Important Questions for CBSE Class 10 Maths Chapter 14 - Statistics 3 Mark Question -CoolGyan

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CBSE Class 10 Maths Chapter-14 Statistics Important Questions

CBSE Class 10 Maths Important Questions Chapter 14 – Statistics

3 Mark Questions

1. The following table shows the weekly wages drawn by number of workers in a factory, find the median of the following data.

Weekly wages (in Rs.)	0-100	100-200	200-300	300-400	400-500
No. of workers	40	39	34	30	45

Ans. We have

Weekly wages (in Rs.)	No. of workers (f)	C.F
0-100	49	40
100-200	39	79
200-300	34	113
300-400	30	143
400-500	45	188

Now and this is in 200-300 class.
Median class= 200-300
Here,
We know that

2. Find the median of the following data:

Marks	Frequency
Less than 10	0
Less than 30	10
Less than 50	25
Less than 70	43
Less than 90	65
Less than 110	87
Less than 130	96
Less than 150	100

Ans. First of all we shall change cumulating series into simple series.
We have

X	F	C.F
0-10	0	0
10-30	10	10
30-50	15	25
50-70	18	43
70-90	22	65
90-110	22	87
110-130	9	96
130-150	4	100

Now , which lies in 70-90 class
Median class = 70-90
Here,
We know that Median, Me =

3. Find the median of the following data.

Wages (in rupees)	No. of workers
More than 150	Nil
More than 140	12
More than 130	27
More than 120	60
More than 110	105
More than 100	124
More than 90	141
More than 80	150

Ans. First of all we shall find simple frequencies.

Wages (in Rupees) (X)	No. of workers (F)	C.F
80-90	9	9
90-100	17	26
100-110	19	45
110-120	45	90
120-130	33	123
130-140	15	138
140-150	2	150

Now which lies in 110-120 class
Median class = 110-120
Here,
We know that Me =

4. Draw a less than Ogive for the following frequency distribution.

Marks	No. of students
0-4	4
4-8	6
8-12	10
12-16	8
16-20	4

Ans. We have

Marks	Frequency (F)	C.F
0-4	4	4
4-8	6	10
8-12	10	20
12-16	8	28
16-20	4	32

Upper class limits	4	8	12	16	20
Cumulative frequency	4	10	20	28	32
Plot the points	(4,4)	(8,10)	(12,20)	(16,28)	(20,32)

Join these points by a free hand curve. We get the required Ogivewhich is as follows:

5. Find the mean age in years from the frequency distribution given below:

Age (in yrs)	15-19	20-24	25-29	30-34	35-39	40-44	45-49	Total
Frequency	3	12	21	15	5	4	2	3

Ans. We have

Class-Interval	Mid value
15-19	17	3	-3	-9
20-24	22	13	-2	-26
25-29	27	21	-1	-21
30-34	32	15	0	0
35-39	37	5	1	5
40-44	42	4	2	8
45-49	47	2	3	6
Total

Let assumed mean ‘a’ = 32, Here h = 5
We know that Mean

6. Find the median of the following frequency distribution:

Wages (in Rs.)	200-300	300-400	400-500	500-600	600-700
No. of Laborers	3	5	20	10	6

Ans. We have

Wages (in Rs.)	No. of laborers (f)	C.F
200-300	3	3
300-400	5	8
400-500	20	28
500-600	10	38
600-700	6	44

Now and this lies in 400-500 class.
Median class = 400-500
Here,
We know that

7. The following tables gives production yield per hectare of wheat of 100 farms of village:

Production yield (in hr)	50-55	55-60	60-65	65-70	70-75	75-80
No. of farms	2	8	12	24	38	16

Change the distribution to a more than type distribution and draw its Ogive.
Ans. More than type Ogive

Production yield (Kg/ha)	C.F
More than or equal to 50	100
More than or equal to 55	98
More than or equal to 60	90
More than or equal to 65	78
More than or equal to 70	54
More than or equal to 75	16

Now, draw the Ogive by plotting the points (50,100), (55,98), (60,90), (65,78), (70,54), (75,16)

8. The A.M of the following distribution is 47. Determine the value of P.

Classes	0-20	20-40	40-60	60-80	80-100
Frequency	8	15	20	P	5

Ans. We have

Class Interval	Mid-value	Frequency
0-20	10	8	80
20-40	30	15	450
40-60	50	20	1000
60-80	70	P	70P
80-100	90	5	450

Since Mean,

Thus, P = 12

9. Thirty women were examined in a hospital by a doctor and the number of heart beats per minute were recorded and summarized as follows. Find the mean heart beats per minute for these women choosing a suitable method.

