Question
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 teacher | Number of states/U.T (fi) |
---|---|
15−20 | 3 |
20−25 | 8 |
25−30 | 9 |
30−35 | 10 |
35−40 | 3 |
40−45 | 0 |
45−50 | 0 |
50−55 | 2 |
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+[2f1−f0−f2f1−f0]×h
Here l is the lower limit, h is the class size, f1 is the frequency of the modal class, f0 is the frequency preceding the modal class, and f2 is the frequency succeeding the modal class.
xi=2Upper limit+Lower limit
Mean,x=A+(∑fi∑fidi×h)
Here A is the assumed mean.
∑fi is the sum of the frequency and ∑fidi is the sum of fidi
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 10 that belongs to the class interval 30−35
Hence, the modal class is 30−35
Let h be the size of the class.
Here, for the given frequency distribution, h=5
Let lbe the lower limit of the median class.
Hence, we have l=30.
Let f1be the frequency of the modal class.
Therefore, we have f1=10
Let us assume that f0be the frequency of class above the calculated modal class.
Then, f0=9 is the required frequency of the class above the calculated modal class.
Let us assume that f2be the frequency of the class below the obtained modal class.
Then, f2=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+[2f1−f0−f2f1−f0]×h
Thus, we have Mode=30+[2×10−9−310−9]×5
=30+20−125
=30+85
=30.6
Thus, we have Mode=30.6
Therefore, we can able to conclude that most of the states/U.T have a teacher-student ratio as 30.6
b) Now we need to calculate the mean.
We need to create a frequency distribution table containing assumed mean A, classmark xi, di andfidi
To find the classmark, we shall apply xi=2Upper limit+Lower limit
Number of students per teacher | xi | Number of states/U.T (fi) | di=hxi−A( di=5xi−32.5 ) | fidi |
---|---|---|---|---|
15−20 | 17.5 | 3 | 517.5−32.5=5−15=−3 | −9 |
20−25 | 22.5 | 8 | 522.5−32.5=5−10=−2 | −16 |
25−30 | 27.5 | 9 | 527.5−32.5=5−5=−1 | −9 |
30−35 | 32.5 | 10 | 532.5−32.5=50=0 | 0 |
35−40 | 37.5 | 3 | 537.5−32.5=55=1 | 3 |
40−45 | 42.5 | 0 | 542.5−32.5=510=2 | 0 |
45−50 | 47.5 | 0 | 547.5−32.5=515=3 | 0 |
50−55 | 52.5 | 2 | 552.5−32.5=520=4 | 8 |
Total | 3+8+9+10+3+2=35 | −9−16−9+3+8=−23 |
Here the assumed mean is A=32.5
Now, we shall apply the formula Mean,x=A+(∑fi∑fidi×h)
Mean,x=32.5+(35−23×5) (Here ∑fi is the sum of the frequency and ∑fidi is the sum of fidi )
⇒Mean,x=32.5+7−23
⇒Mean,x=32.5−3.28
⇒Mean,x=29.22
Hence, the required mean is 29.2
Therefore, we can conclude that the teacher-student ratio on average was 29.2
Note: We can able to conclude that most of the states/U.T have a teacher-student ratio as 30.6 and we can conclude that the teacher-student ratio on an average was 29.2To 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.