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Question: The following distribution gives the state-wise teacher-student ratio in higher secondary schools of...

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.

Number of students per teacherNumber of states/U.T (fi)\left( {{f_i}} \right)
152015 - 2033
202520 - 2588
253025 - 3099
303530 - 351010
354035 - 4033
404540 - 4500
455045 - 5000
505550 - 5522
Explanation

Solution

Here, we are asked to calculate the mode and median for the given frequency distribution table. To find the mode, we need to obtain the modal class. Then we shall substitute the calculated values in the formula of mode. To find the mean, we need to create a table containing the values that we need to apply in the formula.

Formula to be used:
Mode=l+[f1f02f1f0f2]×hMode = l + \left[ {\dfrac{{{f_1} - {f_0}}}{{2{f_1} - {f_0} - {f_2}}}} \right] \times h
Here ll is the lower limit, hh is the class size, f1{f_1} is the frequency of the modal class, f0{f_0} is the frequency preceding the modal class, and f2{f_2} is the frequency succeeding the modal class.
xi=Upper limit+Lower limit2{x_i} = \dfrac{{Upper{\text{ }}limit + Lower{\text{ }}limit}}{2}
Mean,x=A+(fidifi×h)Mean,\overline x = A + \left( {\dfrac{{\sum {{f_i}{d_i}} }}{{\sum {{f_i}} }} \times h} \right)
Here AA is the assumed mean.
fi\sum {{f_i}} is the sum of the frequency and fidi\sum {{f_i}} {d_i} is the sum of fidi{f_i}{d_i}

Complete step-by-step answer:
Here we are asked to find the mode and mean of the given distribution.
First, we shall calculate the mode.
a) The modal class is the class that contains the highest frequency.
From the given data, we can note that the highest frequency is 1010 that belongs to the class interval 303530 - 35
Hence, the modal class is 303530 - 35
Let hh be the size of the class.
Here, for the given frequency distribution, h=5h = 5
Let llbe the lower limit of the median class.
Hence, we have l=30l = 30.
Let f1{f_1}be the frequency of the modal class.
Therefore, we have f1=10{f_1} = 10
Let us assume that f0{f_0}be the frequency of class above the calculated modal class.
Then, f0=9{f_0} = 9 is the required frequency of the class above the calculated modal class.
Let us assume that f2{f_2}be the frequency of the class below the obtained modal class.
Then, f2=3{f_2} = 3 is the required frequency of the class below the obtained modal class.
Now, we shall apply all the values in the formula Mode=l+[f1f02f1f0f2]×hMode = l + \left[ {\dfrac{{{f_1} - {f_0}}}{{2{f_1} - {f_0} - {f_2}}}} \right] \times h
Thus, we have Mode=30+[1092×1093]×5Mode = 30 + \left[ {\dfrac{{10 - 9}}{{2 \times 10 - 9 - 3}}} \right] \times 5
=30+52012= 30 + \dfrac{5}{{20 - 12}}
=30+58= 30 + \dfrac{5}{8}
=30.6= 30.6
Thus, we have Mode=30.6Mode = 30.6
Therefore, we can able to conclude that most of the states/U.T have a teacher-student ratio as 30.630.6
b) Now we need to calculate the mean.
We need to create a frequency distribution table containing assumed mean AA, classmark xi{x_i}, di{d_i} andfidi{f_i}{d_i}
To find the classmark, we shall apply xi=Upper limit+Lower limit2{x_i} = \dfrac{{Upper{\text{ }}limit + Lower{\text{ }}limit}}{2}

Number of students per teacherxi{x_i}Number of states/U.T (fi)\left( {{f_i}} \right)di=xiAh{d_i} = \dfrac{{{x_i} - A}}{h}( di=xi32.55{d_i} = \dfrac{{{x_i} - 32.5}}{5} )fidi{f_i}{d_i}
152015 - 2017.517.53317.532.55=155=3\dfrac{{17.5 - 32.5}}{5} = \dfrac{{ - 15}}{5} = - 39 - 9
202520 - 2522.522.58822.532.55=105=2\dfrac{{22.5 - 32.5}}{5} = \dfrac{{ - 10}}{5} = - 216 - 16
253025 - 3027.527.59927.532.55=55=1\dfrac{{27.5 - 32.5}}{5} = \dfrac{{ - 5}}{5} = - 19 - 9
303530 - 3532.532.5101032.532.55=05=0\dfrac{{32.5 - 32.5}}{5} = \dfrac{0}{5} = 000
354035 - 4037.537.53337.532.55=55=1\dfrac{{37.5 - 32.5}}{5} = \dfrac{5}{5} = 133
404540 - 4542.542.50042.532.55=105=2\dfrac{{42.5 - 32.5}}{5} = \dfrac{{10}}{5} = 200
455045 - 5047.547.50047.532.55=155=3\dfrac{{47.5 - 32.5}}{5} = \dfrac{{15}}{5} = 300
505550 - 5552.552.52252.532.55=205=4\dfrac{{52.5 - 32.5}}{5} = \dfrac{{20}}{5} = 488
Total3+8+9+10+3+2=353 + 8 + 9 + 10 + 3 + 2 = 359169+3+8=23 - 9 - 16 - 9 + 3 + 8 = - 23

Here the assumed mean is A=32.5A = 32.5
Now, we shall apply the formula Mean,x=A+(fidifi×h)Mean,\overline x = A + \left( {\dfrac{{\sum {{f_i}{d_i}} }}{{\sum {{f_i}} }} \times h} \right)
Mean,x=32.5+(2335×5)Mean,\overline x = 32.5 + \left( {\dfrac{{ - 23}}{{35}} \times 5} \right) (Here fi\sum {{f_i}} is the sum of the frequency and fidi\sum {{f_i}} {d_i} is the sum of fidi{f_i}{d_i} )
Mean,x=32.5+237\Rightarrow Mean,\overline x = 32.5 + \dfrac{{ - 23}}{7}
Mean,x=32.53.28\Rightarrow Mean,\overline x = 32.5 - 3.28
Mean,x=29.22\Rightarrow Mean,\overline x = 29.22
Hence, the required mean is 29.229.2
Therefore, we can conclude that the teacher-student ratio on average was 29.229.2

Note: We can able to conclude that most of the states/U.T have a teacher-student ratio as 30.630.6 and we can conclude that the teacher-student ratio on an average was 29.229.2To find the mean and mode, we need to know the required formula. Also, we know how to create a frequency distribution table for calculating the mean.