A.
(c) A is true, but R is false.
(d) A is false, but R is true.

Answer: (a) Both A and R are true, and R is the correct explanation of A.

[ Assertion (A) is TRUE , Reason (R) is TRUE

Because no value occurs most frequently (R), the mode is not defined (A). ]

Assertion-Reasoning :

Assertion (A): The median of the data 3, 3, 5, 7, 9, 11, 11 is 7.

Reason (R): The median is the average of the two middle observations when the number of observations is even.

(a) Both A and R are true, and R is the correct explanation of A.

(b) Both A and R are true, but R is not the correct explanation of A.

(c) A is true, but R is false.

(d) A is false, but R is true.

Answer:  (c) A is true, but R is false.

[ Assertion (A): Data is 3,3,5,7,9,11,11. Here, n=7 (Odd).

Median = 4th term = 7. (A is True)
Reason (R): The median is the middle observation for odd numbers, not the average of two middle terms.  (R is False).

Question:  The mode of a grouped frequency distribution is found by using the formula:

      

In this formula, ​ denotes:
(a) Frequency of the class preceding the modal class
(b) Frequency of the modal class
(c) Frequency of the class succeeding the modal class
(d) Class size

Answer: (b) Frequency of the modal class .

Question: The median of the following data:  15, 8, 26, 24, 15, 18, 20, 24, 15, 19, 15 is:

           (a) 15             (b) 18          (c) 19      (d) 24

Answer:   (b)  18

[ We arrange the data in ascending order:   8, 15, 15, 15, 15, 18, 19, 20, 24, 24, 26.
Here, n = 11 (Odd)

The median

Question:  If the median of the data 6, 8, x, 12, 14 (arranged in ascending order) is 10, then the value of x is:

(a) 8         (b) 10            (c) 12            (d) 14

Answer:   (b)  10 

[ Here, n = 5

Median  ]

Question:  The following data represents the marks scored by 10 students: 12, 20, 15, 18, 10, 15, 25, 15, 30, 18.
If the mode is M and the median is N, then M - N is equal to:

               (a) 0             (b) -1            (c) -1.5           (d) 1.5

Answer:  (c) - 1.5 

We arrange the data in ascending order : 10, 12, 15, 15, 15, 18, 18, 20, 25, 30.

Here, n = 10  (even)

Median  

Mode (M) = Highest frequency = 15 (occurs 3 times).

So,   

Question:  For a moderately skewed distribution, the relationship between Mean, Median, and Mode is given by: Mode = 3(Median) – 2(Mean).
If the Mean of the data 5, 7, 8, 10, 12, 15 is 9.5, what is the estimated Mode?

(a) 8                (b) 9              (c) 10               (d) 11

Answer:  (a)  8

[ Here, Mean = 9.5   , Median 

We have, Mode = 3(Median) – 2(Mean) = 3 × 9 – 2 × 9.5 = 27 – 19 = 8 

Question: A student mistakenly wrote the observation 26 as 62 while calculating the median of 11 observations (already arranged in ascending order). What will be the effect on the median?

(a) It will increase by 36            (b) It will remain the same          (c) It will decrease by 36           (d) It will become double

Answer:   (b) It will remain the same .

[The median only depends on the position of the middle term, not on the extreme values . Since the data is already in ascending order and 26 is not the middle value, changing it to 62 does not affect the median.]

Question: In a frequency distribution , the class mark of a class is 10 and its width is 5 .The lower limit of class is :   

                   (a)  5                         (b)  7.5                      (c)   10                        (d)  12.5

Answer:   (b) 7.5      

 [ Let  and  be the lower term and upper term respectively .

   So, Class mark  

    

    and    Class width   

 

    

  From  , we get     

          The class interval is    ]

Question: The median class of the following data is :

Classes	0 – 10	10 – 20	20 – 30	30 – 40	40 – 50	50 – 60	60 – 70
Frequency	4	4	8	10	12	8	4

           (a) 30 – 40                     (b) 40 – 50                   (c) 20 – 30                      (d) 50 – 60

Answer:  (b) 40 – 50

  [ Here ,  . This observation lies in the class 30 – 40 .

