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Revision Question Bank Statistics 1.

The mean of the following frequency table is 50 but the frequencies f1 and f2 in class intervals 20-40 and 60-80 are missing. Find the missing frequencies. Class interval

0-20

20-40

40-60

60-80

80-100

Total

Frequency

17

f1

32

f2

19

120

Solution : Let the assumed mean be A = 50 and h = 20. Table for step deviation is given below Class interval 0-20 29-40 40-60 60-80 80-100

Frequency (f1)

Mid value

17 f1 32 f2 19

10 30 50 70 90

N   fi  68  f1  f2

Total

ui 

xi  A h -2 -1 0 1 2

f u i

i

Fiui -34 -fi 0 f2 38  4  f1  f2

We have, N   fi  120

given

68 + f1 + f2 = 120  f1 + f2 = 52  But Mean = 50

………(i) [given]

1  A  h   fiui   50 N   4  f1  f2   50  20     50  120  4  f1  f2  50   50 6

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4  f1  f2 0 6  4  f1  f2  0 

 f1  f2  4

2.

………….(ii)

On solving Eqs. (i) and (ii), we get f1 = 28 and f2 =24 If the mean of the following distribution is 6. Find the value of p. xi

2

4

6

10

p+5

fi

3

2

3

1

2

Solution : xi 2 4 6 10 P +5

fi 3 2 3 1 2

Total

 f  11

f x

i

Here, mean = 6,

 Mean 

fixi 6 8 18 10 2P + 10 i i

 f  11 and  f x i

i i

 2p  52

 2p  52

f x

i i

f

i

2p  52 11  2p  52  66  2p  66  52  2p  14 14 p 7 2 Compute the median from the following data.  6

3.

Mid value

115

125

135

145

155

165

175

185

195

Frequency

6

25

48

72

116

60

38

22

3

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Solution : We have, the mid values so, firstly we, should find the upper and lower limits of the various classes. The difference between two consecutive values is h = 125 –115 =10. Lower limit of a class = Mid value – h/ 2 Upper limit = Mid value + h/ 2 Table for cumulative frequency is given below Cumulative Mid value Class groups Frequency frequency 115 125 135

110-120 120-130 130-140

6 25 48

6 31 79

145

140-150

72

151

155 165 175 185 195

150-160 160-170 170-180 180-190 190-200

116 60 38 22 3

267 327 365 387 390

Total

N   fi  390

We have, N=390 390  195 2 The cumulative frequency just greater than N/2. i.e. 195 is 267 and the corresponding class is 150-160. So, 150-160 is the median class. I = 150, f = 116, h = 10, cf = 151 N  cf 195  151 2 Median= l   h  150   10 f 116 44  10 440  150  = 150  116 116 = 150 + 3.79 = 153.76

Now, N fi 

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If the median of the distribution of 60 students given below is 28.5, find the values of x and y. Class

0-10

10-20

20-30

30-40

40-50

50-60

Number of students

5

x

20

15

y

5

Solution : Table for cumulative frequency is given below Class interval

Frequency(f)

Cumulative

0-10

5

5

10-20

X

5+x

20-30

20

25+x

30-40

15

40+x

40-50

Y

40+x+y

50-60

5

45+x + y

Total

 f  60 i

We have, Median = 28.5 Clearly, it lies in the class interval 20-30. So, 20-30 is the median class.

 l = 20, h = 10, / = 20, cf = 5 + x and N = 60 N  cf 2 Median = l  h f  28.20 +

30  5  x   10 20

 28.5  20 

25  x 25  x  8.5  2 2

25  x  17  x  8

We have,

N = 60

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45 + x + j = 60  x + y = 15 On putting x = 8 in x + y = 15, we get y = 7 Hence, x = 8 and 7 = 7 5.

