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Statistics Ex-14.1 Interview Questions Answers

Question 1 :
A survey was conducted by a group of students as a part of their environment awareness programme, in which they collected the following data regarding the number of plants in 20 houses in a locality. Find the mean number of plants per house.

Answer 1 :

Number of Plants

0-2

2-4

4-6

6-8

8-10

10-12

12-14

Number of Houses

1

2

1

5

6

2

3

Which method did you use for finding the mean,and why?

Solution:

In order to find themean value, we will use direct method because the numerical value of fi and xi are small.

Find the midpoint ofthe given interval using the formula.

Midpoint (xi) = (upper limit + lower limit)/2

No. of plants

(Class interval)

No. of houses

Frequency (fi)

Mid-point (xi)

fixi

0-2

1

1

1

2-4

2

3

6

4-6

1

5

5

6-8

5

7

35

8-10

6

9

54

10-12

2

11

22

12-14

3

13

39

Sum f= 20

Sum fixi = 162

The formula to findthe mean is:

Mean = x̄ = ∑fxi /∑f

= 162/20

= 8.1

Therefore, the mean number of plants per house is 8.1

Question 2 :

Consider the following distribution of daily wages of 50 workers of afactory.

Answer 2 :

Daily wages (in Rs.)

100-120

120-140

140-160

160-180

180-200

Number of workers

12

14

8

6

10

Find the mean daily wages of the workers of the factory by using anappropriate method.


Solution:

Find the midpoint ofthe given interval using the formula.

Midpoint (xi) = (upper limit + lower limit)/2

In this case, thevalue of mid-point (xi) is very large, so let us assume the meanvalue, A = 150 and class interval is h = 20.

So, u= (xi – A)/h = ui  = (xi – 150)/20

Substitute and findthe values as follows:

Daily wages

(Class interval)

Number of workers

frequency (fi)

Mid-point (xi)

u= (xi – 150)/20

fiui

100-120

12

110

-2

-24

120-140

14

130

-1

-14

140-160

8

150

0

0

160-180

6

170

1

6

180-200

10

190

2

20

Total

Sum f= 50

Sum fiui = -12

So, the formula tofind out the mean is:

Mean = x̄ = A + h∑fiui /∑f=150+ (20 × -12/50) = 150 – 4.8 = 145.20

Thus, mean daily wageof the workers = Rs. 145.20

Question 3 :

The following distribution shows the daily pocket allowance of childrenof a locality. The mean pocket allowance is Rs 18. Find the missing frequencyf.

Answer 3 :


Daily Pocket Allowance(in c)

11-13

13-15

15-17

17-19

19-21

21-23

23-35

Number of children

7

6

9

13

f

5

4


Solution:

To find out themissing frequency, use the mean formula.

Here, the value ofmid-point (xi)  mean x̄ = 18

 

Class interval

Number of children (fi)

Mid-point (xi)

    fixi    

11-13

7

12

84

13-15

6

14

84

15-17

9

16

144

17-19

13

18 = A

234

19-21

f

20

20f

21-23

5

22

110

23-25

4

24

96

Total

fi = 44+f

Sum fixi = 752+20f

The mean formula is

Mean = x̄ = ∑fixi /∑f=(752+20f)/(44+f)

Now substitute thevalues and equate to find the missing frequency (f)

18 = (752+20f)/(44+f)

18(44+f) = (752+20f)

792+18f = 752+20f

792+18f = 752+20f

792 – 752 = 20f – 18f

40 = 2f

f = 20

So, the missingfrequency, f = 20.

Question 4 : Thirty women were examined in a hospital by a doctor and the number of heart beats per minute were recorded and summarized as follows.

Answer 4 :

 Find the mean heart beats per minutefor these women, choosing a suitable method.

Number of heart beats per minute

65-68

68-71

71-74

74-77

77-80

80-83

83-86

Number of women

2

4

3

8

7

4

2


Solution:

From the given data, let us assume the mean asA = 75.5

x= (Upper limit + Lowerlimit)/2

Class size (h) = 3

Now, find the uand fiui asfollows:

Class Interval

Number of women (fi)

Mid-point (xi)

ui = (xi – 75.5)/h

fiui

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

0

0

77-80

7

78.5

1

7

80-83

4

81.5

3

8

83-86

2

84.5

3

6

Sum fi= 30

Sum fiu= 4

Mean = x̄ = A + h∑fiui /∑f

= 75.5 + 3×(4/30)

= 75.5 + 4/10

= 75.5 + 0.4

= 75.9

Therefore, the mean heart beats per minute forthese women is 75.9

Question 5 : In a retail market, fruit vendors were selling mangoes kept in packing boxes. These boxes contained varying number of mangoes. 

Answer 5 :

The following was the distribution of mangoesaccording to the number of boxes.

Number of mangoes

50-52

53-55

56-58

59-61

62-64

Number of boxes

15

110

135

115

25

Find the mean numberof mangoes kept in a packing box. Which method of finding the mean did you choose?