  The median is 12 , which lies in the class 40 – 50 . ]

Question:  The median class of the following data is :               

Class interval	10 – 20	20 – 30	30 – 40	40 – 50	50 – 60
Frequency	14	11	13	15	17

          (a)  30 – 40                     (b)  10 – 20                     (c) 50 – 60                    (d) 20 – 30           

Answer:  (a)  30 – 40   

[ We construct the table :                           

Class interval	Frequency	C.f
10 – 20 20 – 30 30 – 40 40 – 50 50 - 60	14 11 13 15 17	14 25 38 53 70

   Here,   . This observation lies in the class 30 – 40 . ]  

Question: The model class of the following data is :           

Class interval	15 – 25	25 – 35	35 – 45	45 – 55	55 – 65	65 – 75
Frequency	6	11	7	4	4	2

          (a)  6                              (b)  7                     (c)  4                             (d) 11 

Answer:   (d)  11 

 [ The mode is the most frequency occurring observation . ]

Question: The mean of first 10 odd numbers is :

        (a)  10                        (b) 100                            (c) 20                         (d) 30

Answer:  (a)  10 

[ The first 10 odd numbers are:  1, 3, 5, 7, 9, 11, 13, 15, 17, 19

  ]

Question: The abscissa of the point of intersection of the less than type and of the more than type cumulative frequency curves of a grouped data gives its : 

         (a)  mean             (b)  median             (c)  mode             (d)  all the three above

Answer:   (b) median .

[ The abscissa (x-coordinate) of the point of intersection of the less than type and more than type cumulative frequency curves gives the median of the grouped data. ]

Question: The median of the observations  11 , 12 , 14 , 18 ,  ,  , 30 , 32 , 35 , 41 arranged in ascending  order is 24 , then the value of  :

             (a) 31                             (b) 11                           (c) 18                           (d) 21

Answer:  (d) 21

  [   Here,  .

 A/Q,   Median

   ]

Question:  The following table gives the literacy rate ( in percentage ) of 35 cities .

Literacy rate(in )	45 – 55	55 – 65	65 – 75	75 – 85	85 – 95
Number of cities	3	10	11	8	3

      The upper limit of the median class in the given data is :

                    (a) 55                        (b) 75                        (c)  65                       (d) 85

Answer:  (c) 65

 [ Since,. We observe that the cumulative frequency just greater than  17.5 is 24 and the corresponding class is 60 – 70 . The upper limit is 65 . ]

Question:  If the mean of 15, 12 ,  , 10, 16 , 19 is 12 ,then the value of  .

            (a)  1                         (b)  7                         (c)  2                              (d)  0

Answer:    (d)  0  .

[ We have,  

 ; The value of   is 0 .    ]  

               Answers Following the Questions 

Question: If upper class limit and lower class limit are 55 and 40 , then find the class mark . OR   If the class interval of the data is 40 – 55 , then find the class mark .

 Solution :  We know that ,  Class mark  

Question: If the class mark and lower class limit are 70 and 60 respectively , then find upper class limit .

 Solution :  We know that,  Class mark 

 

Question: If the class mark  and the class size of the data are 40 and 20 respectively , then find the class interval of the data .

 Solution :  Let lower class limit and upper class limit of the data are  and  respectively.

 A/Q,       

 and         

   

  From  we get ,  

  Therefore, the class interval of the data is  30 – 50 . 

Question:  Find the mode of the following marks (out of 10) obtained by 20 students :  4 , 6 , 5 , 9 , 3 , 2 , 7 , 7 , 6 , 5 , 4 , 9 , 10 , 10 , 3 , 4 , 7 , 6 , 9 , 9 

 Solution :  We arrange the given data in the following form : 

            2 , 3 , 3 , 4 , 4 ,4 , 5 , 5 , 6 , 6 , 6 , 7 , 7 , 7 , 9 , 9 , 9 , 9 , 10 , 10

 Here, 9 occurs most frequently [ i.e., four time] . So, the mode is 9 .

Question: Find the mode of the following  data :   51 , 34 , 17 , 58 , 41 , 51 , 17 , 29 , 39 , 29 , 41 , 29

  Solution :  We arrange the given data in the following form :

     17 , 17 , 29 , 29 , 29 , 34 , 39 , 41 , 41 , 51 , 51 , 58

 Here, 29 occurs most frequently [ i.e., three time] . So, the mode is 29 .

