NCERT Class 10 Chapter 14 Statistics CBSE Board Sample Problems Long Answer (For CBSE, ICSE, IAS, NET, NRA 2022)

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Question

The lengths of 50 leaves of a plant are measured correct to the nearest millimeter and the data obtained is represented in the following table

The Nearest Millimeter and the Data Obtained is Represented
Length (in mm)109 - 117118 - 126127 - 135136 - 144145 - 153154 - 162163 - 171
No. of leaves461413643

Find the mean length of the leaves.

Solution

The Nearest Millimeter and the Data Obtained is Represented
Class Interval
108.5 - 117.5

117.5 - 126.5

126.5 - 135.5

135.5 - 144.5

144.5 - 153.5

153.5 - 162.5

162.5 - 171.5

113

122

131

140

149

158

167

4

6

14

13

6

4

3

-3

-2

-1

0

1

2

3

-12

-12

-14

0

6

8

9

Total

Given frequency distribution is not continuous. So first we have to make it continuous.

Here assumed mean (a) = 140; class size = 9

Now,

Hence, mean length of the leaves -137.30mm.

Question

Draw a ‘less than type’ ogive for the following frequency distribution.

Draw a ‘Less Than Type’ Ogive
Class15 - 2020 - 2525 - 3030 - 3535 - 4040 - 45
Frequency13183125155

Solution

Draw a ‘Less Than Type’ Ogive
Draw a ‘Less Than Type’ Ogive
ClassFrequency
Less than 20

Less than 25

Less than 30

Less than 35

Less than 40

Less than 45

13

Question

In a retail market, fruit vendor were selling mangoes in packing boxes. These boxes contained varying number or mangoes. The following was the distribution.

In a Retail Market, Fruit Vendor Were Selling Mangoes in Packing Boxes
No. of mangoes50 - 5253 - 5556 - 5859 - 6162 - 64
No. of boxes1511013511525

Find the mean and median number of mangoes kept in a packing box.

Solution

In a Retail Market, Fruit Vendor Were Selling Mangoes in Packing Boxes
Class intervalMid value ()

(where a = 57)

49.5 - 52.5

52.5 - 55.5

55.5 - 58.5

58.5 - 61.5

61.5 - 64.5

51

54

57 = a

60

63

-2

-1

0

1

2

15

110

135

115

25

-30

-110

0

115

50

15

125

260

375

400

Mean

For median: Median

Here

Median class is 55.5 - 58.5

Median

Question

The following table gives production yield of rice per hectare in some farms of a village:

Production Yield (In Kg/Hectare)
Production yield (in kg/hectare)10 - 2020 - 3030 - 4040 - 5050 - 60
No. of farms3912206

Draw a more than type ′ ogive. Also, find median from the curve.

Solution

Production Yield (In Kg/Hectare)
Production yield (in kg/hectare)

Class interval

FrequencyProduction yield more than or eual to
10 - 20

20 - 30

30 - 40

40 - 50

50 - 60

3

9

12

20

6

10

20

30

40

50

50

47

38

26

6

Frequency

Question

Draw a More Than Type Ogive and from It Find Median
Life time (in hours)More than or equal to 240More than or equal to 280More than or equal to 320More than or equal to 360More than or equal to 400More than or equal to 440More than or equal to 480
Number of bulbs100958777472210

Draw a more than type ogive and from it find median. Verify it by actual calculations.

Solution

Draw a More Than Type Ogive and from It Find Median
More than or equal toNumber of bulbs
240

280

320

360

400

440

480

100

95

87

77

47

22

10

Draw a More Than Type Ogive and from It Find Median

Median from curve is 396.

Median from Curve is 396
Class intervalFrequencyCumulative frequency
240 - 280

280 - 320

320 - 360

360 - 400

400 - 440

440 - 480

480 - 520

5

8

10

30

25

12

10

5

13

23

53

78

90

100

Median class
100

Median

Question

Change the following distribution to a ‘more than type’ distribution. Hence draw the ‘more than type’ ogive for this distribution.

Change the Following Distribution to a ‘More Than Type’ Distribution
Class Interval20 - 3030 - 4040 - 5050 - 6060 - 7070 - 8080 - 90
Frequency108122462515

Solution

Class Interval Cumulative Frequency
Class IntervalCumulative Frequency
More than or equal to 20100
More than or equal to 3090
More than or equal to 4082
More than or equal to 5070
More than or equal to 6046
More than or equal to 7040
More than or equal to 8015

Plot of points (20.100) , (30.90) , (40,82) , (50.70) , (60,46) , (70.40) and (80.15)

Join the points to get a curve

Question

The mean of the following distribution is 18. Find the frequency f of the class 19 — 21.

Find the Frequency F of the Class 19 — 21
Class11 - 1313 - 1515 - 1717 - 1919 - 2121 - 2323 - 25
Frequency36913f54

Solution

Find the Frequency F of the Class 19 — 21
Classxf
11 - 1312336
13 - 1514684
15 - 17169144
17 - 191813234
19 - 2120f20f
21 - 23225110
23 - 2524496

Mean

Question

If the median of the following frequency distribution is 32.5. Find the values of and .

If the Median of the Following Frequency Distribution is 32
Class0 - 1010 - 2020 - 3030 - 4040 - 5050 - 6070 - 80Total
Frequency59123240

Solution

ClassFrequency Cumulative Frequency
ClassFrequencyCumulative Frequency
0 - 10
10 - 205
20 - 309
30 - 4012
40 - 50
50 - 603
60 - 702
40

Median median class is 30 - 40.

Now

Also,

Question

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

Daily Income of 50 Workers of a Factory
Daily Income400 - 420400 - 420400 - 420400 - 420400 - 420
Number of workers1212121212

Convert this distribution to less than a type of cumulative frequency distribution and draw its ogive.

Solution

Daily Income Number of Workers Cumulative Frequen
Daily IncomeNumber of workersCumulative Frequency
400 - 4201212
420 - 4401426
440 - 460834
460 - 480640
480 - 5001050

Correct Table

Drawing an ogive with coordinates

Question

Calculate Mean of the following distribution

Calculate Mean of the Following Distribution
C. I10 - 1920 - 2930 - 3940 - 4950 - 5960 - 6970 - 7980 - 8990 - 99
Frequency59131716111298

Solution

Calculate Mean of the Following Distribution
10 - 1914.55-40-420
20 - 2924.59-30-3-27
30 - 3934.513-20-2-26
50 - 5044.517-10-1-17
40 - 4954.516000
60 - 6964.51110111
70 - 7974.51220224
80 - 8984.5930327
90 - 9994.5840432
100

Question

If the mean of x and is M, find the mean of and

Solution

Mean of and is