Number of heart beats per minute	No. of women
65-68	2
68-71	4
71-74	3
74-77	8
77-80	7
80-83	4
83-86	2

Ans. Let assumed mean ‘a’ = 75.5. We have

No. of heart beats per minute	No. of women ()	Class Mark i.e mid value
65-68	2	66.5	-3	-6
68-71	4	69.5	-2	-8
71-74	3	72.5	-1	-3
74-77	8	75.5=a	0	0
77-80	7	78.5	1	7
80-83	4	81.5	2	8
83-86	2	84.5	3	6

We know that
Mean [By step Deviation Method]

10. Following distribution shows the marks obtained by a class of 100 students:

Marks	10-20	20-30	30-40	40-50	50-60	60-70
Frequency	10	15	30	32	8	5

Change the distribution to less than type distribution and draw its Ogive.
Ans. Less than type Ogive

Marks	Marks	Frequency	Cumulative Frequency
10-20	Less than 20	10	10
20-30	Less than 30	15	25
30-40	Less than 40	30	55
40-50	Less than50	32	87
50-60	Less than 60	8	95
60-70	Less than 70	5	100

Now, draw the Ogiveby plotting (20,10), (30,25), (40,55), (50,87), (60,95), (70,100)

11. Following table shows the daily pocket allowances given to the children of a multi-story building. The mean of the pocket allowances is Rs. 18. Find out the missing frequency.

Class Interval	11-13	13-15	15-17	17-19	19-21	21-23	23-25
Frequency	3	6	9	13	?	5	4

Ans. Let the missing frequency = f, we have

Class interval		Mid-value		f_iu_i
11-13	3	12	-3	-9
13-15	6	14	-2	-12
15-17	9	16	-1	-9
17-19	13	18	0	0
19-21	f	20	1	F
21-32	5	22	2	10
23-25	4	24	3	12

Let assumed mean a = 18, Here h = 2
We know that mean

Hence, missing frequency = 8

12. A survey regarding the heights (in cm) of 51 girls of Class X of a school was conducted and the following data was obtained. Find the median height.

Height (in cm)	No. of girls
Less than 140	4
Less than 145	11
Less than 150	29
Less than 155	40
Less than 160	46
Less than 165	51

Ans. We have,

Class Intervals	Frequency (f)	C.F
Below 140	4	4
140-145	7	11
145-150	18	29
150-155	11	40
155-160	6	46
160-165	5	51

Here,which is in the class 145-150
Here,
Median

Median height of the girls = 149.03

13. Calculate the mean for the following distribution:

Class Interval	0-4	4-8	8-12	12-16	16-20	20-24	24-28	28-32
Frequency	2	5	8	16	14	10	8	3

Ans. By stepdeviation Method
Let assumed mean a = 14

Class interval	Mid-value	Frequency	Deviation	Product
0-4	2	2	-12	-24
4-8	6	5	-8	-40
8-12	10	8	-4	-32
12-16	14	16	0	0
16-20	18	14	4	56
20-24	22	10	8	80
24-28	26	8	12	96
28-32	30	3	16	48
Total

We know that Mean

14. The percentage of marks obtained by 100 students in an examination are given below:

Marks	30-35	35-40	40-45	45-50	50-55	55-60	60-65
Frequency	14	16	18	23	18	8	3

Determine the median percentage of marks.
Ans.

Class interval	Mid-value	Frequency	Deviation	Product
0-4	2	2	-12	-24
4-8	6	5	-8	-40
8-12	10	8	-4	-32
12-16	14	16	0	0
16-20	18	14	4	56
20-24	22	10	8	80
24-28	26	8	12	96
28-32	30	3	16	48
Total

We know that Mean

Therefore , which lies in the class 45-50
l₁(The lower limit of the median class) = 45
c(The cumulative frequency of the class preceding the median class) = 48
f(The frequency of the Median class)= 23
h(The class size) = 5
Median

So, the median percentage of marks is 45.4

15. Draw a less than Ogive for the following frequency distribution:

Marks	0-4	4-8	8-12	12-16	16-20
No. of students	4	6	10	8	4

Ans. We have

Marks	Frequency	C.F
0-4	4	4
4-8	6	10
8-12	10	20
12-16	8	28
16-20	4	32

Upper class limits	4	8	12	16	20
Cumulative Frequency	4	10	20	28	32
Plot the points	(4, 4)	(8, 10)	(12, 20)	(16, 28)	(20, 32)

Joint these points by a free hand curve; we get the required Ogive which is as follows:

16. The A.M of the following frequency distribution is 53. Find the value of P.

Classes	0-20	20-40	40-60	60-80	80-100
Frequency	12	15	32	P	13

Ans. We have

Class Interval	Mid-value	Frequency
0-20	10	12	120
20-40	30	15	450
40-60	50	32	1600
60-80	70	P	70P
80-100	90	13	1170

Since Mean

Thus, P = 28