Compute the median for the following cumulative frequency distribution. Less than

20

30

40

50

60

70

80

90

100

cf

0

4

16

30

46

66

82

92

100

Solution : We are given less than cumulative frequency distribution. So, we first construct a frequency table from the given less than cumulative frequency distribution and then we will make necessary computations to compute median. Table for cumulative frequency is given below Frequency (f)

frequency (cf)

20-30 30-40 40-50 50-60 60-70 70-80 80-90 90-100

4 12 14 16 20 16 10 8

4 16 30 46 66 82 92 100

Total

 f  100

Class intervals

i

Here, N   fi  100 Now,

N  50 2

We observe that, the cumulative frequency just N greater than

N = 50 is 66 and the 2

corresponding a class is 60-70. So, 60-70 is the median class. l = 60, f= 20, cf = 46and h = 10

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N  cf 2 Median = l  h f  Median  60 

6.

50  46 40  10  60   62 20 20

The following data gives the distribution of total household expenditure (in Rs) of manual workers in a city. Expenditure (in Rs) Frequency

1000-1500 1500-2000 24

2000-2500

2500-3000

3000-3500

3500-4000

33

28

30

22

40

4000-4500 4500-5000 16

7

Find the mode from the above data. Solution : We observe that, the class 1500-2000 has the maximum frequency 40. So, it is the modal class such that l= 1500, h = 500, f1 = 40, f0= 24 and f2 = 33 Mode  l 

= 1500  =1500+ 7.

f1  f0  500 2f1  f0  f2

40  24  500 80  24  33

16  500  1500  374.826  1847.826 23

Draw an ogive for the following frequency distribution by less than method. Marks

0-10

10-20

20-30

30-40

40-50

50-60

Number of students

7

10

23

51

6

3

Solution : Firstly we prepare the cumulative frequency distribution table by less than method as given below.

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Cumulative Mark Marks less than students frequency 0-10 7 10 7 10-20 10 20 17 20-30 23 30 40 30-40 51 40 91 40-50 6 50 97 50-60 3 60 100 Other than the given class interval, we assume a class 10-0 before the first class interval Number of students

0-10 with zero frequency. Now, we mark the upper class limits (including the imagined class) along X-axis on a suitable scale and the cumulative frequencies along Y-axis on a suitable scale.

Thus, we plot the points (0,0), (10,7), (20,17), (30,40), (40, 91), (50, 97) and (60, 100). Now, we join the plotted points by a free hand curve to obtain the required ogive. 8.

The median of the distribution given below is 14.4. Find the values of x and y, if the total frequency is 20. Class interval Frequency

0-6

6-12

12-18

18-24

24-30

4

X

5

Y

1

Solution : Table for cumulative frequency in given below

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Class interval Frequency 0-6 4 6-12 X 12-18 5 18-24 Y 24-30 1 Since, N = 20 10 + x + y = 20  x + y = 20 –10 ………(i)  x + y = 10 Also, we have Median = 14.4 which lies in the class interval 12-18 The median class is 12-18, such that l = 12, f = 5, cf = 4 +x and h = 6 N   2  cf    h f    

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Cumulative frequency 4+0=4 4 + x = (4 +x) 5 + (4+x) = 9+ x Y +(9+x) = 9 + x + y 1 + (9 + x + y) = 10 + x + y

 10   4  x    14.4  12    6 5    4x   14.4  12  2  6    h  5 

 24  6x   2.4  12     5  24  6x   12  2.4  9.6 5  24  6x   9.6  5  6x  48  24  24 24 4 6 From eqs.(i), x + y = 10 4 + y = 10  x

[put x =4]

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y = 10–4 =6  Thus, x =4 and y = 6 Find the mean, mode and median of the following data. Class 0-10 10-20 20-30 30-40 40-50 50-60 60-70

Frequency 3 4 7 15 10 7 4

Solution : Let the assumed mean ‘A’ = 35 Here, h =10  ui 

x i  35 10

Table for step deviation and their product with corresponding frequency in given below Class interval 0-10 10-20 20-30 30-40 40-50 50-60 60-70

xi

fi

Cf

5 15 25 35 45 55 65

3 4 7 15 10 7 4

3+0=3 4+3=7 7+7=14 15+14=29 10+29=39 7+39=46 4+46=50

 f  50

Total Here,

i

 f  50 ,  u f x i

i

i

i

ui 

x i  35 10 -3 -2 -1 0 1 2 3

fiui -9 -8 -7 0 10 14 12

u f x i

i

i

 12

 12

  uifi  (i) = A   h f   i   12    35     10  50  www.pioneermathematics.com S.C.O. - 326, Sector 40–D, CHD. Phone: 9815527721, 4617721

