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Normal Distribution

Theory and Formula



Question


Q1. Skewness of Normal distribution is ____________ .

Solution.
zero
Solution

Q2. In Normal distribution mean deviation is ___________ .

4/5
S.D
Solution

Q3. The point of inflexion of a normal distribution are 30 and 50 , then the values of parameters are .

Solution.
Point of inflexion = μ + σ

μ + σ = 50 -----------(1)

μ - σ = 30 -----------------(2)

On solving both equation we get μ = 40 and S.D = 10
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. x and y are two independent normal variate with respective mean as 15 and 7 and the S.D as 3 and 4 . find the parameter values of variable z(z = x + y).

Solution.
x ∿ N (22,25)
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. Mean deviation about median of a S.N.V is

Solution.
0.8

M.D =
4/5
S.D   and S.D = 1 in S.N.V

So M.D =
4/5
= 0.8
Solution

Q8. In normal distribution mean deviation about mean is 18, then what will be the value of mean deviation about median .

Solution.
18, because mean = median = mode
Solution

Q9. If two quartiles of N(μ, σ2) are 10 and 18.6 respectively, find the standard deviation of the distribution .

Solution.
Q.D =
Q3 - Q1/2
Q.D = 0.675 S.D

Q.D =
18.6 - 10/2
  = 4.3

S.D =
4.3/0.675
= 6.37
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