Question

Mean-5, S.D.-2.6, Median- 5, Q.D 1.5 then Coefficient of QD is

Answer 1

Given :   Mean-5, S.D.-2.6, Median- 5, Q.D 1.5  

To find :  Coefficient of QD  

Solution:

Mean = 5

Median = 5

normal Distribution

QD = Quartile deviation = 1.5

QD = (Q3 - Q1)/2

=> 1.5 = (Q3 - Q1)/2

=> Q3 - Q1 = 3

Median = 5  => Q2  = 5

Q1 = 5 - QD  = 5 - 1.5 = 3.5

Q3 = 5 + QD = 5 + 1.5 = 6.5

Q3 + Q1  = 6.5 + 3.5 = 10

coefficient of quartile deviation   =  (Q₃ - Q₁)/(Q₃ + Q₁)

= 3/10

= 0.3

coefficient of quartile deviation   =   0.3

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Answer 2

Step-by-step explanation:

Coefficient of QD= [(Q3-Q1)/(Q3+Q1 )]100

QD=1.5 (given)

QD= (Q3-Q1)/2

1.5=(Q3-Q1)/2

1.5×2= Q3-Q1

3=Q3-Q1

Median=5 (given)

Median=Q2

Median=(Q3+Q1)/2 ...[Q2 =(Q3+Q1)/2]

5×2=Q3+Q1

10=Q3+Q1

Then,

Coefficient of QD= [3/10] ×100

=30