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12  25  2.5  37.4 5 (ii) To find mode Here, greatest frequency = 15  Modal class = 30-40 l = 30, f1 =15, f0 = 7, f2 = 10 and h = 10  f f  l   1 0 h  2f1  f0  f2 

= 35 

  15  7 = 35     10 2 15  7  10      8  = 35     10  30  17  8 80 = 35   10  35  13 13 = 35+6.15=41.15 (iii) To find median Here, N = 50 N 50  25 Now,  2 2  median class is 30-40. such that cf = 14, f =15 and h = 10 N   2  cf  l h f      25  14   11   10  30  = 30     15   10  15    22  30  7.3  37.3 = 30  3 10. Draw a cumulative frequency curve for the following frequency distribution by less than

method. Age (in year) Number of persons

0-9 5

10-19 15

20-29 20

30-39 23

40-49 17

50-59 11

60-69 9

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Solution : The given frequency distribution is not continuous. So, we first make it continuous and prepare the cumulative frequency distribution as under Cumulative Frequency 0.5-9.5 5 9.5 5 9.5-19.5 15 19.5 20 19.5-29.5 20 29.5 40 29.5-39.5 23 39.5 63 39.5-49.5 17 49.5 80 49.5-59.5 11 59.5 91 59.5-69.5 9 69.5 100 Now, we plot points (9.5, 5), (19.5,20), (29.5,40), (39.5,63), (49.5,80), (59.5,91) and Age(in year)

Frequency

Age less than

(69.5,100) and join them by a free hand smooth curve to obtain the required ogive.

Solutions: In your E-mail ID Provided By You

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Chapter Test {Statistics} M: Marks: 40 1.

M: Time: 40 Min.

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

[4]

0-6

6-10

10-14

14-20

20-28

28-38

38-40

11

10

7

4

4

3

1

Solution: Table for mid value and their product with corresponding frequency is given below Number of days 0-6 6-10 10-14 14-20 20-28 28-38 38-40 Total

Number of students 11 10 7 4 4 3 1  fi  40

Mid value

fi x i

3 8 12 17 24 33 39

33 80 84 68 96 99 39  fixi  499

 f  40 and  f x  499 fx  Mean, x   f Here,

i

i i

i i i

=

499 40

= 12.48 Hence, mean number of days is 12.48. 2.

If the median of the distribution given below is 29.5, then find the values of x and y.

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Class interval 0-10 10-20 20-30 30-40 40-50 50-60 Total

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Frequency 5 a 10 15 b 5 50

Solution: Here, N = 50 Now, construct a cumulative frequency table from the given data. Class interval 0-10 10-20 20-30 30-40 40-50 50-60 Total

Frequency 5 A 10 15 b 5 N = 50

Cumulative Frequency 5 5+a 15+a 30+a 30 + a + b 35 = a + b N = 35 + a + b

Since, median is 28.5, so median class is 20-30. Here, N = 50, l = 20, h = 10, f = 20 and cf = 5 + a N    cf  2   Median = l  h   f

 50   2  5  a    29.5 = 20 + 10 × 20  29.5  20 

 29.5 

25  5  a 2

40  20  a 2

 59 = 40 + 20 – a

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Also, 50 = 35 + a + b  15 = 1 + b

[put a = 1]

 b = 14

3.

The following frequency distribution gives the monthly consumption of electricity of 68 consumers of a locality. Find the median of the data. Monthly consumption

65-85 85-105

Number of consumers

4

105-125

125-145

13

20

5

[4] 145-165 165-185 185-205 14

8

4

Solution: Table for cumulative frequency is given below Monthly consumption (in unit)

Number of consumers  fi 

Cumulative frequency (cf)

65-85 85-105 105-125 125-145 145-165 165-185 185-205 Total

4 5 13 20 = f 14 8 4 N = 68

4 9 22 cf 42 56 64 68

Here, N = 68 Now,

N 68   34 2 2

Thus, 34 lies in the cumulative frequency 42. So, median class is 125-145. Here,

l = 125,

f = 20,

cf = 22

h = 145 – 125 = 20

N   2  cf   Median = l   h f    

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 34  22  = 125     20  20 

= 125 

12  20 20

= = 125+12 = 137 Hence, the median of given data is 137. 4.

