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Question
Question 2
Determine the mean of the following distribution
MarksNumber of studentsBelow 10 5Below 20 9Below 30 17Below 40 29Below 50 45Below 60 60Below 70 70Below 80 78Below 90 83Below 100 85
Determine the mean of the following distribution
MarksNumber of studentsBelow 10 5Below 20 9Below 30 17Below 40 29Below 50 45Below 60 60Below 70 70Below 80 78Below 90 83Below 100 85
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Solution
Here, we observe that 5 students have scored marks below 10, i.e. it lies between class interval 0-10 and 9 students have scored marks below 20.
So, (9-5)=4 students lies in the class interval 10-20. Continuing in the same manner, we get the complete frequency distribution table for give data.
MarksNumber of studentsClass marksui=xi−ah=xi−45hfiui0−10 5 5 −4−2010−20 9−5=4 15 −3−1220−30 17−9=8 25 −2−1630−40 29−17=12 35 −1−1240−50 45−29=16 a=45 0050−60 60−45=15 55 11560−70 70−60=10 65 22070−80 78−70=8 75 32480−90 83−78=5 85 42090−100 85−83=2 95 510 N=∑fi=85∑fiui=29
Here, (assumed mean) a=45
and (class width) h=10
By step deviation method,
Mean (¯x)=a+∑fiui∑fi×h=45+2985×10=45+5817 =45+3.41=48.41
So, (9-5)=4 students lies in the class interval 10-20. Continuing in the same manner, we get the complete frequency distribution table for give data.
MarksNumber of studentsClass marksui=xi−ah=xi−45hfiui0−10 5 5 −4−2010−20 9−5=4 15 −3−1220−30 17−9=8 25 −2−1630−40 29−17=12 35 −1−1240−50 45−29=16 a=45 0050−60 60−45=15 55 11560−70 70−60=10 65 22070−80 78−70=8 75 32480−90 83−78=5 85 42090−100 85−83=2 95 510 N=∑fi=85∑fiui=29
Here, (assumed mean) a=45
and (class width) h=10
By step deviation method,
Mean (¯x)=a+∑fiui∑fi×h=45+2985×10=45+5817 =45+3.41=48.41
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