In an examination 150 students appeared and their marks( out of 200) are given in the following distribution . Find the missing frequencies x and y, where it is given that mean mark is 103. Marks 0-25 25-50 50-75 75-100 100-125 125-150 150-175 175-200 No of std. 2 10 x 30 y 15 12 21
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Class interval Mid value Xi Frequency Fi XiFi
0 - 25 12.5 2 25
25 - 50 37.5 10 375
50 - 75 62.5 x 62.5x
75 - 100 87.5 30 2625
100 - 125 112.5 y 112.5y
125 - 150 137.5 15 2062.5
150 - 175 162.5 12 1950
175 - 200 187.5 21 3937.5
___ ________________
90+x+y 10975 + 62.5x + 112.5y
90 + x + y = 150
x + y = 60 or x = 60 - y ------- eqn (1)
Mean = ∑XiFi/∑Fi
103 = 10975 + 62.5x + 112.5y/150
103 x 150 = 10975 + 62.5x + 112.5y
15450 - 10975 = 62.5x + 112.5y
4475 = 62.5x + 112.5y
substituting value of x from eqn (1), we get
4475 = 62.5 (60-y) + 112.5y
4475 = 3750 - 62.5y + 112.5y
725 = 50y
y = 14.5
∴ x = 60 - y
x = 60 - 14.5
x = 45.5
missing frequencies are x = 45.5 and y = 14.5
0 - 25 12.5 2 25
25 - 50 37.5 10 375
50 - 75 62.5 x 62.5x
75 - 100 87.5 30 2625
100 - 125 112.5 y 112.5y
125 - 150 137.5 15 2062.5
150 - 175 162.5 12 1950
175 - 200 187.5 21 3937.5
___ ________________
90+x+y 10975 + 62.5x + 112.5y
90 + x + y = 150
x + y = 60 or x = 60 - y ------- eqn (1)
Mean = ∑XiFi/∑Fi
103 = 10975 + 62.5x + 112.5y/150
103 x 150 = 10975 + 62.5x + 112.5y
15450 - 10975 = 62.5x + 112.5y
4475 = 62.5x + 112.5y
substituting value of x from eqn (1), we get
4475 = 62.5 (60-y) + 112.5y
4475 = 3750 - 62.5y + 112.5y
725 = 50y
y = 14.5
∴ x = 60 - y
x = 60 - 14.5
x = 45.5
missing frequencies are x = 45.5 and y = 14.5
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Answer:
above is absolutely correct
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