Question: Find the median of the following data :  41 , 39 , 48 , 52 , 46 , 62 , 54 , 40 , 96 , 98 ,60

 Solution :   First of all we arrange the data in ascending  order :

           39 , 40 , 41 , 46 , 48 , 52 , 54 , 60 , 62 , 96 , 98          ;    

Here   ( odd)

     Therefore, the median

Question: Find the median of the following data :  48 , 29 , 84 , 32 , 78 , 95 , 72 , 53

 Solution : First of all we arrange the data in ascending  order , as follows :

               29 , 32 , 48 , 53 , 72  , 78 , 84 , 95        ;      Here   is even

 The median 

Question: If the mode of a data is 45 and mean is 27 , the find the median of the data.

 Solution :   We know that , 3 Median = Mode + 2 Mean

 

  

 Therefore, the median of the data is 33 . 

             SECTION = B

Question: If the mean of the following data is 18.75 . Find the value of p. [2005 ]

	10	15	p	25	30
	5	10	7	8	2

Solution:  We have ,

10            15             P            25            30	5 10 7 8 2	50 150 7p 200 60
Total	32	7p + 460

Mean 

 

Therefore, the value of p is 20 .

Question: The following data gives the information on the observed lifetimes (in hours) of 225 electrical components :

Lifetimes (in hours)	Frequency
0 – 20 20 – 40 40 – 60 60 – 80 80 – 100 100 – 120	10 35 52 61 38 29

Determine the modal lifetimes of the components .

Solution: We construct the table :

Lifetimes (in hours)	Frequency
0 – 20    20 – 40    40 – 60    60 – 80    80 – 100   100 – 120	10         35         52         61         38         29

   Here,  

Using the formula ,

  Mode 

  

Therefore, the modal lifetimes of the components is 65.63 (approx.) .

Question:  The lengths of 40 leaves of a plat are measured correct to the nearest millimeter, and  the data obtained is represented in the following table :

Length (in mm)	Number of leaves
118 – 126 127 – 135 136 – 144 145 – 153 154 – 162 163 – 171 172 – 180	3 5 9 12 5 4 2

Find the median length of the leaves .  

Solution :  We have ,

      The lower class limit of 127 – 135 is 127  and the upper class limit of 118 – 126 is 126 .

       Difference 

New Lower limit of a class  

New upper limit of a class

So, New the class interval are :  

We construct the table,

Length (in mm)	Number of leaves	C.F.
117.5 – 126.5 126.5 – 135.5 135.5 – 144.5 144.5 – 153.5 153.5 – 162.5 162.5 – 171.5 171.5 – 180.5	3 5 9 12 5 4 2	3 8 17 29 34 38 40

Here,   . This observation lies in the class 144.5 – 153.5 . 

  Here,   

We have,   Median  

 m.m.

Question: A student noted the number of cars passing through a spot on a road for 100 periods each of 3 minutes and summarised  it in the table given below . Find the mode of the data :

Number of cars	Frequency
0 – 10 10 – 20 20 – 30 30 – 40 40 – 50 50 – 60 60 – 70 70 – 80	7 14 13 12 20 11 15 8

Solution: We construct the table :

Number of cars	Frequency
0 – 10 10 – 20 20 – 30 30 – 40 40 – 50 50 – 60 60 – 70 70 – 80	7 14 13 12 20 11 15 8

  Here,  

Using the formula ,    Mode 

 

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

x	15	17	19	20+p	23
f	2	3	4	5p	6

Solution: We have,

x	f	fx
15         17         19     20+p        23	2 3 4 5p 6	30 51 76 100p + 5p² 138
Total	5p + 15	5p² + 100p + 295

Mean  

 

 

  

The value of p is 1 

Question: Find the mean of the following data :

Class interval	10 – 25	25 – 40	40 – 55	55 – 70	70 – 85	85 – 100
Number of students	2	3	7	6	6	6

 Solution: We construct the table :

Class interval	Class marks	Number of students
10 – 25        25 – 40        40 – 55        55 – 70        70 – 85       85 – 100	17.5              32.5              47.5              62.5              77.5              92.5	2                          3                          7                          6                          6                          6	35            97.5          332.5          375.5          465.5          555.0
Total

                             Mean  

Question: Consider the following distribution of daily wages of 50 workers of a factory .

