Math Class X 1 Question Bank

Question Bank Statistics

1. Construct a histogram for the following frequency distribution.

Class-interval 5 – 12 13 – 20 21 – 28 29 – 36 37 – 44 45 – 52 Frequency 4 12 26 15 6 18

Solution. The given frequency distribution is in inclusive form, so, we first convert it to the exclusive form.

Adjustment factor = 13 – 122

= 0.5

∴ We subtract 0.5 from each lower limit and add 0.5 to each upper limit.

The required histogram is shown alongside :

2. Draw an ogive for the following distribution.

Marks 0-10 10-20 20-30 30-40 40-50 50-60 60-70 70-80 80-90 90-100No. of Students

6 9 11 23 28 32 21 14 4 2

Class-interval 4.5–12.5 12.5–20.5 20.5–28.5 28.5–36.5 36.5–44.5 44.5–52.5 Frequency 4 12 26 15 6 18

Math Class X 2 Question Bank

Solution. We first prepare the cumulative frequency table as under.

Marks No. of students Cumulative frequency 0 – 10

10 – 20 20 – 30 30 – 40 40 – 50 50 – 60 60 – 70 70 – 80 80 – 90 90 – 100

6 9

11 23 28 32 21 14 4 2

6 15 26 49 77

109 130 144 148 150

Now, plot the following points on the graph paper. (0, 0), (10, 6), (20, 15), (30, 26), (40, 49), (50, 77), (60, 109), (70,

130), (80, 144), (90, 148), (100, 150)

Math Class X 3 Question Bank

3. Draw a more than cumulative frequency curve for the following

data: Class interval 0 – 10 10 – 20 20 – 30 30 – 40 40 – 50 50 – 60 60 – 70 Frequency 4 5 11 14 11 10 6

Solution : We first prepare a more than frequency table. Class More

than 0 More

than 10 More

than 20More

than 30More

than 40 More

than 50More

than 60Cumulative frequency

61 57 52 41 27 16 6

The more than cumulative frequency curve is shown below :

Math Class X 4 Question Bank

4. The marks obtained by a set of students in an examination are given

below : Marks 5 10 15 20 25 30 No. of student 6 4 6 12 x 4

Given that the mean mark of the set is 18, calculate the numerical value of x.

Solution. Marks x No. of students f f × x

5 10 15 20 25 30

6 4 6

12 x 4

30 40 90

240 25x 120

Total 32 + x 520 + 25x

Mean = 520 + 25 18 = 35

fx xf x

⇒+

∑∑

⇒ 576 + 18x = 520 + 25x

⇒ 7x = 576 – 520 = 56 ⇒ x = 8

Math Class X 5 Question Bank

5. Using short-cut method, find the mean height from the following

frequency distribution : Height (in cm) 58 60 62 65 66 68 No .of plants 15 14 20 18 8 5

Solution. Let the assumed mean, A be 65. Height (in cm)

xi No. of plants

fi di = xi – A = xi – 65

fidi

58 60 62 65 66 68

15 14 20 18 8 5

–7 –5 –3 0 1 3

–105 –70 –60 0 8

15 Total if∑ = 80 i if d∑ = –212

∴ Mean = A + i i

i

f df

∑∑

= 65 + –21280

⎛ ⎞⎜ ⎟⎝ ⎠

= 65 – 2.65 = 62.35

Hence, mean height = 62.35 cm.

Math Class X 6 Question Bank

6. Using step-deviation method, calculate the mean of the following

data : Marks 0-10 10-20 20-30 30-40 40-50 50-60 60-70 70-80 80-90 90-100No. of Students

2 8 15 21 4 52 37 21 15 5

Solution. Here, class size c = 10. Take assumed mean, A = 45. Marks Class-mark

xi No. of

students fi ui =

– Aixc

fiui

0 – 10 10 – 20 20 – 30 30 – 40 40 – 50 50 – 60 60 – 70 70 – 80 80 – 90 90 – 100

5 15 25 35 45 55 65 75 85 95

2 8

15 21 4

52 37 21 15 5

– 4 – 3 – 2 – 1 0 1 2 3 4 5

– 8 – 24 – 30 – 21

0 52 74 63 60 25

Total Σfi = 180 Σfiui = 191

∴ Mean = A + c × i i

i

f uf

∑∑

= 45 + 10 × 191180

= 45 + 10.61 = 55.61

Math Class X 7 Question Bank

7. The mean of the following frequency distribution is 56, but the frequencies f1 and f2 in classes 20 – 40 and 80 – 100 respectively are missing. Find the missing frequencies.

