Chapter 14. STATISTICS

Chapter 14 . Statistics

Exercise 14.1 complete solution Exercise 14.2 complete solution Exercise 14.3 complete solution Exercise 14.4 complete solution

1. Class mark 2. There is a empirical relationship between the three measures of central tendency :        3 Median = Mode + 2 Mean 3. (a) If  is odd, then the median   observation and,  (b) If n is even, then the median observations. 4. The mean for grouped data can be found by : (i) the direct method : Mean (ii) the assume mean method : Mean (iii) the step deviation method : Mean with the assumption that the frequency of a class is centred at its mid-point, called its class mark. 5. The mode for grouped data can be found by using the formula: Mode  where  lower limit of the modal class,    size of the class interval (assuming all class sizes to be equal),    frequency of the modal class,    frequency of the class preceding the modal class,    frequency of the class succeeding the modal class. 6. The median for grouped data is formed by using the formula: Median  where  lower limit of median class,   number of observations,   cumulative frequency of class preceding the median class,    frequency of median class,    class size (assuming class size to be equal) 7. The median of grouped data can be obtained graphically as the x-coordinate of the point of intersection of the two ogives for this data.

 Class 10 Maths Chapter 14. STATISTICS Exercise 14.1 Solutions

1. A survey was conducted by a group of students as apart of their environment  awareness  program , in which they collected the following data regarding the number of plants in 20 house in a locality . Find the mean number of plants per house .

Number of plants	0 – 2	2 – 4	4 – 6	6 – 8	8 – 10	10 – 12	12 – 14
Number of houses	1	2	1	5	6	2	3

Which method did you use for finding the mean , and why ?   

Solution: We construct the table ,

Numbers of plants	Class mark	Number of houses
0 – 2       2 – 4       4 – 6       6 – 8      8 – 10     10 – 12     12 – 14	1         3         5         7         9        11        13	1         2         1         5         6         2         3	1            6           5         35         54         22         39
Total

      Mean

Because, class mark and frequency are sufficiently small . So, the direct method is an appropriate choice .

2. 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 – 120 120 – 140 140 – 160 160 – 180 180 – 200	110 130 150 170 190	– 40  – 20 0 20 40	12 14 8 6 10	– 480  – 280 0 120 400
Total

Using the assumed mean method ,

   Mean

    

   

  

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

3. 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 .

4. Thirty women were examined in a hospital by a doctor and the number of heart beats per minute were recorded and summarized as follows . Find the mean heart beats per minute for these women , choosing a suitable method .

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

Solution: We construct the table ,

Number of heart beats per minute			Number of women
65 – 68 68 – 71 71 – 74 74 – 77 77 – 80 80 – 83 83 – 86	66.5 69.5 72.5 75.5 78.5 81.5 84.5	– 9  – 6  – 3 0 3 6 9	2 4 3 8 7 4 2	– 18  – 24  – 9 0 21 24 18
Total

Using the assumed mean method , we have

  Mean

                    

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

5.  In a retail market , fruit vendors were selling mangoes kept in packing boxes . These boxes contained varying number of mangoes . The following was the distribution of mangoes according to the number of boxes .

Number of mangoes	50 – 52	53 – 55	56 – 58	59 – 61	62 – 64
Number of boxes	15	110	135	115	25

Find the number of mangoes kept in a packing box .Which method of finding the mean did you choose ?

Solution: We construct the table :

No. of mangoes
50 – 52 53 – 55 56 – 58 59 – 61 62 – 64	51 54 57 60 63	– 6  – 3 0 3 6	15 110 135 115 25	– 90  – 330 0 345 150
Total			400	75

Here ,   ,     and

Using the assumed mean method ,

Mean

              

             

             

               

6. The table below shows the daily expenditure on food of 25 households in a locality .

Daily expenditure (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 a suitable method .

Solution: We construct the table , we have

Daily exp. (Rs)			No. of house -holds
100 – 150 150 – 200 200 – 250 250 – 300 300 – 350	125 175 225 275 325	– 2  – 1 0 1 2	4 5 12 2 2	– 8  – 5 0 2 4
Total

     Here ,

Using  the step-deviation method , we have

Mean

    

    

    

Therefore, the daily expenditure on food is Rs 211 .