Daily wages (in Rs)	100 – 120	120 – 140	140 – 160	160 – 180	180 – 200
Number of workers	12	14	8	6	10

 Find the mean daily wages of the workers of the factory by using an appropriate method .

Solution:  We construct the distribution table ,

Daily wages (in Rs)			No. of workers
100 – 120	110	– 40	12	– 480
120 – 140	130	– 20	14	– 280
140 – 160	150	0	8	0
160 – 180	170	20	6	120
180 – 200	190	40	10	400
Total

Using the assumed mean method ,

   Mean

  Therefore, the wages of the workers of the factory is Rs 145.20 .

Question: The following distribution shows the daily pocket allowance of children of a locality . The mean pocket allowance is Rs. 18 . Find the missing frequency  .

Daily pocket allowance (in Rs)	Number of children
11 – 13              13 – 15              15 – 17              17 – 19              19  - 21              21 – 23             23 – 25	7           6           9          13                       5            4

Solution:  We construct the table ,

Daily pocket allowance (in Rs)		Number of children
11 – 13 13 – 15 15 – 17 17 – 19 19  - 21 21 – 23 23 – 25	12 14 16 18 20 22 24	7           6           9        13                  5         4	84            84          144          234                    110           96
Total

                Mean 

               

 

  

Therefore, the missing frequency is 20 .

                  SECTION = C

Question:  The table below shows the daily expenditure on food of 25 household in a locality.  

Daily expenditure (in Rs.)	100 – 150	150 – 200	200 – 250	250 – 300	300 – 350
Number of households	4	5	12	2	2

  Find the mean daily expenditure on food by the assumed mean method .

Solution:  We constructed the table : 

Daily exp. (in Rs.)	No. of households	Class mark
100 – 150 150 – 200 200 – 250 250 – 300 300 – 350	4 5 12 2 2	125 175      225  275 325	– 100 – 50 0 50 100	– 400 – 250 0 100 200
Total

 Using the assumed mean method :

Mean 

 Required the daily expenditure on the food is Rs. 211 .

Question: The distribution below shows the number of wickets taken by bowlers in one-day cricket matches . Find the mean number of wickets by choosing a suitable method . What does the mean signify ?  [Using the assumed mean method .]

Number of wickets	20 – 60	60 – 100	100 – 150	150 – 250	250 – 350	350 – 450
Number of bowlers	7	5	16	12	2	3

Solution: We constructed the table :

No. of wickets	No. of bowlers
20 – 60          60 – 100        100 – 150        150 – 250        250 – 350        350 – 450	7                  5                16                12                 2                 3	40      80     125      200     300     400	– 160             – 120              – 75                  0              100              200	– 1120            – 600         – 1200                  0             200             600
Total

   Using the assumed mean method :

  Mean  

  Therefore, the number of wickets taken by these 45 bowlers in one day cricket is 152.89 .

Question:  Thirty  women were examined in a hospital by a doctor and the number of heart beats per minutes were recorded and summarized as follows . Find the mean heart per minute for these women , choosing a suitable method [Using the step deviation method] .

Number of heart beats per min.	65 – 68	68 – 71	71 – 74	74 – 77	77 – 80	80 – 83	83 – 86
Number of women	2	4	3	8	7	4	2

Solution: We constructed the table :    

No. of heart beats per min.	No. of women
65 – 68 68 – 71 71 – 74 74 – 77 77 – 80 80 – 83 83 – 86	2 4 3 8 7 4 2	66.5 69.5 72.5 75.5 78.5 81.5 84.5	– 3 – 2 – 1 0 1 2 3	– 6 – 8 – 3 0 7 8 6
Total

      Using the step-deviation method :     Here   ,  

             Mean 

    Therefore, the number of heart beats per minute for these women is 75.9 .

Question: Month pocket money of 50 students of a class are  given in the following distribution :

Monthly pocket money  (Rs.)	0 – 50	50 – 100	100 – 150	150 – 200	200 – 250	250 – 300
Number of student	2	7	8	30	12	1

    Find modal class and also give class mark of the modal class .

