Q.1     A survey was conducted by a group of students as a part of their environment awareness 
programmer, 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.

Which method did you use for finding the mean, and why ?
Sol.      Calculation of Mean

          Hence, mean x¯=1nΣfixi
                                   = 120×162=8.1 plants
We have used direct method because numerical values of xi and fi are small.

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Q.2     Consider the following distribution of daily wages of 50 workers of a factory.

          Find the mean daily wages of the workers of the factory by using an appropriate method.
Sol.      Let the assumed mean, A =150; class-interval, h = 20
so         ui=xi−Ah=xi−15020
We construct the table :

               x¯=A+hΣfiuiΣfi=150+20×−1250 = 150 – 4.8 = 145.2               
               Hence, mean = Rs 145.20 .

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Q.3     The following distribution shows the daily pocket allowance of children locality mean. The mean pocket allowance is Rs 18. Find the missing frequency f.

Sol.       Let the assumed mean, A = 16, class interval h = 2
             So       ui=xi−Ah=xi−162

We have, x¯ = 18, A = 16 and h = 2
Therefore, x¯=A+h(1NΣfiui)
⇒18=16+2(2f+24f+44)
⇒2=2(2f+24f+44)
⇒f+44=2f+24
⇒2f – f = 44 – 24
             ⇒f=20
             Hence, the missing frequency is 20.

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Q.4      Thirty women were examined in a hospital by a doctor and the number of heart beats per minute were recorded and summarised as follows. Find the mean heart beats per minute for these women, choosing a suitable method.

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Sol.       Let the assumed mean, A = 75.5, class interval
h = 3, so     ui=xi−Ah=xi−75.53

x¯=A+hΣfiuiΣfi=75.5+3×430
                = 75.5 + 0.4 = 75.9
Hence, the mean heart beats per minute for these women is 75.9.

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Q.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 mangoes according to
the number of boxes.

Find the mean number of mangoes kept in a packing box. Which method of finding the mean did you choose ?
Sol.      Here, the class intervals are formed by the exclusive method. If we make the series an inclusive one the mid-values remain  same. So there is no need to convert the series.
Let the assumed mean be A = 60 and h = 3
so      ui=xi−Ah=xi−603
Calculation of Mean

          Therefore x¯=A+hΣfiuiΣfi=60+3×−375400
                              = 60 – 2.8125 = 57.1875 = 57.19(nearly)
Hence, average number of mangoes per box = 57.19 (nearly)

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Q.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.
Sol.       Let the assumed mean, A = 225, class interval, h = 50
so      ui=xi−Ah=xi−22550
We construct the following table :
                x¯=A+hΣfiuiΣfi
= 225 +50×−725
= 225 – 14 = 211
Hence, the mean daily expenditure of food is Rs 211.

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Q.7     To find the concentration of SO2 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 :

           Find the mean concentration of SO2 in the air.
Sol.       Calculation of mean by direct method

               Hence, x¯=∑fixi∑fi = 2.9630 = 0.099 ppm

Q.8     A class teacher has the following absentee record of 40 students of a class for the whole term. Find the mean number of days a student was absent.

 

Sol.      Here, the class size varies, and xi ‘s are small. Let us apply the direct method here.

           
             Therefore, mean, x¯=ΣfixiΣfi  = 49940 = 12.475
             Hence,      mean = 12.48 days

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

Sol.       Let the assumed mean be A = 70 and h = 10
so      ui=xi−7010

Therefore,  x¯=A+h×ΣfiuiΣfi
                                    = 70 + 10 ×−235
                                    = 70 – 0.57 = 69.43
Thus, mean literacy rate is 69.43% .

 

Q.1      The following table shows the ages of the patients admitted in a hospital during a year :

             
          Find the mode and the mean of the data given above. Compare and interpret the two measures of central tendency.

Sol.      The class 35 – 45 has the maximum frequency, therefore, this is the modal class.

Here ℓ = 35, h = 10, f1 = 23, f0 = 21, f2 =14
Now, let us substitute these values in the formula
Mode =ℓ+(f1−f02f1−f0−f2)×h
                      = 35 + 23−212×23−21−14×10
                      =35+246−21−14×10
                       = 35 + 211×10
                      = 35 + 1.8 = 36.8
Calculation of Mean
Let assumed mean, A = 30, class interval h = 10
so      ui=xi−Ah=xi−3010

Therefore, x¯=A+h×ΣfiuiΣfi = 30 + 10 ×4380
                               = 30 + 5.37 = 35.37
Hence, Mode = 36.8 years, Means = 35.37 years.
Maximum number of patients admitted in the hospital are of age group 36.8 years (approx.) while on an
average the age of a patient admitted to the hospital is 35.37 years.

