Question

If the two quartiles of a normal distribution are 47.30 and 52.70 respectively, what is the mode of the distribution? Also find the mean deviation about median of this deviations​

Answer 1

Answer:

3.2

Step-by-step explanation:

we know that MD = .8 SD

& QD = .675 SD

A.T.Q

Q1= 47.3. &. Q3 = 52.7

QD = 52.7-47.3/2 = 2.7

Therefore SD = 2.7/.675 = 4

MD = .8×4 = 3.2

Answer 2

Concept Introduction:

When calculating the average departure from the mean value of a given data set, statisticians employ a metric known as the mean deviation.

Solution:

Given, that the two quartiles of a normal distribution are 47.30 and 52.70 respectively

We have to find out the mean deviation

We all know that, mean deviation = $0.8$ SD

quartile deviation = $0.675$ SD

quartile deviation = $( 52.7 - 47.3 ) / 2 = 2.7$

So, SD = $2.7 / 0.675 = 4$

mean deviation = $0.8$ × $4 = 3.2$

Final Answer:

The final answer is $3.2$

#SPJ2