Solution:  We construct the table :

Monthly pocket money (Rs.)	Class marks	No. of student
0 – 50 50 – 100 100 – 150 150 – 200 200 – 250 250 – 300	25 75 125 175 225 275	2 7 8 30 12 1

    Here the maximum class frequency is 30 . So, the modal class is 150 – 200 .      

        Here,       ,  ,    ,   ,   

      Mode 

 Therefore, the mode is 177.5

Question:  The distribution below gives the weights of 30 students of a class . Find the median weight of the students .                                                                                                                  

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:  We construct the table :

Weight (in kg)	Frequency	Cumulative frequency
40 – 45 45 – 50 50 – 55 55 – 60 60 – 65 65 – 70 70 – 75	2            3            8            6            6            3            2	2                   5                 13                 19                 25                 28                 30

   Since,  . This observation lies in the class 55 – 60 .  

  Here ,   ,    ,   and  

   Median  

  Therefore, the weight of the students is 56.67 .   

Question: If median of the distribution given below is 28.5 , find the values of  and . [SEBA 2016,2021,2023]

Class interval	Frequency
0 – 10 10 – 20 20 – 30 30 – 40 40 – 50 50 – 60	5                          20            15                          5
Total	60

Solution: We construct the distribution table :

Class interval	Frequency	C.F.
0 – 10	5	5
10 – 20
20 – 30	20
30 – 40	15
40 – 50
50 – 60	5
Total	60

   Given ,   

    

     

The median is 28.5 , which lies in the class 20 – 30 .

Here,  

Using the formula ,  Median 

   

Putting the value of  in  , we get       

     The value of   and  .

Question: A life insurance agent found the following data for distribution of ages of 100 policy holders . Calculate the median age , if policies are given only to persons having age 18 years onwards but less than 60 year .

Age (in years)	Number of policy holders
Below 20 Below 25 Below 30 Below 35 Below 40 Below 45 Below 50 Below 55 Below 60	2 6 24 45 78 89 92 98 100

Solution:  We construction the distribution table :

Age  (in years)		Number of policy holders (c.f.)
15 – 20 20 – 25 25 – 30 30 – 35 35 – 40 40 – 45 45 – 50 50 – 55 55 – 60	2 4 18 21 33 11 3 6 2	2 6 24 45 78 89 92 98 100

Since, . This observation lies in the class 35 – 40 . Then,

Here,     

We have,

Median  

  years

                   SECTION = D

Question:  The following frequency distribution gives the monthly consumption of electricity of 68 consumers of a locality . Find the median , mean and mode of the data and compare them .

Monthly consumption (in units)	Number of consumers
65 – 85 85 – 105 105 – 125 125 – 145 145 – 165 165 – 185 185 – 205	4 5 13 20 14 8 4

Solution: We the frequency distribution table :

Class interval				C.F.
65 – 85	75	– 60	4	4	– 240
85 – 105	95	– 40	5	9	– 200
105 – 125	115	– 20	13	22	– 260
125 – 145	135	0	20	42	0
145 – 165	155	20	14	56	280
165 – 185	175	40	8	64	320
185 – 205	195	60	4	68	240
Total			68		140

Here,  

This observation lies in the class 125 – 145 . 

        

Using median the formula :

Median    units

Using assumed mean method :   Here,

   Mean   units

Using  mode the formula :  Here,  

 Mode 

   units

The three measures are approximately the same in this case .

Question:  If the mode of the following distribution is 57.5 , then find the value of   :

Class interval	30 – 40	40 – 50	50 – 60	60 – 70	70 – 80	80 – 90	90 – 100
Frequency	6	10	16		10	5	2

Solution:  We construct the table :

Class interval	30 – 40	40 – 50	50 – 60	60 – 70	70 – 80	80 – 90	90 – 100
Frequency	6	10	16		10	5	2

  Here , the maximum class frequency is 16  and the modal class is  50 – 60 .

       ,     ,    ,   

       Mode 

A/Q,  

        

                   

                   

                    

 Therefore, the value of x is 14 . 

Question:  Find the missing frequencies in following frequency distribution table , if median is 32 . 

