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# Statistics Ex-14.1 Interview Questions Answers

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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.

 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 fi = 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.

 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) ui = (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 fi = 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.

 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.

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 fiui = 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.

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.

 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 ui = 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:

 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 SO2 (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.

 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.

 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

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