* Empirical relationship between three measurements of the central tendency
Mode = 3 Median – 2 Mean
* If class interval is discontinuous then make it continuous by subtracting 0.5 from lower limit and adding 0.5 to upper limit.
* Model Class:– A class interval having maximum frequency is known as model class.
* Median Class : – A class interval in which commulative frequency is greater than and nearest to 'n/2' is called median class.
Q. 1. The median and mode respectively of a frequency distribution are 26 and 29, Then what is its mean ?
Sol.
Median (Md) = 26
Mode (Mo)= 29
Mean (M) =?
Mode = 3Median –2Mean
29 = 3(26) –2M
2M = 78 –29
2M = 49
M = 49/2
M = 24.5
Q.2 If the difference of mode and median of a data is 24, then what is the difference of median and mean ?
Sol.
Let
Mean = x
Median = y
Mode = z
We have,
Condition (1)
Mode – Median = 24
z–y = 24
z = 24 + y ....(1)
Condition (1)
z = 3y–2x ......(2)
24 + y = 3y–2x ......(2)
24 + y = 3y–2x
(2y–2x = 24)÷2
y–x = 12
Median – Mean = 12
Q. 3 The mean of the following distribution is 48 and sum of all the frequency is 50. Find the missing frequencies x and y .
We prepare following table to find mean.
Q.4 If the mean for the following frequency distribution is 26, find the value of x and y :
Class Number of Students
10–20 2
20–30 3
30–40 x
40–50 y
Total 40
Sol.
Class fi. xi. fi×xi
00–10 6 5 30
10–20 x 15 15x
20–30 9 25 225
30–40 n 35 35y
40–50 11 45 495
Total 40
Sum of frequencys
x + y + 26 = 40
x + y = 14
x = 14 – y ........(1)
Mean = fi × xi / fi
26 = (750 +15x + 35y)/40
1040 = 750 +15x + 35y
[15x + 35y = 1040] ÷5
3x + 7y = 58 ........(2)
Placing the value of 'x' from equation (1) in equation (2)
3x + 7y = 58
3(14–n) + 7y = 58
42 – 3n + 7y = 58
4y = 58 –42
4y = 16
y = 4
Placing the value of 'x' in equation (1)
x = 14 – y
x = 14 – 4
x = 10
y = 4
Q.5 Find the Mode of the following grouped frequency distribution.
We observe that the class 12-15 has maximum frequency 23. Therefore, 12-15 is the modal class.
We have, l = 12,
h = 3,
f0 = 10
f1 = 23, and
f2 = 21
Q. 6 If the Mode for the following frequency distribution is 25 and the sum of The frequencies is 110, find the value of x and y :
Class Number of Students
10–15 15
15–20 x
20–25 40
25–30 y
25–30 15
Total 110
Sol.
Class Number of Students
10–15 15
15–20 x
20–25 40
25–30 y
25–30 15
Total 110
Sum of frequencys
x + y + 70 = 110
x + y = 110–70
x + y = 40
x = 14 – y ........(1)
Mean = fi × xi / fi
26 = (750 +15x + 35y)/40
1040 = 750 +15x + 35y
[15x + 35y = 1040] ÷5
3x + 7y = 58 ........(2)
Placing the value of 'x' from equation (1) in equation (2)
3x + 7y = 58
3(14–n) + 7y = 58
42 – 3n + 7y = 58
4y = 58 –42
4y = 16
y = 4
Placing the value of 'x' in equation (1)
x = 14 – y
x = 14 – 4
x = 10
y = 4
Q. 7 If the median for the following frequency distribution
is 28.5, find the value of x and y :
Sol.
We prepare following cumulative frequency table to find median class.
Q.8 The median of the following data is 525. Find the values of x and y if the total frequency is 100.
Here median is 525, which lies between class 500-600. Thus median class is 500-600.
l = 20
N/2=200/2=50
f = 20
cf = 36 + x
h= 100
Q. 8 Draw more than ogive for the following distribution. Find the median from the curve.
From graph, N
/2 = 100 /2 = 50
Hence, Median = 25
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