Statistics Chapter 14 Extra Questions and Solutions For Class 10 CBSE Mathematics

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Statistics Chapter 14 Extra Questions For Class 10 CBSE Mathematics

1 . Find the mean of the following distribution:

 X : 19 21 23 25 27 29 31
F: 13 15 16 18 16 15 13

Solution.

X F FX
19 13 247
21 15 315
23 16 368
25 18 450
27 16 432
29 15 435
31 13 403
  N = 106 Σ FX = 2620

MEAN = Σ FX / N = 2620 / 106 = 25

Therefore, Mean is 25.

2 . If the mean of the following is 20.6 , find value if p ?

X 10 15 p 25 35
F 3 10 25 7 5

Solution.

X F FX
10 3 30
15 10 150
p 25 25p
25 7 175
35 5 175
  N : 50 Σ FX : 530 + 25p

Now,

Mean = Σ FX / N = ( 530 + 25p ) / 50

It is given that mean is 20.6 ,

20.6 = ( 530 + 25p ) / 50

( 20.6 * 50 ) = 530 + 25p

P = 500 / 25 = 20

Therefore, the value of p is 20.

YOU ARE READING: Statistics Chapter 14 Extra Questions For Class 10 CBSE Mathematics

3. If the mean of the following is 15 , find value of p i.e. the missing frequency ?

X 5 10 15 20 25
F 6 p 6 10 5

Solution.

X F FX
5 6 30
10 p 10p
15 6 90
20 10 200
25 5 125
  N: p+27 Σ FX : 445 + 10p

Mean = Σ FX / N =  ( 445 + 10p) / ( p + 27)

Now, Mean = 15

Mean = (445 + 10p) / ( p + 27 )

15 = (445 + 10p)/( p+ 27)

15p + 405 = 445 + 10p

5p = 40

p = 8.

So, the value of the following missing frequency p is 8.

4 . Mean is 12.58 of the respective, find the value of p ?

X 5 8 10 12 p 20 25
F 2 5 8 22 7 4 2

Solution.

X F FX
5 2 10
8 5 40
10 8 80
12 22 264
p 7 7p
20 4 80
25 2 50
  N : 50 Σ FX = 524 + 7p

And , Mean = Σ FX / N = ( 524 + 7p) / 50

Mean = 12.58,

12.58 = ( 524 + 7p) / 50

12.58 * 50 = 524 + 7p

629 – 524 = 7p

7p = 105

p = 15

So, the value of p is 15 .

YOU ARE READING: Statistics Chapter 14 Extra Questions For Class 10 CBSE Mathematics

5. Find missing frequency p , where mean of distribution is 7.68 ?

X 3 5 7 9 11 13
F 6 8 15 p 8 4

Solution.

X F FX
3 6 18
5 8 40
7 15 105
9 p 9p
11 8 88
13 4 52
  N : 41 + p Σ FX : 303 + 9p

Mean = Σ FX / N = ( 303 + 9p) / (41 + p)

Here, mean = 7.68

7.68 = (303 + 9p)/(41 + p)

1.32p = 11.88

p= 9

Therefore, the value of p is 9 whose mean is 7.68.

6.  The number of telephone calls received at an exchange per interval for 250 successive one- minute intervals are given in the following frequency table:

CALLS (X) 0 1 2 3 4 5 6
INTERVALS(F) 15 24 29 46 54 43 39

Compute the mean number of calls per interval ?

Solution.

Let the assumed mean (A)  is 3

X F U = X -A FU
0 15 -3 -45
1 24 -2 -48
2 29 -1 -29
3 46 0 0
4 54 1 54
5 43 2 86
6 39 3 117
  N : 250   Σ FU : 135

Mean number of calls = A + (Σ FU/ N)

                                       =  3 + ( 135 / 250 )

                                       = 885 / 250

                                        = 3.54

So, the mean number of calls per interval is 3.54 .

7.  The following table gives the number of branches and number of plants in the garden of a school.

Branches (X) 2 3 4 5 6
Plants ( F) 49 43 57 38 13  

Compute the average number of branches per plant?

Solution.

Let the assumed mean (A) is 4

X F U= X – A FU
2 49 -2 -98
3 43 -1 -43
4 57 0 0
5 38 1 38
6 13 2 26
  N: 200   Σ FU : -77

Average no of branches per plant = A + (Σ FU/N)

                                                            = 4 – 77/200  =  (800 – 77)/200 = 3.615

So, the average number of branches per plant is 3.615.

YOU ARE READING: Statistics Chapter 14 Extra Questions For Class 10 CBSE Mathematics

8.  The following table gives the distribution of total household expenditure (in rupees) of manual workers in a city. Find the average expenditure (in rupees) per household?

Expenditure X : Frequency F : Expenditure X: Frequency F:
100 – 150 24 300 – 350 30
150 – 200 40 350 – 400 22
200 – 250 33 400 – 450 16
250 – 300 28 450 – 500 7

Solution.

