Q1.
There are two groups having 15 and 13 observations with H.M. 75 and 65 respectively. Find the combined H.M.
Solution.
formula of combined H.M.=n1/(n1/H1) + n2/(n2/H2)
=15/(15/75) + 13/(13/65)
70
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.
Q3 is equal to __________ .
Solution.
75th percentile .
Solution
Q4.
40th percentile is equal to_______ .
Solution.
4th Decile .
Solution
Q5.
If A.M between two numbers is 81 and G.M between is 27 . The H.M is ___________ .
Solution.
(A.M)(H.M) = (GM)2
(81) (H.M) = 27 x 27
3(H.M) = 27 ⇒ H.M = 9
Solution
Q6.
M.D of a given data is 240 , if 1st year all value of data is increased by 40, and in 2nd year each value is hiked
by 10% then find new M.D .
Solution.
For first year M.D will remain same 40, but in second year M.D will hiked by 10% , then new M.D will be 44.
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.
Quartiles can be determined graphically using .
Solution.
Ogive curve (or Cumulative frequency curve) .
Solution
Q9.
If same amount is added to all the values of an individual series then the mean deviation will be
(a) changed
(b) unchanged (c) 1 (d) none of these
Solution.
unchanged
Note : change of origin has no impact on measurement of dispersion .
Solution
Q10.
The S.D of first n natural number is ________ .
Solution.
n = 10
S.D = √(n2 - 1)/12
= √(100 - 1)/12
= √(99/12)
= √(33/4)
Solution
