Example of Mean Deviation about Mean for Discrete Frequency Distribution Video Lecture | Mathematics for GRE Paper II

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watched a distance part 1 - Part A so we
have this values and we have to find me
division about me so the first thing
that we'll do is we'll find out me
correct
hey this is f IX I to find the value of
me we have to find xif I first and then
do sufficient so this is X I this is Fi
let's find X I F Y so X I is 2 F ways to
know this guy's X I am I so this becomes
2 into 2 4 similarly head also exercise
PI fi is a this becomes 8 into 546 into
10 becomes 16 18 to 7 becomes 18 to 7 is
56 10 into 8 becomes 80 and 12 with the
five income 16 so I we have found X is 5
now we have X IFI and we have fi
knowledge to the submission you'll add
all this value 4 plus 40 is go to 44 44
plus 6 is equal to 104 104 plus 56 come
to 160 160 plus 80 is equal to 240 and
plus 16 spending 300 so this value is
300 alleged summation here
similarly for fi into summation here a
plus 2 10 10 plus 10 20 27 27 plus a 35
35 plus 5 foot this is 40 now we have to
use 5x that that is mene mene is nothing
but sigma f IX i see my fi x is nothing
but clear n okay but we have added CB
upon X IFI and we've added this this is
nothing but addition of all the tunnel
this is 3300
divided by Sigma fi Sigma F is you are
doubt this value at 40 40 because I am
saying I is 1 to N I'm adding all the
values
so this becomes 300 by 40 that this 7.5
so first target is the chief we have got
the mean mean is nothing but all right
here 7.5 once we have the mean next
thing is the part we have to find the
distance of all these values distance of
all the excise from me that is good to
find X by minus X dash or I can say X I
minus 7.5 Azam more this is the distance
so from 2 if I want to find 2 minus 7.5
more but I will get I'll get five point
five so this becomes 5 still ready for
five I want to find the distance of
seven point five that is 5 minus 7 point
5 more and that value comes out to be 2
point 5 so this value is two point five
it
similarly for six I'll find distance 6
minus 7 point 5 mod that value is
nothing but 1 point 5 so this value is 1
point 5 so for a we have to find
distance 8 minus 7.5 define the mode
that is nothing but point 5 this value
is 0.5
similarly for 10 10 minus 7 point 5 will
find the mod and this value comes out to
be 2 point 5 so this guy is similarly
for 12 we have to find the distance from
7.5 we'll take the mod this becomes 4
point 5 this value is this is the value
of distance of excise from medium at a 7
point 5 so we have got the X I - mean
mods all the others you will add this
value so if you add because addition is
you know addition is not it but actually
if you multiply in add so we have to
find fi excise so what will you know I
will find fi into
X I minus X more complement this value
we're fine
because this is something we have to
find it similar to the C we had excise
so we want xif I know I should have X I
will use x i- mean so now we'll find fi
into exile by the scene so we'll find
this value into this step so 5.5 into 2
is 11
similarly 8 into 2.5 is nothing but 2010
into 1.5 is nothing but 15 7 if the
point where is nothing but 3 point 5
correct eighty-two point where it is
nothing but 20 and 5 into 4 point 5 is
only one twenty two point five this is
the value of f IX eyes knowing to find
some also you see the formula formula is
summation of f IX i minus x it is me so
will sound this 11 plus 20 is 31 31 plus
15 is 46 46 plus three point five is
forty nine point five plus twenty is
sixty nine point five seventy seventy to
ninety this value is 19 correct so we
have 93 and n is nothing but Sigma fi
excise as you know this and this is M so
I'm going to find the mean division get
it is formula you say is nothing but
Sigma of F I into X by minus X more x
mean mod divided by Sigma F I correct in
both the case I is equal to one to n is
equal to one
correct this is all okay so this comes
out to be 92 and this comes out to be 14
Sigma fi you know this so the answer is
92 by 40 and that value will come out to
be 9 to 4 into 280 there is something
2.3 a boxcar and we have got the value
very simple what we have done here we
have exercise we have f5 we found X IJ
why I did this because we wanted to find
the meat first we got XA not bigger than
me mean is nothing but Sigma X IFI by a
firefighters so we got mean as 7.5 now
once we have the mean 7.5 we found a
deviation of all these points from the
mean that this we find mod of X I - mean
have we got this values then the formula
we had was mean division is Sigma fi
into mod of X I - me while here also if
you see because see logically we are
finding meat only
so if you see both the form with almost
same this is a file to X I this is Fi
into x i- in because the both case we
are fighting me we are dealing and then
we have mining deviation of the beam
this is deviation
this is deviation this is the values to
find the value it will say fi x i to
find it means the mean about division is
a fi into x i- mean so then we found a
fi into x i- me more the values we added
it and we got them thank you
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