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30. The marks obtained by 100 students of a class in an examination are given below;0-10=210-20=520-30=630-40=840-50=1050-60=2160-70=470-80=2Hence find the Mean and Median |
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Answer» Step-by-step explanation: Here we have, the cumulative frequency distribution. So, FIRST we convert it into an ordinary frequency distribution. we observe that are 80 students GETTING marks greater than or equal to 0 and 77 students have SECURED 10 and more marks. Therefore, the number of students getting marks between 0 and 10 is 80-77= 3. Similarly, the number of students getting marks between 10 and 20 is 77-72= 5 and so on. Thus, we obtain the FOLLOWING frequency distribution. Marks Number of students 0-10 3 10-20 5 20-30 7 30-40 10 40-50 12 50-60 15 60-70 12 70-80 6 80-90 2 90-100 8 Now, we compute mean arithmetic mean by taking 55 as the assumed mean. Computative of Mean Marks (x i
) Mid-value (f i
) Frequency u i
10 x i
−55
f i
u i
0-10 5 3 -5 -15 10-20 15 5 -4 -20 20-30 25 7 -3 -21 30-40 35 10 -2 -20 40-50 45 12 -1 -20 50-60 55 15 0 0 60-70 65 12 1 12 70-80 75 6 2 12 80-90 85 2 3 6 90-100 95 8 4 32 Total ∑f i
=80 ∑f i
u i
= -26 We have, N= sumf i
=80,∑f i
u i
=−26, A= 55 and h= 10 ∴ X =A+h[ N 1
∑f i
u i
] ⇒ X =A+h[ N 1
∑f i
u i
] ⇒ X =55+10× 80 −26
=55−3.25=51.75Marks. I HOPE that it's help you please mark me as brainlist |
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