7. To find out the concentration of   in the air (in parts per million, i.e., ppm) , the data was collected for 30 localities in a certain city and is presented below :

Concentration of  (in ppm)	Frequency
0.00 – 0.04 0.04 – 0.08 0.08 – 0.12 0.12 – 0.16 0.16 – 0.20 0.20 – 0.24	4 9 9 2 1 2

Find the mean concentration of  , in the air .

Solution : We construct the table :

Concentration of
0.00 – 0.04 0.04 – 0.08 0.08 – 0.12 0.12 – 0.16 0.16 – 0.20 0.20 – 0.24	0.02 0.06 0.10 0.14 0.18 0.22	4 9 9 2 1 2	0.08 0.54 0.90 0.28 0.18 0.44
Total		27	2.42

Using the direct method , we have

Mean

Therefore, the concentration of  in the air is 0.09 ppm .

8. A class teacher has the following absentee record of 40 students of a class for the whole term . Find the mean number of days a student was absent .

Number of days	Number of students
0 – 6 6 – 10 10 – 14 14 – 20 20 – 28 28 – 38 38 – 40	11               10                7                4                4                3                1

Solution: We construct the table :

No. of days
0 – 6 6 – 10 10 – 14 14 – 20 20 – 28 28 – 38 38 – 40	3 8 12 17 24 33 39	11 10 7 4 4 3 1	33 80 84 68 96 99 39
Total		40	499

Using the direct method , we have

Mean

     

Therefore, the number of days is 12.475 .  

9. The following table gives the literacy rate (in percentage) of 35 cities . Find the mean literacy rate .

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

Solution: We construct the table :

Literacy rate
45 – 55 55 – 65 65 – 75 75 – 85 85 – 95	50 60 70 80 90	– 20  – 10 0 10 20	3 10 11 8 3	– 60  – 100 0 80 60
Total			35	– 20

Using the assumed mean method ,

Mean

    

Therefore , the mean literacy rate is 69.4%  . 

 Class 10 Maths Chapter 14. STATISTICS Exercise 14.2 Solutions 

1. The following table shows the ages of the patients admitted in a hospital during a year :

Age (in years)	Number of patients
5 – 15 15 – 25 25 – 35 35 – 45 45 – 55 55 – 65	6 11 21 23 14 5

Find the mode and the mean of the data given above . Compare and interpret the two measures of central tendency .

Solution : We construct the table :

Age	Class mark	No. of patients
5 – 15 15 – 25 25 – 35 35 – 45 45 – 55 55 – 65	10 20 30 40 50 60	6 11 21 23 14 5	60 220 630 920 700 300
Total		80	2830

     Here,  

Using the formula ,

  Mode

            

              years

Using the direct method ,

    Mean years

Therefore, the maximum number of patients admitted in the hospital are of the age 36.82 years , while on an average the age of a patient admitted to the hospital is 35.37 years .

2. 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.) .

3. The following data gives the distribution of total monthly household expenditure of 200 families of a villages . Find the modal monthly expenditure of the families . Also , find the mean monthly expenditure :

Expenditure (in Rs)	Number of families
1000 – 1500 1500 – 2000 2000 – 2500 2500 – 3000 3000 – 3500 3500 – 4000 4000 – 4500 4500 – 5000	24 40 33 28 30 22 16 7

Solution : We construct the distribution table :

Expenditure	Class mark
1000 – 1500 1500 – 2000 2000 – 2500 2500 – 3000 3000 – 3500 3500 – 4000 4000 – 4500 4500 – 5000	1250 1750 2250 2750 3250 3750 4250 4750	– 2000  – 1500  – 1000  – 500 0 500 1000 1500	24 40 33 28 30 22 16 7	– 48000  – 60000  – 33000  – 14000              0     11000     16000     10500
Total			200	– 117500

   Here,  

Using the formula ,

Mode

Using direct method :  Here ,

Mean

Therefore, the modal monthly expenditure is Rs 1847.83 and the mean monthly expenditure is Rs 2662.5  .

4. The following distribution gives the state-wise teacher-student ratio in higher secondary schools of India . Find the mode and mean of this data . Interpret the two measures .

