Question

Weights of female cats of a certain breed are normally distributed with mean 4.1 kg and standard deviation 0.6 kg.

Answer 1

The proportion is 0.4401.

We find z-scores that correspond with both ends of this interval.  Z-scores are given using the formula:

$z=\frac{X-\mu}{\sigma}$

For the lower end, 

$z=\fracc{3.7-4.1}{0.6}=\frac{-0.4}{0.6}=-0.67$

The proportion of scores to the left of this number is 0.2514.

For the higher end, the z-score is

$z=\frac{4.4-4.1}{0.6}=\frac{0.3}{0.6}=0.5$

The proportion of scores to the left of this number is 0.6915.

To find just the proportion of scores between these two, we subtract;

0.6915-0.2514 = 0.4401.