Question

For a normal distribution, if the mean is 680 and standard deviation is 70 then quartile deviation value is​

Answer 1

Answer:

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Answer 2

Concept:

The Quartile Deviation is half of the difference between the upper and lower quartiles in a frequency distribution.

$Quartile$ $Deviation=$ $\frac{(Q3-Q1)}{2}$

Quartiles: The first quartile(Q1) and third quartile(Q3) can be calculated by using the mean µ and the standard deviation σ.

Q1 = µ − ( .675)σ

Q3 = µ + (  .675)σ

Given:

mean (μ) = 680

standard deviation (σ) = 70

To find:

The value of quartile deviation.

Solution:

Calculate the first quartile Q1

Q1 = µ − ( .675)σ

$Q1=680-(.675)*70$

$Q1=632.75$

Calculate the third quartile Q3

Q3 = µ + ( .675)σ$QD=\frac{94.5}{2}$

$Q3=680+(.675)*70$

$Q3=727.25$

Calculate Quartile Deviation (QD)

$QD=\frac{(Q3-Q1)}{2}$

$QD=\frac{727.25-632.75}{2}$

$QD=\frac{94.5}{2}$

$QD=47.25$

Hence the quartile deviation is 47.25.