Class interval	0 – 10	10 – 20	20 – 30	30 – 40	40 – 50	50 – 60	Total
Number of students	10		25	30		10	100

Solution:  We construct the frequency distribution table :

Class interval	No . of students	C.F.
0 – 10           10 – 20           20 – 30           30 – 40           40 – 50           50 – 60	10                                  25                30                                  10	10
Total

                     

          

    The median is 32 , which lies in the class 30 – 40 .

   Here ,   ,     ,   , 

     Using the formula :    

      Median

 Putting the value of   in  , we get       

Question:  Determine the missing frequency x , from the following data , when mode is 67 .

Class interval	40 – 50	50 – 60	60 – 70	70 – 80	80 – 90
Frequency	5		15	12	7

Solution:  We construct the distribution frequency table of the given data as follows :

Class interval	Frequency
40 – 50                   50 – 60                   60 – 70                   70 – 80                   80 – 90	5                                    15                 12                   7

              Here , ,   ,     ,   and   

    Using mode formula :

      Mode

  

 

 

    

               Therefore, the value of x is 8  .

              SECTION = E 

Question:  The median of the following distribution is 78.5 . If the sum of the frequencies is 60, find the values of and  .

Class Interval	50 - 60	60 - 70	70 - 80	80 - 90	90 - 100	100 - 110
Frequency	5	x	20	25	y	5

Solution:  We construct the table : 

Class interval	Frequency	Cumulative frequency
50 – 60 60 – 70 70 – 80 80 – 90 90 – 100 100 – 110	5                       20          15                        5	5

                     Here ,    

                  

                    

The median is 78.5 , which lies in the class 70 – 80 .

Here,    ,   ,    ,  

We have,   Median  

 

  

     

   Putting  in equation i , we have    

Question:  If the mean of the following frequency distribution is 145 , find the missing frequencies  and  .

Class interval	0 – 50	50 – 100	100 – 150	150 – 200	200 – 250	250 – 300	Total
Frequency	8	12		25		5	80

Solution: We construct the table :  

Class interval	Frequency ()	Class mark ()
0 – 50      50 – 100    100 – 150    150 – 200    200 – 250      250 – 300	8             12                            25                             5	25             75           125           175           225           275	125                        900                       125                       4375                       225                       1375
Total

       Here ,              

                             

                               

                              Mean

                                   

                                    

                                   

                                    

                                   

                                   

                     From  and  , we get  

                                                            

                                               

                            From , we have  

                                                       

                    Therefore, the value of      and   .

Question: The following distribution gives the daily income of 50 workers of a factory. 

Daily income (in Rs.)	100 – 120	120 – 140	140 – 160	160 – 180	180 – 200
Number of workers	12	14	8	6	10

   Find the mean , mode  and  median of the given distribution .

Solution:  We construct the table :

Daily income (in Rs.)				C.f.
100 – 120           120 – 140           140 – 160           160 – 180           180 – 200	12      14        8        6      10	110    130    150    170    190	1320         1820         1200         1020         1900	12         26        34                     40         50
Total

    Using the formula of mean  :

   Mean  

  Using the formula of mode :   Here , ;    ;     ;   ;   

 Mode  

 Using the formula of median :  Since,   ,   will be in 120 – 140 . This observation lies in the class 120 – 140 . So ,The frequency of  the median class is 14 .

   Here,     ;  ;   ;  

    Median 

Question: The median of the following data is 525 . Find the value of and  , if the total frequency is 100 .

Class interval	0 – 100	100 – 200	200 – 300	300 – 400	400 – 500	500 – 600	600 – 700	700 – 800	800 – 900	900 – 1000
Frequency	2	5		12	17	20		9	7	4

Solution:  We construct the table :

Class interval	frequency	Cumulative  frequency
0 - 100 100 - 200 200 - 300 300 - 400 400 - 500 500 - 600 600 - 700 700 - 800 800 - 900 900 – 1000	2 5   12 17 20 9 7 4	2 7

 Given,    and   

 

   

 The median is  , which lies in the class   

  Here ,    ,    ,  ,  

We have, 

 Median

         

 

  

    

  Putting  in equation  we have,   

  Required the value of  and are and   respectively. 

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Posted 5 years ago