Number of students per teacher	Number of states/U.T.
15 – 20 20 – 25 25 – 30 30 – 35 35 – 40 40 – 45 45 – 50 50 – 55	3 8 9 10 3 0 0 2

Solution : We construct the distribution table :

No. of students per teacher
15 – 20 20 – 25 25 – 30 30 – 35 35 – 40 40 – 45 45 – 50 50 – 55	17.5 22.5 27.5 32.5 37.5 42.5 47.5 52.5	3 8 9 10 3 0 0 2	52.5 180 247.5 325 112.5 0 0 105
Total		35	1022.5

  Here,  

Using the formula ,

  Mode

Using the direct method ,

    Mean

Therefore, the most states/U.T. have a student teacher ratio of 30.6 and on an average, this ratio is 29.2 .

5. The given distribution shows the number of runs scored by some top batsmen of the world in one-day international cricket matches .

Runs scored	Number of batsmen
3000 – 4000 4000 – 5000 5000 – 6000  6000 – 7000 7000 – 8000 8000 – 9000 9000 – 10000 10000 – 11000	4 18 9 7 6 3 1 1

Find the mode of the data .

Solution : We construct the table :

Runs scored	No. of batsmen
3000 – 4000 4000 – 5000 5000 – 6000 6000 – 7000 7000 – 8000 8000 – 9000 9000 – 10000 10000 – 11000	4 18 9 7 6 3 1 1

Using the formula ,     Here,  

  Mode

 

6. 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

 Class 10 Maths Chapter 14. STATISTICS Exercise 14.3 Solutions

1. 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 :

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

Here,

This observation lies in the class 125 – 145 . Then,

     

Using the formula ,

Median

              

                units

Using assumed mean method ,

Here,

   Mean

 

   units

Using the formula ,

Here,  

          

 Mode

 units

The three measures are approximately the same in this case .

2. If median of the distribution given below is 28.5 , find the values of  and .

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 10 – 20 20 – 30 30 – 40 40 – 50 50 – 60	5                   20        15                  5	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  .

3. 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

Here,   

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

    

Using the formula ,

Median  

              

              

              

              

                years

4. 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 .

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 . Then,

    

Using the formula ,

Median  

              

              

              

              

                m.m.

5. The following table gives the distribution of the life time of 400 neon lamps :

Life time (in hours)	Number of lamps
1500 – 2000 2000 – 2500 2500 – 3000 3000 – 3500 3500 – 4000 4000 – 4500 4500 – 5000	14 56 60 86 74 62 48

Find the median life time of a lamp .

Solution : We construct the distribution table :

Life time  (in hours)	Number of lamps	C.F.
1500 – 2000 2000 – 2500 2500 – 3000 3000 – 3500 3500 – 4000 4000 – 4500 4500 – 5000	14 56 60 86 74 62 48	14 70 130 216 290 352 400

Here,   

This observation lies in the class 3000 – 3500 . Then,

     

Using the formula ,

Median

            

            

            

            

             hours .

6. 100 surnames were randomly picked up from a local telephone directory and the frequency distribution of the number of letters in the English alphabets in the surname

was obtained as follows :

Number of letters	Number of surnames
1 – 4 4 – 7 7 – 10 10 – 13 13 – 16 16 – 19	6 30 40 16 4 4

Determine the median number of letters in the surnames . Find the mean number of letters in the surname ? Also ,find the modal size of the surnames .

Solution : We construct the table :

Number of letters			C.F.
1 – 4 4 – 7 7 – 10 10 – 13 13 – 16 16 – 19	2.5 5.5 8.5 11.5 14.5 17.5	6 30 40 16 4 4	6 36 76 92 96 100	15 165 340 184 58 70
Total		100		832

Here,   

This observation lies in the class 7 – 10 . Then,

     

Using the formula ,

Median

              

              

              

              

               

Using the direct method,

  Mean 

                

Using the mode formula,

Here,

             

 Mode

            

            

            

            

            

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

Weight (in kg)	Number of students
40 – 45 45 – 50 50 – 55 55 – 60 60 – 65 65 – 70 70 – 75	2 3 8 6 6 3 2

Solution: We construct the distribution table :

Weight (in kg)	Number of students	C.F.
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

Here,   

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

   

Using the formula ,

Median  

             

             

             

             

              kg .