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

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CBSE Class 10 Maths Chapter-14 Statistics Important Questions

## 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 fiui 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