Question

The length of a fish species has a normal distribution with a mean of 16cm and a standard deviation of 6cm.

A) what is the IQR of the lengths of these fish?

B) What is the length of the range in which the middle 95% of the fish lie?

C) The upper and lower deciles are the points that mark out the highest 10% and the lowest 10% of the population. Find the upper and lower deciles for the fish length distribution.

Answer 1

a)here as we know that for first and 3rd quartile ; critical z is - 0.67 and 0.67

therfore 1st quartile =mean +z*std deviation =16-0.67*6=11.98

3rd quartile =mean +z*std deviation =16+0.67*6=11.98=20.02

hence IQR =Q3-Q1 =20.02-11.98 =8.04

b)

\for 95% interval critical values z= -/+ 1.96

hence as above 95% range =2*z*std deviation =2*1.96*6=23.52

c)

for deciles ; crtiical value z =-/+1.28

upper decile=mean +z*std deviation =16+1.28*6=23.68

lower decile=mean +z*std deviation =16-1.28*6=8.32