Question 2Determine the mean of the following distribution- - - - - Below 10 5- Below 20 9- Below 30 17- Below 40 29- Below 50 45- Below 60 60- Below 70 70- Below 80 78- Below 90 83- Below 100 85- -

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Finding Mean for Grouped Data When Class Intervals Are Given

Question 2Det...

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Question 2
Determine the mean of the following distribution
$\begin{matrix} M a r k s & N u m b e r o f s t u d e n t s B e l o w 10 & 5 B e l o w 20 & 9 B e l o w 30 & 17 B e l o w 40 & 29 B e l o w 50 & 45 B e l o w 60 & 60 B e l o w 70 & 70 B e l o w 80 & 78 B e l o w 90 & 83 B e l o w 100 & 85 \end{matrix}$

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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.
$\begin{matrix} M a r k s & N u m b e r o f s t u d e n t s & C l a s s m a r k s & u_{i} = \frac{x_{i} - a}{h} = \frac{x_{i} - 45}{h} & f_{i} u_{i} 0 - 10 & 5 & 5 & - 4 & - 20 10 - 20 & 9 - 5 = 4 & 15 & - 3 & - 12 20 - 30 & 17 - 9 = 8 & 25 & - 2 & - 16 30 - 40 & 29 - 17 = 12 & 35 & - 1 & - 12 40 - 50 & 45 - 29 = 16 & a = 45 & 0 & 0 50 - 60 & 60 - 45 = 15 & 55 & 1 & 15 60 - 70 & 70 - 60 = 10 & 65 & 2 & 20 70 - 80 & 78 - 70 = 8 & 75 & 3 & 24 80 - 90 & 83 - 78 = 5 & 85 & 4 & 20 90 - 100 & 85 - 83 = 2 & 95 & 5 & 10 N = \sum f_{i} = 85 & \sum f_{i} u_{i} = 29 \end{matrix}$
Here, (assumed mean) a=45
and (class width) h=10
By step deviation method,
Mean $(¯ x) = a + \frac{\sum f_{i} u_{i}}{\sum f_{i}} \times h = 45 + \frac{29}{85} \times 10 = 45 + \frac{58}{17} = 45 + 3.41 = 48.41$

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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