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Q.2      The following data gives the information on the observed lifetimes (in hours) of 225 electrical components :

            Determine the modal lifetimes of the components.
Sol.        The class 60 – 80 has the maximum frequency, therefore, this is the modal class.
Here ℓ = 60 , h = 20, f1 = 61, f0 = 52 and f2 = 38 .
Now, let us substitute these values in the formula
Mode = ℓ+(f1−f02f1−f0−f2)×h
                        =60+61−522×61−52−38×20
                         = 60+(61−52122−52−38)×20
                         = 60+932×20= 60 + 5.625 = 65.625
Thus, the modal lifetimes of the components is 65.625 hours.

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Q.3      The following data gives the distribution of total monthly household expenditure of 200 families of a village. Find the modal monthly expenditure of the families. Also, find the mean
monthly expenditure :

Sol.      The class 1500 – 2000 has the maximum frequency,therefore, this is the modal class.
Here ℓ = 1500, h = 500, f1=40,f0=24 and f2=33
Now, let us substitute these values in the formula
Mode = ℓ+(f1−f02f1−f0−f2)×h
                      =1500+(40−242×40−24−33)×500
                      = 1500+40−2480−24−33×500
                      = 1500+1623×500
                      = 1500 + 347.83 = 1847.83
Therefore, modal monthly expenditure = Rs. 1847.83
Let the assumed mean be A = 3250 and h = 500.

             We have, N = 200, A = 3250, h = 500 and Σfiui=−235
Therefore, mean = A+h(1NΣfiui)
                                     = 3250 + 500 ×−235200
                                     = 3250 – 587.5 = 2662.5 
            Hence, the average expenditure is Rs 2662.50.

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Q.4      The following distribution gives the state – wise teacher- student ratio in higher secondary schools of India. Find the mode and mean of this data. interpret the two measures.

Sol.      The class 30 – 35 has the maximum frequency, therefore, this is the modal class.
Here ℓ = 30, h = 5, f1 =10, f0 = 9 and f2 =3
Now, let us substitute these values in the formula
Mode = ℓ+(f1−f02f1−f0−f2)×h
                      =30+(10−92×10−9−3)×5
                      = 30 + 10−920−9−3×5
                      = 30 +18×5
                      = 30 + 0.625
                      = 30.625 = 30.6(approx.)
Calculation of mean :

               x¯=ΣfiuiΣfi
                   = 1022.535 = 29.2
Therefore, mean = 29.2
Thus, most states/U.T. have a student teacher ratio of 30.6 and on an average, the ration is 29.2.

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Q.1     The following distribution gives the daily income of 50 workers if a factory.

Convert the distribution above to a less than type cumulative frequency distribution and draw its ogive.
Sol.       Converting the given distribution to a less than type cumulative frequency distribution, we get

Let us now plot the points corresponding to the ordered pairs (120, 12), (140, 26), (160, 34), (180, 40) and (200, 50) on graph paper and join them by a free hand smooth curve.

               The curve thus obtained is  less than type ogive.

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Q.2     During the medical check-up of 35 students of a class, their weights were recorded as follows :

Draw a less than type ogive for the given data. Hence obtain the median weight from the graph and verify the result by using the formula.
Sol.       Here the values 38, 40, 42, 44, 46, 48, 50 and 52 are the upper limits of the respective class-intervals. To represent the data in the table graphically, we mark the upper limits on the class intervals on x-axis and
their corresponding cumulative frequencies on the y-axis, choosing a convenient scale. Let us now plot the points corresponding to the ordered pairs given by (38, 0), (40, 3), (42, 5), (44, 9),(46, 14), (48, 28), (50, 32) and (52, 35) on a graph paper an join them by a free hand smooth curve.

The curve thus obtained is  less than type ogive To locate n2=352= 17.5 on the y-axis. From this point draw a line parallel to the x-axis cutting the curve at a point. From this point, draw a perpendicular to the x-axis. The point of intersection of this perpendicular with the x-axis gives the median of the data. In this case it is 46.95. Let us make the following table in order to find mode by using the formula

The class 46 – 48 has the maximum frequency, therefore, this is modal class
Here, ℓ = 46, h = 2, f1 = 14, f0 = 5 and f2 = 4
      Now, Mode = ℓ+(f1−f02f1−f0−f2)×h
                        = 46+14−528−5−4×2 = 46+919×2
                        = 46 + 0.95 = 46.95
Thus, mode is verified.

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Q.3      The following tables gives production yield per hectare of wheat of 100 farms of a village.

Change the distribution to a more than type distribution and draw its ogive.
Sol.      Converting the given distribution to a more than type distribution, we get

.
Now, draw the ogive by plotting the points (50, 100), (55, 98), (60, 90), (65, 78), (70, 54) and (75, 16) on
this graph paper and join them by a free hand smooth curve.

The curve thus obtained is  more than type ogive.