For the following frequency distribution, find the value of f1 and f2, if mean is given to be 50.

[4]

Class interval

0-20

20-40

40-60

60-80

80-100

Total

Frequency

f1

28

f2

24

19

120

Solution: Let assumed mean A = 50 and h = 20 Table for step deviation and their product with corresponding frequency is given below Class interval

Frequency (fi) f1 28 f2 24 19

0-20 20-40 40-60 60-80 80-100 Total

120 = 71 + f1 +f2

Mid value (xi) 10 30 50 70 90

ui 

xi  A h –2 –1 0 1 2

ui x i

–2f1 –28 0 24 38

u x

i i

 34  2f1

We have, 71 + f1 + f2 =120  f1 +f2 =120 – 71

 f1  f2  49

..(i)

  ui x i  Mean, x  A  h     f  i  

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 50 = 50 + 20 ×

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34  2f1  120

 0

34  2f1  34  2f1  0 6

 f1 

34  17 2

On putting the value of f1 in Eq. (i), we get 17+f2 =49  2 =49 – 17= 32 Hence, 5.

f1 =17 and f2 =32

Given below is the frequency distribution of the heights of students in a school. Height (in cm) Number of students

160-162

163-165

166-168

169-171

172-174

15

118

142

127

18

[4]

Solution: Clearly, we have to find the mode of data. The given data is an inclusive series. So, we convert it into an exclusive form, as given below Class interval 159.5-162.5 162.5-165.5 165.5-168.5 168.5.171.5 171.5.174.5

Frequency 15 118 142 127 18

Clearly, the class interval 165.5-168.5 has maximum frequency, so it is the modal class.

 l = 1655, f1 = 142, f0 = 118, f2  127 and h  3

  f1  f0    Now, Mode, Mo  l  h   2f1  f0  f2    

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  {142  118}   = 165.5 + 3   2  142  118  127      

 3  24  = 1655 +    284  245   3  24  = 165.5     39 

= 165.5 

24 13

= 165.5+1.85 = 167.35 Thus, mode= 167.35 cm Hence, the average height of maximum number of students is 167.35 cm. 6.

The following table gives the height of trees. Height Less than 7 Less than 14 Less than 21 Less than 28 Less than 35 Less than 42 Less than 49 Less than 56

[4]

Number of trees 26 57 92 134 216 287 341 360

Draw less than ogive’ and ‘more than ogive’ Solution: Given distribution is cumulative frequency distribution of less than type. Height

Less than 7

Less than 14

Less than 21

Less than 38

Less than 35

Less than 42

Less than 49

Less than 56

Number of trees

26

57

92

134

216

287

341

360

Now, we mark the upper class limits along X- axis and cumulative frequency along y-axis. Thus, we plot the points (7, 26), (14, 57), (21, 92), (28, 134), (35, 216), (42, 287), (49, 341) and (56, 360). Join all these points by a free hand smooth curve to www.pioneermathematics.com S.C.O. - 326, Sector 40–D, CHD. Phone: 9815527721, 4617721

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obtain an ogive of less than type. The cumulative frequency distribution of more than type is Height 0-7 7-14 14-21 21-28 28-35 35-42 42-49 49-56

Frequency 26 57 – 26 = 31 92 – 57 = 35 134 – 92 = 92 216 – 134 = 82 287 – 216 = 71 341 – 287 = 54 360 – 341 = 19

More than type

Cumulative frequency

More than or equal to 0

360

More than or equal to 7

334

More than or equal to 14

303

More than or equal to 21

268

More than or equal to 28

226

More than or equal to 35

144

More than or equal to 42

73

More than or equal to 49

19

Now, we mark the lower class limits along X-axis and cumulative frequency along Y-axis. Thus, we plot the points (0, 360), (7,334), (14, 303), (21, 268), (28, 226), (35, 144), (42, 73) and (49, 19).

