Finding the mean of the grouped data.

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In the formula: \overline { x } =a+\frac { \sum { { f }_{ i }{ d }_{ i } } }{ \sum { { f }_{ i } } } for finding the mean of the grouped data, d’is are deviations from a (assumed mean) of
(a) lower limits of the classes
(b) upper limits of the classes
(c) mid-points of the classes
(d) frequencies of the classes

Solution:

The formula \overline { x } =a+\frac { \sum { { f }_{ i }{ d }_{ i } } }{ \sum { { f }_{ i } } } is the finding of mean of the grouped data, d’is are mid-points of the classes

In the formula: \overline { x } =a+c\left( \frac { \sum { { f }_{ i }{ u }_{ i } } }{ \sum { { f }_{ i } } } \right) , for finding the mean of grouped frequency distribution, ui =
(a) \frac { { y }_{ i }+a }{ c }
(b) c({ y }_{ i }-a)
(c) \frac { { y }_{ i }-a }{ c }
(d) \frac { a-{ y }_{ i } }{ c }

Solution:

In \overline { x } =a+c\left( \frac { \sum { { f }_{ i }{ u }_{ i } } }{ \sum { { f }_{ i } } } \right) ,
for finding the mean of grouped frequency, ui is \frac { { y }_{ i }-a }{ c } (c)

While computing mean of grouped data, we assumed that the frequencies are
(a) evenly distributed over all the classes
(b) centred at the class marks of the classes
(c) centred at the upper limits of the classes
(d) centred at the lower limits of the classes

Solution:

For computing mean of grouped data,
we assumed that frequencies are centred at class marks of the classes. (b)

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