Let assumed mean (A) is 275

Class interval mid value X D = X – A U =(X-275)/50 Frequency F FU
100 – 150 125 -150 -3 24 -72
150 -200 175 -100 -2 40 -80
200 – 250 225 -50 -1 33 -33
250 – 300 275 0 0 28 0
300 – 350 325 50 1 30 30
350 – 400 375 100 2 22 44
400 – 450 425 150 3 16 48
450 – 500 475 200 4 7 28
        N: 200 Σ FU = -35

A = 275 and h =50

Mean = A + h * (Σ FU / N)

           = 275 + 50 * ( -35 /200)

          =  266.25

9.  A survey was conducted by a group of students as a part of project regarding the number of plants in 200 houses in a locality. Find the mean number of plants per house.Which method did you use for finding the mean, and why?

Plants 0-2 2 – 4 4 – 6 6 – 8 8- 10 10 -12 12 -14
House 1 2 1 5 6 2 3

Solution.

From the given data we have to find the class interval we know that,

Class marks (x) = (upper class limit + lower class limit)/2

Plants Houses X FX
0-2 1 1 1
2-4 2 3 6
4-6 1 5 5
6-8 5 7 35
8- 10 6 9 54
10 -12 2 11 22
12 -14 3 13 39
  N : 20   Σ FU = 162

Mean = Σ FU / N

            = 162 / 20 = 8.1

Thus , the mean number of plant per house is 8.1 . Here we will use the Direct Method as values of class marks X and F are very small.

10. Consider the following distribution of daily wages of workers of a factory.Find the mean daily wages of the workers of the factory by using an appropriate method?

Daily wages (in ₹)   100-120 120-140 140-160 160-180 180-200
Number of workers: 12 14 8 6 10

Solution.

Let the Assumed Mean (A) is 150

Class interval Mid value X D = X -150 U = (X-150)/20 F FU
100-120 110 -40 -2 12 -24
120-140 130 -20 -1 14 -14
140-160 150 0 0 8 0
160 -180 170 20 1 6 6
180 – 200 190 40 2 10 20
        N: 50 Σ FU = -12

A = 150 and h = 20

Mean = A + h * (Σ FU /N)

            = 150 + 20 * ( -12 /50 )

            =  150 – 4.8

            =  145.20 .

YOU ARE READING: Statistics Chapter 14 Extra Questions For Class 10 CBSE Mathematics

11. Find the mean of the frequency distribution :

Class interval 0-6 6-12 12-18 18-24 24-30
frequency 6 8 10 9 7

Solution:

Let the Assumed Mean (A) is 15.

Class interval Mid value X D = X – A U= X- A/ 6 F FU
0 -6 3 -12 -2 6  -12
6 – 12 9 -6 -1 8  -8
12 – 18 15 0 0 10  0
18 – 24 21 6 1 9   9
24 – 30 27 12 2 7 14
        N : 40 Σ FU =3

Here, A = 15 and h = 6

Mean = A + h x (Σ FU /N)

= 15 + 6 x (3/40) = 15 + 0.45 == 15.45

12. Following are the lives in hours of 15 pieces of the components of aircraft engine. Find the median: 

715, 724, 725, 710, 729, 745, 694, 699, 696, 712, 734, 728, 716, 705, 719.

Solution.

Arranging the given data in ascending order, we have

694, 696, 699, 705, 710, 712, 715, 716, 719, 721, 725, 728, 729, 734, 745.

Here the number of terms is an old number i.e., N = 15

We use the following procedure to find the median.

Median = (N + 1)/2 th term

= (15 + 1)/2 th term

= 8th term

So, the 8th term in the arranged order of the given data should be the median.

Therefore, 716 is the median of the data.

13. The following is the distribution of height of students of a certain class in a certain city. Find the median height.

Height in cm 160-162 163-165 166-168 169-171 172-174
students 15 118 142 127 18

Solution.

Class Interval exclusive Class Interval inclusive Class Interval Frequency Cumulative Frequency
160-162 159.5-162.5 15 15
163-165 162.5-165.5 118 133 (F)
166-168 165.5-168.5 142 (f) 275
169-171 168.5-171.5 127 402
172-174 171.5-174.5 18 420
     N : 420  

Here, we have N as 420. The cumulative frequency just greater than N/2 is 275 then 165.5 – 168.5 is the median class. Now,

L = 165.5, f = 142, F = 133 and h = (168.5 – 165.5) = 3

=165.5 + 1.63 = 167.13

Statistics

YOU ARE READING: Statistics Chapter 14 Extra Questions For Class 10 CBSE Mathematics

14.  Calculate the median from the following data:

Marks 0-10 10-20 20-30 30-40 40-50 50-60 60-70 70-80
Students 15 35 60 84 96 127 198 250

Solution.

Marks below Students Class Interval frequency Cumulative Frequency
10 15 0-10 15 15
20 35 10-20 20 35
30 60 20-30 25 60
40 84 30-40 24 84
50 96 40-50 12 96
60 127 50-60 31 127
70 198 60-70 71 198
80 250 70-80 52 250
      N: 250  

Here , N = 250  so, N/2 = 125.