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Find the value of p, if the mean of the following distribution is 7.5. X Y

3 6

5 8

7 15

9 p

[4]

11 8

13 4

Solution: Calculation of Mean xi 3 5 7 9 11 13

fi 6 8 15 p 8 4 N   fi  41  p

fixi 18 40 105 9p 88 52  fixi  303  9p

We have,  fi  41  p,  fixi  303  9p

 Mean =  7.5 

 fi x i  fi

303  9p 41  p

 7.5 × (41 + p) = 303 + 9p  307.5 + 7.5p = 303 + 9p  9p – 7.5p = 307.5 – 303  1.5p = 4.5  p = 3

8.

Find the mean, mode and median for the following data.

[4]

Class interval

0-10

10-20

20-30

30-40

40-50

Frequency

8

16

36

34

6

Solution: Table for cumulative frequency and product with mid values and frequency is given below

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Class interval

Frequency ( fi )

Mid value  xi 

Cumulative frequency

0-10 10-20 20-30 30-40 40-50

8 16 36 34 6

5 15 25 35 45

8 24 60 94 100

 f  100 N = f1 =100 and  f x

Total Here,

x

fi x i

f

i



40 240 900 1190 270

f x

i

i i

fi x i

i i

 2640

= 2640

2640 = 26.4 100

Here, N = 100 

N 100   50 2 2

Cumulative frequency just greater than 50 is 60 and the corresponding class is 20-30.

 l =20, h = 30 – 20=10, f = 36 and cf = 24 N    cf   Median, M e  l  h  2  f     50  24   10  26   = 20  10    20   36   36   65  180  65 245    27.2 =  20    9  9 9 

and mode = 3 × median – 2 x mean = 3 × 27.2 – 2 × 26.4

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= 81.6-52.8 = 28.8 9.

A survey was conducted by a group of students as a part of their environment awareness programme, in which they collected the following data regarding the number of plants in 20 houses in a locality.

[4]

Number 0-2 2-4 4-6 6-8 8-10 of plants Number 1 2 1 5 6 of houses (i) Which mathematical concept will be used in the question?

10-12

12-14

2

3

(ii) Find the average (mean) number of plants per house. (iii) Which method did you use for finding the mean and why? (iv) What value (s) (quality) of the student is represented in the environment awareness programme.

[4]

Solution : (i) Mean (ii) 8.1 plants Calculation of Mean Class-Interval 0–2 2–4 4–6 6–8 8–10 10–12 12–14

Mid-values (xi) 1 3 5 7 9 11 13

Frequency (fi)

fixi

1 2 1 5 6 2 3

1 6 6 35 54 22 39

N   fi  20

i fi xi  162

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Mean =

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 fi x i  fi

 Mean =

162 20

= 8.1 (iii) We have used direct method, Because the numerical values of x and f are small. (iv) Making people aware of the importance of plant in nature. 10. A survey was conducted by a group of students regarding the number of family members (family size) in 30 families of a locality. The data obtained was as follows: Family size 1-5 5-9 9-13 Number of 5 4 3 families (i)Find the median family size for the above data.

13-17

17-21

Total

7

4

23

(ii) What value is indicated from the data?

[4]

Solution: (i) Computation of Median Class 1–5 5–9 9–13 13–17 17–21

Frequency 5 4 3 7 4

Cumulative frequency 5 9 12 19 23

We have N = 23



N 23   11.5 2 2

The cumulative frequency just greater than N/2 12 and the corresponding class is 9–13. Thus, 9–13 is the median class such that l = 9, f = 3, f = 9 and h = 4 www.pioneermathematics.com S.C.O. - 326, Sector 40–D, CHD. Phone: 9815527721, 4617721

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N F 2 h  Median = l  f

= 9

11.5  9 ×4 3

=9+

25 4 30

= 9

10  9  3.33 3

= 12.33

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