Class-interval

0 - 20 20 - 40 40 - 60 60 - 80 80 - 100 100 - 120 Total

Frequency 16 f1 25 16 f2 10 90 Solution. Let the assumed mean A be 50. Here, c = 20

Class-interval Class-mark xi

Frequency fi

ui = – Aixc

= – 5020

ix

fiui

0 – 20 20 – 40 40 – 60 60 – 80 80 – 100

100 – 120

10 30 50 70 90

110

16 f1 25 16 f2 10

–2 –1 0 1 2 3

–32 –f1 0

16 2f2 30

Σfi = 67 + fi + fi Σfiui = 14 – fi + 2f2

We have Σfi = 90 = 67 + f1 + f1 ⇒ f1 + f2 = 23 ... (i)

Now mean = A + c × i i

i

f uf

∑∑

⇒ 56 = 50 + 20 × 1 214 – + 90f f⎛ ⎞

⎜ ⎟⎝ ⎠

⇒ 6 = 2 1 214 – + 9f f⎛ ⎞

⎜ ⎟⎝ ⎠

⇒ 14 – f1 + 2f2 = 27 ⇒ 2f2 – f1 = 13 ... (ii) Adding (i) and (ii), we get 3f2 = 36 ⇒ f2 = 12 From (i), f1 = 23 – 12 = 11 Hence, f1 = 11 and f2 = 12.

Math Class X 8 Question Bank

8. The following table gives marks scored by students in an

examination.

Calculate the mean marks. Solution. We may prepare a frequency table as under :

Marks Class- marks xi

CumulativeFrequency

Frequency fi

ui = – Aixc

= – 22.55

ix

fi ui

0 – 5 5 – 10 10 – 15 15 – 20 20 – 25 25 – 30 30 – 35 35 – 40

2.5 7.5

12.5 17.5 22.5 27.5 32.5 37.5

3 10 25 49 65 73 78 80

3 7

15 24 16 8 5 2

–4 –3 –2 –1 0 1 2 3

–12 –21 –30 –24

0 8

10 6

Σfi = 80 Σfi ui = –63

Mean = A + c × i i

i

f uf

∑∑

= 22.5 + 5 × – 6380

⎛ ⎞⎜ ⎟⎝ ⎠

= 22.5 – 6316

⎛ ⎞⎜ ⎟⎝ ⎠

= 22.5 – 3.9375 = 18.5625 = 18.56.

Marks less than 5

less than 10

less than 15

less than 20

less than 25

less than 30

less than 35

less than 40

No. of students

3 10 25 49 65 73 78 80

Math Class X 9 Question Bank

9. Calculate the mean, the median and the mode of the following

numbers : 3, 1, 5, 6, 3, 4, 5, 3, 7, 2 Solution : Arranging the data in ascending order of their magnitude,

we have : 1, 2, 3, 3, 3, 4, 5, 5, 6, 7

(i) Mean = Sum of the observationsNo. of observations

1 + 2 + 3 + 3 + 3 + 4 + 5 + 5 + 6 + 7 39 =

10 10= = 3.9.

(ii) Here, number of observations, N = 10 (even)

∴ Median = 1 10 10th observations + + 1 th observations2 2 2

⎡ ⎤⎛ ⎞⎜ ⎟⎢ ⎥⎝ ⎠⎣ ⎦

= 1 7(3 + 4) = = 3.52 2

(iii) Since, 3 occurs maximum number of times. ∴ Mode = 3. 10. The marks scored by 19 students in a test are given below. 31, 22, 36, 27, 25, 26, 33, 24, 37, 32, 29, 28, 36, 27, 35, 35, 32, 26,

28 Find (i) median (ii) lower quartile (Q1) (iii) upper quartile (Q3) (iv)

inter-quartile range. Solution : Arranging the given data in ascending order of their

magnitudes, we have : 22, 24, 25, 26, 26, 27, 27, 28, 29, 29, 31, 32, 32, 33, 35, 35, 36, 36,

37. Here, number of observations, N = 19 (odd)

Math Class X 10 Question Bank

(i) Median = N + 1

2⎛ ⎞⎜ ⎟⎝ ⎠

th observation = 19 + 12

th observation

= 10th observation = 29

(ii) Lower quartile (Q1)= N + 14

⎛ ⎞⎜ ⎟⎝ ⎠

th observation = 19 + 14

th

observation = 5th observation = 26.

