Math Assignment Ch-14 Class X

Mathematics Assignment

Class X, Chapter 14

Extra questions of statistics class 10, chapter 14, Important and board based questions based on chapter 14 of class 10

Chapter: 14(Statistics)
Formulas used to solve the Assignment on statistics

Methods of finding mean in grouped data:-

1. Direct method : $\overline{x}=\frac{\sum f_{i}x_{i}}{\sum f_{i}}$

2. Assumed mean method : $\overline{x}=a+\frac{\sum f_{i}d_{i}}{\sum f_{i}}$

3. Step-Deviation method : $\overline{x}=a+h\frac{\sum f_{i}u_{i}}{\sum f_{i}}$

Where,
a	Assumed mean
di	x_i-a
ui	d_i/h or (x_i-a)/h
h	height of C.I.

Class mark= $\frac{Upper\: Limit+Lower\: Limit}{2}$

Range = Highest value - Lowest value

Class size = Upper limit - Lower limit

MODE of Data in Statistics

Mode: = $l+\frac{f_{1}-f_{0}}{2f_{1}-f_{0}-f_{2}}\times h$

Where l = lower limit of the modal class,

h = size of the class interval

f₁ = frequency of the modal class

f₀ = frequency of the class preceding the modal class,

f₂ = frequency of the class succeeding the modal class.

Median of Data in Statistics

Median: = $l+\left ( \frac{\frac{N}{2}-Cf}{f} \right )\times h$

Median Class:- Class corresponding to the $Cf\geq \frac{N}{2}$ is called median class

Where l= lower limit of median class

N = Sum of all observations

f = frequency of the median class.

Cf= cumulative frequency of class preceding the median class

h= height of median class

Empirical Formula of Data in Statistics

Empirical Formula: Relation between mean, mode , median;-

Mode=3 Median - 2 Mean

Empirical formula does not provide the same result in different situations. So, students should use it only when asked otherwise avoid it.

NOTE :- For finding median, height of class interval need not to be equal or class interval may be irregular.

* If median class is the first class –interval then cumulative frequency (Cf) of the preceding class interval should be taken as zero.

* For finding Mean class interval need not to be continuous.

* For finding Mode and Median class interval should be continuous.

Assignment on Statistics
Solve the following questions

1. Find the mean of the following data :- Ans:- 3.52

Absent Students	No. of days
0	1
1	4
2	10
3	50
4	34
5	15
6	4
7	2
Total	120

2. Find the mean by using all 3 methods separately:- Ans:-25.86

Marks	No. of Students
0-10	20
10-20	24
20-30	40
30-40	36
40-50	20
Total

3. If mean = 6 then find p. Ans:- 7

X	Y
2	3
4	2
6	3
10	1
P+5	2

4. If mean =56, then find f1 & f2 . Ans:- f1=11& f2=12

Class Interval	Frequency
0-20	16
20-40	f1
40-60	25
60-80	16
80-100	f2
100-120	10
Total	90

5. If mean=50 then find f1 & f2. Ans:- f1 =28 & f2 = 24

Class Interval	Frequency
0-20	17
20-40	f1
40-60	32
60-80	f2
80-100	19
Total	120

6. If mean =18 then find p. Ans:- 1,-3

X	Y
13	8
15	2
17	3
19	4
20+p	5p
23	6

7. Find the mean by using all 3 methods. Ans:- 165.502

Class Interval	Frequency
110-120	2
120-130	5
130-140	11
140-150	21
150-160	26
160-170	34
170-180	36
180-190	28
190-200	16

8.Find mean. Ans: 12.38

Do we use here step-deviation method ? Why or Why not

Class Interval	Frequency
0-6	11
6-10	10
10-14	7
14-20	4
20-28	4
28-38	3
38-40	1

9. Find arithmetic mean by using suitable method. Ans:-196.8

Class Interval	Frequency
0-80	22
80-160	35
160-240	44
240-320	25
320-400	24

10. For the given data find mean & mode.

Ans : Mean=26 & Mode=22.85

Class Interval	Frequency
More than 0	60
More than 10	56
More than 20	40
More than 30	20
More than 40	10
More than 50	3

11. Find mode. Ans:-18

Class Interval	Frequency
0-5	3
5-10	7
10-15	15
15-20	30
20-25	20
25-30	10
30-35	5

12. Find median. Ans: 40

Class Interval	Frequency
0-10	5
10-30	15
30-60	30
60-80	8
80-90	2

13. Find mode. Ans:-63.75

Height	No. of Plants
More than 30	34
More than40	30
More than 50	27
More than 60	19
More than 70	8
More than 80	2

14. Find median. Ans:-29.42

Class Interval	Frequency
1-10	10
11-20	21
21-30	51
31-40	45
41-50	26

15. If median=46 then find x & y. Ans: x = 33.5 & y = 45.5

Class Interval	Frequency
10-20	12
20-30	30
30-40	X
40-50	65
50-60	Y
60-70	25
70-80	18
Total	229

16. If median =35 then find f1 & f2.

Ans: F1=35 & f2= 25

Class Interval	Frequency
0-10	10
10-20	20
20-30	f1
30-40	40
40-50	f2
50-60	25
60-70	15
Total	170

17. Find median ages. Ans:- 35.76

Ages	No. of men
Less than 20	2
Less than 25	6
Less than 30	24
Less than 35	45
Less than 40	78
Less than 45	89
Less than 50	92
Less than 55	98
Less than 60	100

18. Calculate Mean, mode, median.

Ans : Median=52 , Mean= 51.63 & Mode = 53.33

Marks	No of Students
More than 0	80
More than 10	76
More than 20	71
More than 30	65
More than 40	57
More than 50	43
More than 60	28
More than 70	15
More than 80	10
More than 90	8
More than 100	0

19. Find median by using less than ogive & actual calculation.

Ans:- 27.5

Marks	No. of Students
Less than 10	4
Less than 20	10
Less than 30	30
Less than 40	40
Less than 50	47
Less than 60	50

20. Draw less than ogive & more than type ogive and find median. Ans:-34.33

Class Interval	Frequency
0-10	4
10-20	8
20-30	11
30-40	15
40-50	12
50-60	6
60-70	3

21. Mode of the following data is 45, find x and y ,given n= 50

Ans x=10, y = 2

Class Interval	Frequency
10-20	4
20-30	8
30-40	X
40-50	12
50-60	10
60-70	4
70-80	Y

22. Find the unknown entries a, b, c, d, e, f in the following distribution if total no. of students is 50

class interval	frequency	c.f.
150-155	12	a
155-160	b	25
160-165	10	c
165-170	d	43
170-175	e	48
175-180	2	f

23. If mode = 24.5 and mean is 29.75 then find median by using empirical formula. [Ans: 28]

24. Find the mode and median of the data. Ans[Mode= 9.56, Median= 12.19]

Class Interval	Frequency
5-8	40
8-11	90
11-14	44
14-17	58
17-20	53
20-23	10

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