Averages
Question
Q1.
The variance of 4 are 6 are
Solution.
S.D =
lb-al/2
⇒
variance =
lb-al2/4
variance =
(6-4)2/4
=
4/4
= 1
Solution
Q2.
There are two groups having 13 and 12 observations with H.M. 130 and 120 respectively. Find the combined H.M.
Solution.
formula of combined H.M.=n1/(n1/H1) + n2/(n2/H2)
=13/(13/130) + 12/(12/120)
125
Solution
Q3.
Two variables x and y are given by y = 7x + 40 . If mode of x is 3 then mode of y is ____________ .
Solution.
Y = 7(3) + 40 = 21 + 40 = 61
so mode of y = 61
Note : Mode has impact of change of origin and change
of scale both .
Solution
Q4.
The relation between x and y is 4x - 5y = 40, if mean deviation of x = 15, find M.D of y.
Solution.
4x - 5y = 40
5y = 4x - 40 ⇒ y =
4/5
x - 8
y =
4/5
(15) - 8
M.D of y = 12
Solution
Q5.
What is the H.M of 1, 1/2, 1/3 -------1/n ?
Solution.
A.M of 1,2,3,-------n =
∑n/n
=
n+1/2
So, H.M is
2/n+1
Solution
Q6.
The standard deviation of first eight natural numbers is __________ .
Solution.
S.D of 1
st n natural number is
√(
n2-1/12
)
= √(
64-1/12
) =
= √(
63/12
) =
√(
21/4
) =
=
1/2
√21
Solution
Q7.
The A.M of two numbers is 20 and G.M is 16 then what will these two numbers .
Solution.
Let a and b are two numbers
A.M = (a + b)/2 , G.M = √ab
a = 32 , or 8
b = 8 or 32 .
Note : Using options one can solve these type of questions .
Solution
Q8.
If x and y are two variables, having relation 2x + 3y = 30 and mode of x is 5, find the mode of y.
Solution.
3y = 30 - 2x
y =
30 - 2x/3
=
30 - 10/3
=
20/3
Solution
Q9.
The average of 30 results is 20 and the average of other 20 results is 30. what is the average of all the results.
(a)24 (b)48 (c)25 (d)50
Solution.
Required average
= (20 x 30 + 20 x 30)/(30 x 20)
= 600 + 600/50
= 1200/50 = 24
Solution
Q10.
Which result holds for a set of distinct positive observations, among AM, GM and HM ?
Solution.
A.M > G.M > H.M
Note : If all the observations are equal then A.M = G.M = H.M
Solution