The cumulative frequency just greater than N/ 2 is 127 then median class is 50 – 60 such that L = 50, f = 31, F = 96, h = 60 -50 = 10

= 50 + 9.35 = 59.35

15.  Calculate the missing frequency from the following distribution, it being given that the median of the distribution is 24?

Age in years 0-10 10-20 20-30 30-40 40-50
No of persons 5 25 ? 18 7

Solution.

Let the unknown frequency be x.

Class Interval Frequency Cumulative Frequency
0-10 5 5
10-20 25 30 (F)
20-30 x (f) 30 +x
30-40 18 48 + x
40-50 7 55 + x
  N : 170  

Given, median = 24.

Then, median class = 20 – 30; L = 20, h = 30 -20 = 10, f = x, F = 30

4x = 275 + 5x – 300

4x – 5x = – 25

– x = – 25

x = 25

Therefore, the Missing frequency is 25 where the median is given as 24.

16. The following table gives the distribution of the life time of 400 neon lamps.Find the median life?

Life Time : in hours Number of Lamps
1500-2000 14
2000-2500 56
2500-3000 60
3000-3500 86
3500-4000 74
4000-4500 62
4500-5000 48

Solution.

Life time in hours Number of lamps F Cumulative Frequency CF
1500-2000 14 14
2000-2500 56 70
2500-3000 60 130 (CF)
3000-3500 86 (F) 216
3500-4000 74 290
4000-4500 62 352
4500-5000 48 400
  N: 400  

It’s seen that, the cumulative frequency just greater than n/2 (400/2 = 200) is 216 and it belongs to the class interval 3000 – 3500 which becomes the Median class = 3000 – 3500. Frequency (F) of median class = 86 and Cumulative frequency (CF) of class preceding median class = 130 and, the Class size (h) = 500.

We use the following formula to solve it:

= 3406.98

Therefore, the median lifetime of lamp is 3406.98 hours.

Statistics

YOU ARE READING: Statistics Chapter 14 Extra Questions For Class 10 CBSE Mathematics

17. Find the missing frequencies  for the following distribution if the mean is 1.46?

No of accidents: 0 1 2 3 4 5 Total
Frequency No of days : 46 ? ? 25 10 5 200

Solution:

No of Accidents (X) No of Days (F) FX
0 46 0
1 x x
2 y 2y
3 25 75
4 10 40
5 5 25
  N : 200 Sum : x + 2y + 140

It’s given that, N = 200,Therefore

⇒ 46 + x + y + 25 + 10 + 5 = 200

⇒ x + y = 200 – 46 – 25 – 10 – 5

⇒ x + y = 114 —- (i)

And also given, Mean = 1.46

⇒ Sum/ N = 1.46

⇒ (x + 2y + 140)/ 200 = 1.46

⇒ x + 2y = 292 – 140

⇒ x + 2y = 152 —- (ii)

Subtract equation (i) from equation (ii), we get

x + 2y – x – y = 152 – 114

⇒ y = 38

Now, on putting the value of y in equation (i), we find x = 114 – 38 = 76

Therefore, the missing frequencies are 76 and 38 .

Statistics

18. Find the mode of the following data:

        (i) 3, 5, 7, 4, 5, 3, 5, 6, 8, 9, 5, 3, 5, 3, 6, 9, 7, 4 

        (ii) 15, 8, 26, 25, 24, 15, 18, 20, 24, 15, 19, 15 

Solution.

         (i) 3, 5, 7, 4, 5, 3, 5, 6, 8, 9, 5, 3, 5, 3, 6, 9, 7, 4 

Value (X) : 3 4 5 6 7 8 9
Frequency (F): 4 2 5 2 2 1 2

Thus , the mode = 5 since it occurs the maximum number of times.

        (ii) 15, 8, 26, 25, 24, 15, 18, 20, 24, 15, 19, 15 

Value (X) : 8 15 18 19 20 24 25
Frequency (F): 1 4 1 1 1 2 1

Thus, the mode = 15 since it occurs the maximum number of times.

19. The shirt size worn by a group of 200 persons, who bought the shirt from a store, are as follows. Find the model shirt size worn by the group?

Shirt Size: 37 38 39 40 41 42 43 44
No of person: 15 25 39 41 36 17 15 12

Solution:

Shirt Size: 37 38 39 40 41 42 43 44
No of person: 15 25 39 41 36 17 15 12


From the data its observed that, Model shirt size = 40 since it was the size which occurred for the maximum number of times.

20. Find the mode of the following distribution.

Class Interval : 25-30 30-35 35-40 40-45 45-50 50-55
Frequency : 25 34 50 42 38 14

Solution.

Class Interval : 25-30 30-35 35-40 40-45 45-50 50-55
Frequency: 25 34 50 42 38 14

It’s seen that the maximum frequency is 50. So, the corresponding class i.e., 35 – 40 is the modal class. And,

l = 35, h = 40 – 35 = 5, f = 50, f1 = 34, f= 42

So, we use the following formula:

= 38.33

YOU ARE READING: Statistics Chapter 14 Extra Questions For Class 10 CBSE Mathematics