(iii) Upper quartile (Q3) = ( )3 N + 14

th observation

= ( )3 19 + 14

th observation

= 15th observation = 35. (iv) Inter-quartile range = Q3 – Q1 = 35 – 26 = 9. 11. From the following frequency distribution, calculate : (i) median (ii) lower quartile (iii) upper quartile (iv) semi-inter-

quartile range. Variate 13 15 18 20 22 24 25 Frequency 6 4 11 9 16 12 2

Solution : First we prepare a cumulative frequency table.

Variate Frequency Cumulative frequency 13 15 18 20 22 24 25

6 4

11 9

16 12 2

6 10 21 30 46 58 60

∴ N = 60 (even)

Math Class X 11 Question Bank

(i) Median = 1 N Nth observations + + 1 th observations2 2 2

⎡ ⎤⎛ ⎞⎜ ⎟⎢ ⎥⎝ ⎠⎣ ⎦

= 12

[30th observation + 31st observation]

= 12

(20 + 22) = 21.

(ii) Lower quartile (Q1) = N th4

observation = 60 th4

observation

= 15th observation = 18.

(iii) Upper quartile (Q3) = 360 th4

observation

= 45 th observation = 22.

(iv) Semi-inter quartile range = 3 1Q – Q 22 – 18 = 2 2

= 2.

12. The marks obtained by 120 students in a mathematics test are given below :

Marks 0-10 10-20 20-30 30-40 40-50 50-60 60-70 70-80 80-90 90-100No. of Students

5 9 16 22 26 18 11 6 4 3

Draw an ogive for the given distribution on a graph paper. Use a suitable scale for your ogive. Use your ogive to estimate.

(i) the median (ii) the lower quartile (iii) the number of students who obtained more than 75% in the test. (iv) the number of students who did not pass in the test if the pass

percentage was 40.

Math Class X 12 Question Bank

Solution. First, we prepare a cumulative frequency table :

Marks No. of students Cumulative frequency 0 – 10

10 – 20 20 – 30 30 – 40 40 – 50 50 – 60 60 – 70 70 – 80 80 – 90 90 – 100

5 9

16 22 26 18 11 6 4 3

5 14 30 52 78 96

107 113 117 120

Now, plot the points (0, 0), (10, 5), (20, 14), (30, 30), (40, 52), (50, 78), (60, 96), (70, 107), (80, 113), (90, 117), (100, 120).

Join these points by a free hand curve, which gives the required ogive. (as shown in the graph)

Here, N = 120.

(i) N2

= 60

Let A be a point on y-axis representing 60. Through, A, draw a horizontal line of meet the ogive at B.

Through B, draw BC ⊥ x-axis. The abscissa of C represents 43.5. Hence, median = 43.5.

Math Class X 13 Question Bank

(ii) N2

= 30

Let D be a point on the y-axis representing 30. Through A, draw a horizontal line to meet the ogive at E. Through E, draw EF ⊥ x-axis. The abscissa of F is 30.

Hence, lower quartile = 30. (iii) Let G be a point on the x-axis representing 75. Through G, draw

a vertical line to meet the ogive at H. Through H, draw a horizontal line to meet the y-axis at I. The

ordinate of y is 110. Hence, the required number of students is 120 – 110 = 10.

Math Class X 14 Question Bank

(iv) Let J be a point on the x-axis representing 40. Through J, draw a

vertical line to meet the ogive at K. Through K, draw a horizontal line to meet the y-axis at L. The ordinate of L is 52.

Hence, required number of students = 52. 13. The daily profits in rupees of 100 shops in a departmental store are

distributed as follows :

Profit per shop (in Rs)

0 - 100 100 - 200 200 - 300 300 - 400 400 - 500 500 - 600

No. of shops 12 18 27 20 17 6 Draw a histogram of the above data on a graph paper and hence

estimate the mode. Solution. The required histogram is shown below :

Mark the upper corners of the highest rectangle and the corners of

the adjacent rectangles as A, B, C and D as shown in the graph. Join AC and BD to intersect at P. Draw PM ⊥ x-axis. The abscissa

of M is 255. Hence, mode = 255