Q1.
x and y are two independent normal variate with mean 6 and 10 respectively . The S.D as 3 and 4 respectively . If a random variable z is defined
as z = x + y , then the distribution of z is also a normal with parameters______ .
Solution.
z ∿ N (16, 5)
Solution
Q2.
If two quartiles of a normal distribution are 30 and 70 respectively , find the mode .
Solution
As we know that in normal distribution
mean = mode = median =
Q1 + Q3/2
Mode =
100/2
= 50
Solution
Q3.
If lower and upper quartiles of a normal distribution are 30 and 70 respectively , find mean deviation about mean .
Solution.
Quartile Deviation =
Q3 - Q1/2
= 0.675 (S.D)
20 = 0.675 (S.D)
S.D = 29.63
So, mean deviation about mean =
4/5
(S.D)
=
4/5
x 29.63 = 23.704
Solution
Q4.
Q1 = 10 and Q3 = 30 , find point of inflexion .
Solution.
As we know point of inflexion are are μ ± (S.D)
μ =
Q1 + Q3/2
= 20
Q3 - Q1/2
= 0.8 (S.D) ⇒ 10 = 0.8(S.D) ⇒ S.D = 12.5
point of inflexion are 7.5 and 32.5
Solution
Q5.
The p.d.f of a normal variate x is given as
f(x) =
1/4√(2π)
e
-(x - 12)2/32
find the point of inflexion .
Solution.
Point of inflexion are μ ± (S.D) Now comparing this with f(x) =
1/(S.D)√(2π)
e
-(x - μ)2/2.(S.D)2
S.D = 4 , μ = 12
So, point of inflexion are 8 and 16
Solution
Q6.
The p.d.f of a normal variate x is given as
f(x) =
1/√(50π)
e
-(x - 8)2/50
Find mean deviation about mean .
Solution.
Now comparing this with f(x) =
1/(S.D)√(2π)
e
-(x - μ)2/2.(S.D)2
M.D =
4/5
S.D
S.D = 5 (on comparing)
M.D = 4
Solution
Q7.
Find Quartile deviation for a standard normal distribution .
Solution.
Q.D = 0.675 (S.D)
In the case of S.N.V standard deviation is 1
So, Q.D = 0.675
Solution
Q8.
Area covered between μ ± 3(S.D) is
(a) 0.99 (b) 0.95 (c) 0.9973 (d) 0.9454
Solution.
(c) 0.9973
Solution
Q9.
Which statement is false for normal distribution
(a) Curve is bell shaped
(b) Its skewness is zero
(c) It is bi-modal
(d) It is biparametric
Solution.
(c) It is bi-modal
Note : Normal Distribution is unimodal, it has only one peak .
Solution
Q10.
Relation between mean deviation about mean and standard deviation of normal distribution is .
Solution.
M.D : S.D = 12 : 15
M.D/S.D
=
4/5
or 5 M.D = 4 S.D
Solution