Solution:

Since, the given data is not continuous so weadd 0.5 to the upper limit and subtract 0.45 from the lower limit as the gapbetween two intervals are 1

Here, assumed mean (A) = 57

Class size (h) = 3

Here, the step deviation is used because thefrequency values are big.

Class Interval

Number of boxes (fi)

Mid-point (xi)

di = xi – A

fidi

49.5-52.5

15

51

-6

90

52.5-55.5

110

54

-3

-330

55.5-58.5

135

57 = A

0

0

58.5-61.5

115

60

3

345

61.5-64.5

25

63

6

150

Sum fi = 400

Sum fidi = 75

The formula to find out the Mean is:

Mean = x̄ = A +h ∑fidi /∑f

= 57 + 3(75/400)

= 57 + 0.1875

= 57.19

Therefore, the mean number of mangoes kept ina packing box is 57.19

Question 6 : The table below shows the daily expenditure on food of 25 households in a locality. Find the mean daily expenditure on food by a suitable method.

Answer 6 :


Daily expenditure(in c)

100-150

150-200

200-250

250-300

300-350

Number of households

4

5

12

2

2


Solution:

Find the midpoint of the given interval usingthe formula.

Midpoint (xi) = (upper limit +lower limit)/2

Let is assume the mean (A) = 225

Class size (h) = 50

Class Interval

Number of households (fi)

Mid-point (xi)

di = xi – A

u= di/50

fiui

100-150

4

125

-100

-2

-8

150-200

5

175

-50

-1

-5

200-250

12

225

0

0

0

250-300

2

275

50

1

2

300-350

2

325

100

2

4

Sum fi = 25

Sum fiui = -7

Mean = x̄ = A +h∑fiui /∑fi

 =225+50(-7/25)

= 225-14

= 211

Therefore, the mean daily expenditure on foodis 211

Question 7 : To find out the concentration of SO_2 in the air (in parts per million, i.e., ppm), the data was collected for 30 localities in a certain city and is presented below:

Answer 7 :


Concentration of SO2 ( in ppm)

Frequency

0.00 – 0.04

4

0.04 – 0.08

9

0.08 – 0.12

9

0.12 – 0.16

2

0.16 – 0.20

4

0.20 – 0.24

2

Find the meanconcentration of SO2 in the air.


Solution:

To find out the mean, first find the midpointof the given frequencies as follows:

Concentration of SO(in ppm)

Frequency (fi)

Mid-point (xi)

fixi

0.00-0.04

4

0.02

0.08

0.04-0.08

9

0.06

0.54

0.08-0.12

9

0.10

0.90

0.12-0.16

2

0.14

0.28

0.16-0.20

4

0.18

0.72

0.20-0.24

2

0.20

0.40

Total

Sum fi = 30

Sum (fixi) = 2.96

The formula to find out the mean is

Mean = x̄ = ∑fixi /∑fi

= 2.96/30

= 0.099 ppm

Therefore, the mean concentration of SO2 inair is 0.099 ppm.

Question 8 :

A class teacher hasthe following absentee record of 40 students of a class for the whole
term. Find the mean number of days a student was absent.

Answer 8 :


Number of days

0-6

6-10

10-14

14-20

20-28

28-38

38-40

Number of students

11

10

7

4

4

3

1


Solution:

Find the midpoint of the given interval usingthe formula.

Midpoint (xi) = (upper limit +lower limit)/2

Class interval

Frequency (fi)

Mid-point (xi)

fixi

0-6

11

3

33

6-10

10

8

80

10-14

7

12

84

14-20

4

17

68

20-28

4

24

96

28-38

3

33

99

38-40

1

39

39

Sum fi = 40

Sum fixi = 499

The mean formula is,

Mean = x̄ = ∑fixi /∑fi

= 499/40

= 12.48 days

Therefore, the mean number of days a studentwas absent = 12.48.


Question 9 : The following table gives the literacy rate (in percentage) of 35 cities. Find the mean
literacy rate.

Answer 9 :


Literacy rate (in %)

45-55

55-65

65-75

75-85

85-98

Number of cities

3

10

11

8

3


Solution:

Find the midpoint of the given interval usingthe formula.

Midpoint (xi) = (upper limit +lower limit)/2

In this case, the value of mid-point (xi)is very large, so let us assume the mean value, A = 70 and class interval is h= 10.

So, u= (xi-A)/h =u= (xi-70)/10

Substitute and find the values as follows:

Class Interval

Frequency (fi)

(xi)

di = xi – a

ui = di/h

fiui

45-55

3

50

-20

-2

-6

55-65

10

60

-10

-1

-10

65-75

11

70

0

0

0

75-85

8

80

10

1

8

85-95

3

90

20

2

6

Sum fi  = 35

Sum fiui  = -2

So, Mean = x̄ = A+(∑fiui /∑fi)×h

= 70+(-2/35)×10

= 69.42

Therefore, the mean literacy part = 69.42


Selected

 

Statistics Ex-14.1 Contributors

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