1).given observations 4,9,11,14,37 mean deviation about median is
a).11, b).8.5, c).7, d).4.75.
2).if X ~ n(3,36) & Y~ N(5,64) then X + Y ~ (8,A).the value of A is
a)100, b).10, c).64, d).36.
1).given observations 4,9,11,14,37 mean deviation about median is
a).11, b).8.5, c).7, d).4.75.
2).if X ~ n(3,36) & Y~ N(5,64) then X + Y ~ (8,A).the value of A is
a)100, b).10, c).64, d).36.
Q1 ANS- A,11
Q2 ANS- A,100
clarify the answer please........................................
QNO-1= MEDIAN=(N+1)/2TH TERM IS MEDIAN=(5+1)/2=3RD TERM IS MEDIAN,WHICH IS 11. ans
QNO-2=X+Y THEN ALLVARIABLE WILL BE IN SUM=E.G 3+5=8 AS IT IS 36+64=100 ans
2nd question i can't understand..
VARIABLES OF X= (3,36)
VARIABLES OF Y = (5,64)
X+Y=SUM OF VARIABLES
X+Y=3+5 & 36+64=(8,100)
100 IS VALUE OF A
ok 2nd question i understand but 1st question is mean deviation about median.
is answe for first ques is 7.6 exactly
please clarify the answer
median is 11
4-11= 7 ( ignoring -ve sign)
9-11=2
11-11 =0
14-11=3
37-11 = 26
now take simple average of (7+2+0+3+26)/5 = 7.6
tell ... is answer ryt
yes.answer is 7.
7.6 is the right answer
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